replace operates on vectors, not data.frames. Try this: > replace(df$Group.1,df$Group.1 == 1, "male") [1] "male" "2" And to change the data, you can do: > df$Group.1 <- replace(df$Group.1,df$Group.1 == 1, "male") > df Group.1 x 1 male 4000 2 2 3890 What you are trying to do looks like a...
You can dispense with most of your pattern - much of it is unnecessary. Try this: <img.*?> With the unnecessary brackets removed, the important change is adding ? to make it a reluctant quantifier - one that matches as little as possible....
php,regex,replace,php-shorttags,shorttags
Just make the = symbol as optional one. preg_replace('~<\?=?(?!php\b)~', '<?php ', $str); OR preg_replace('~<\?=?(?!php\b)(\w)~', '<?php \1', $str); DEMO...
excel,function,excel-vba,replace
This is the literal translation of what you shown : Sub MacroClear() Dim wbD As Workbook, _ wbC As Workbook, _ wsD As Worksheet, _ wsC As Worksheet, _ Dic() As String 'Replace the names in here with yours Set wbD = Workbooks("Dictionnary") Set wbC = Workbooks("FileToClean") Set wsD =...
Try using the unicode character code, \u2022, instead: message.replace(/\u2022/, "<br />\u2022"); ...
javascript,jquery,string,if-statement,replace
The problem is with the check of tagcounter <= 0. When you are checking that for the first time, tagcounter is still 1. You decrement it later. Hence the "+" never gets appended. Hence when you are trying to remove "+Light" it doesn't work because the "+" is not there...
no need to iterate body element try this: want to change to with that js? i have used to make it function addTitleToSurveyUrls() { var elements = document.getElementsByTagName('a'); for(var el in elements) { var element = elements[el]; var href = element.getAttribute("href"); if(href.indexOf('survey_')>-1) { element.setAttribute('title', 'Some TITLE HERE'); } } }...
You could construct a simple loop and print using cat. for (i in 1:length(day)) { cat("Today is", day[i], "and it is a", weather[i], "day.\n") cat(day[i], "is a", weather[i], "day\n") } # Today is Monday and it is a sunny day. # Monday is a sunny day # Today is Tuesday...
.net,regex,vb.net,string,replace
It is impossible to achieve in a single pass. I suggest: Adding parentheses: (\p{Lu}+)\s*(\p{Lu}+)(.*) (replace with $1($2$3)) See demo Then, you can use a regex to find all spaces between numbers and replace them with a comma: (?<=\d+)\s+(?=\d+). See demo (see Context tab). Here is a working VB.NET code: Dim...
You can use gsub as follows: gsub(".*experiences.*", "exp", string, perl=TRUE) # As @rawr notes, set perl=TRUE for improved efficiency This regex matches strings that have any characters 0 or more times (i.e. .*) followed by "experiences", followed by any characters 0 or more times. In this case, you are still...
regex,bash,perl,replace,command-line
replace %20 after last forward slash of href attributes of non .jpg links You can use the following to match: %20(?=(?:(?!\.jpg">)[^>\/])*>) And replace with - See DEMO...
#addones lstcol <- list('cola', 'colb') lstvec <- list(veca, vecb) myfunc <- function(COL, VEC) { dataframe[[COL]][dataframe$ID %in% VEC] <<- 1 } for(i in 1:length(lstcol)) { myfunc(lstcol[[i]], lstvec[[i]]) } dataframe # ID cola colb colc cold cole colf #1 XXXYYY 1 0 0 0 0 0 #2 XXYYXX 1 0 0 0...
The replace method isn't changing the contents of your string; Strings are immutable. It's returning a new string that contains the changed contents, but you've ignored the returned value. Change example.replace(oldNumber, newNumber); with example = example.replace(oldNumber, newNumber); ...
regex,vb.net,replace,replaceall
Let me explain why your regex does not work. \\ - Matches \ (?=[0]) - Checks (not matches) if the next character is 0 (?<![\\]) - Checks (but not matches) if the preceding character (that is \) is not \. The last condition will always fail the match, as \...
@PetSerAl comment is correct, here is code to test it: $sep = "," "AAA BBB" -Replace '(?<A>\w+)\s+(?<B>\w+)',"`${A}$sep`${B}" Output: AAA,BBB Explanation: Powershell will evaluate the double quoted string, escaping the $ sign with a back tick will ensure these are not evaluated and a valid string is provided for the -Replace...
One way to do this is to have a function for replacing text. The function would get the match object from re.sub and insert id captured from the string being replaced. import re s = 'ref_layerid_mapping="x4049" lyvis="off" toc_visible="off"' pat = re.compile(r'ref_layerid_mapping=(.+) lyvis="off" toc_visible="off"') def replacer(m): return "ref_layerid_mapping=" + m.group(1) +...
You can try with group by: select Firstname , Lastname , case when count(*) > 1 then '**ERROR**' else age end from Table group by Firstname , Lastname , age; Or in case you want to return all rows with duplicates: select t.Firstname , t.Lastname , case when (select count(*)...
I'm pretty sure that you don't need a regular expression for this. Just use var str2 = str.Replace(":\"{", ":{").Replace("}\"", "}"); ...
php,replace,binary,hex,preg-replace
If the string is at fixed position, you can write data directly: $position=hexdec("00010edb"); // You have to pre-calculate it once $data="some data"; // Replacement if ($f=fopen("your_file", "r+")) { fseek($f, $position); fwrite($f, $data); fclose($f); } else { echo "Can't open file"; } ...
javascript,regex,replace,capture-group
I would do like, > "Apple Orange".replace(/(?:^|\s)([A-Z])|./g, "$1") 'AO' Don't complex the things. Just capture all the uppercase chars which exists just after to a space or at the start. And then match all the remaining characters. Now replace all the matched chars by $1. Note that all the matched...
php,arrays,string,wordpress,replace
The Verbose Way: You're close...But I'd simplify a bit and forego the str_replace(). I'd do something like the following: $shipping_method = $order->get_shipping_method(); if ( $shipping_method === 'Free Shipping' ) { $shipping_method = 'Free'; } elseif ( $shipping_method === 'FedEx_GROUND' ) { $shipping_method = 'FedEx'; } elseif ( $shipping_method === 'UPS...
javascript,jquery,regex,replace
join your array and create a single regex, then you can use replace with a callback function (or use the single line replace like in sln's answer) to create the spans Note this assumes all emoji names are between : var emojies = ["smile", "smiley", "grin", "joy"]; //Create a single...
Assuming you want to replace the word 'apple' with 'banana' (exact match) in the contents of the files and not on the names of the files (see my comment above) and that you are using the bash shell: #!/bin/bash COUNTER=0 for file in *.txt ; do COUNTER=$(grep -o "\<apple\>" $file...
regex,powershell,replace,split
Leveraging the power in PowerShell we can turn the output of pnputil into an object array that will make it much easier to parse the data you are looking for (since it appears you are looking for something specific). Each entry is a group of variables with a blank line...
Use re.sub. Just match all the chars upto I then replace the matched chars with I. re.sub(r'.*I', 'I', stri) ...
The leading spaces are for multiple reasons: When you remove expletives, they are surrounded by spaces, that are not removed. You could capture your expletives using the following regex instead: \baaaa[^a-zA-Z]? This regex will capture a word boundary, followed by the expletive (replaced with aaaa here), and then an optional...
You don't need bash and unnecessary piping through multiple system utilities, perl -i -pe' BEGIN { @chars = ("a" .. "z", "A" .. "Z", 0 .. 9); push @chars, split //, "[email protected]#$%^&*()-_ []{}<>~\`+=,.;:/?|"; sub salt { join "", map $chars[ rand @chars ], 1 .. 64 } } s/put your...
You can try: :%s/\vNumberInt\((\d+)\)/\1/ Using very magic mode (\v), do grouping in the number between parentheses and use it in the replacement part as backreference (\1): It yields: "bz":1, "something": "something else NumberInt(1)" "bz":1, "something": "something else NumberInt(2)" "bz":1, "something": "something else NumberInt(3)" "bz":1, "something": "something else NumberInt(4)" "bz":1, "something":...
python,replace,out-of-memory,large-files
You need to read one bite per iteration, analyze it and then write to another file or to sys.stdout. Try this code: mesh = open("file.mesh", "r") mesh_out = open("file-1.mesh", "w") c = mesh.read(1) if c: mesh_out.write("{") else: exit(0) while True: c = mesh.read(1) if c == "": break if c...
This might be done more simply using str_replace() $search = array('[b]', '[/b]', '[i]', '[/i]'); $replace = array('<b>', '</b>', '<i>', '</i>'); str_replace($search, $replace, $string ); ...
In order to only match your lm words and not affect the following ones, you need to restrict the pattern. I suggest turning .* to [^[:blank:]]* to match 0 or more characters other than spaces or tabs: sed 's/\([^_]lm[01?]_[^[:blank:]]*\)\([ (]\)/\1_out\2/g' text_to_edit (EDIT) Or, if the only characters there are between...
string,parsing,batch-file,search,replace
Not sure about the requirements, but, maybe awk "{print $1}" 2.txt | findstr /g:/ /l /b 1.txt Without awk cmd /q /c"(for /f "usebackq" %%a in ("2.txt") do echo(%%a)" | findstr /g:/ /l /b 1.txt Only awk awk "{a[$1]=!a[$1]} !a[$1]" 2.txt 1.txt ...
You need to use a capturing group ((.*)) to match the text between the quotes and use that in your replacement. Note that if you have to handle escaped quotes inside the string or anything complex, you'll want to use a real parser rather than complicating the regex more (I'd...
javascript,jquery,ajax,replace
You need to preserve $(this) in a variable for the callback like var $link = $(this); $.ajax({ type : "POST", cache : false, url : $(this).attr('href'), data : $(this).serialize(), success : function(data) { $('.loading-icon').hide(); $link.attr('href').replace(/delete/, 'add'); $link.removeClass('remove-favourite').addClass('add-favourite'); } }) this itself will have changed because the execution context of the...
You have to use a regex object like this: C# string input = "This is t3xt with n4mb3rs."; Regex rgx = new Regex("[0-9]"); string result = rgx.Replace(input, "Z"); VB Dim input As String = "This is t3xt with n4mb3rs." Dim rgx As New Regex("[0-9]") Dim result As String = rgx.Replace(input,...
regex,linux,shell,unix,replace
You can do it using awk as: awk '/\[x/{f=1} {if(f)printf "%s",$0; else print $0;} /y\]/{print ""; f=0}' Output: [x data1 data2 data3 data4 y] [a data5 data 6 data7 data 8 b> [x data y] You can also simplify to: awk '/\[x/,/y\]/{ORS=""; if(/y\]/) ORS="\n";}{print}' Output: [x data1 data2 data3 data4...
replace doesnt use a regular expression. Just use myString = myString.replace("[merchant]", "newText"); ...
javascript,node.js,methods,replace
I don't think you want to loop. Try doing a global regex with the incrementing number in the replace pattern: function(doc) { var number = 1; return doc.replace(/(<a\s+[^>]*id="r)("[^>]*>[^<]*<\/a>)/gi, function(m, a, b) { return a + (number++) + b; }); }; This looks for any <A> tags with id="r", keeps their...
Python code, assuming your substitutions are stored in subs.csv: import csv subs = dict(csv.reader(open('subs.csv'), delimiter='\t')) source = csv.reader(open('all_snp.map'), delimiter='\t') dest = csv.writer(open('all_snp_out.map', 'wb'), delimiter='\t') for row in source: row[1] = subs.get(row[1], row[1]) dest.writerow(row) The line row[1] = subs.get(row[1], row[1]): row[1] is the Affx column, and it replaces it with a...
Try setting input maxlength to 5 ; utilizing .val(function(index, val)) ; RegExp /\d{3}|[^\d{2}\.]|^\./ $("#valueSconto").on("input", function() { $(this).val(function(i, val) { return val.replace(/\d{3}|[^\d{2}\.]|^\./g, ""); }); }); <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"> </script> <input type="text" id="valueSconto" maxlength="5" /> ...
python,string,replace,floating-point,integer
(?<=\d)\.0+\b You can simple use this and replace by empty string via re.sub. See demo. https://regex101.com/r/hI0qP0/22 import re p = re.compile(r'(?<=\d)\.0+\b') test_str = "15.0+abc-3" subst = "" result = re.sub(p, subst, test_str) ...
sql,sql-server,regex,tsql,replace
You could try the output keyword, that won't tell you exactly what was taken out but it will tell you the state of the record before you changed it. And yes, it would work better for you if the information was properly normalized. UPDATE table SET a = replace..... OUTPUT...
Try match where 'm1' is the matrix match(vector1, m1[,1]) #[1] 2 1 2 3 Or unname(setNames(as.numeric(m1[,2]), m1[,1])[vector1]) #[1] 2 1 2 3 ...
You don't need to cast it to NSString to use stringByReplacingCharactersInRange you just need to change the way you create your string range as follow: let binaryColor = "000" let resultString = binaryColor.stringByReplacingCharactersInRange( Range(start: binaryColor.startIndex, end: advance(binaryColor.startIndex, 1)) , withString: "1") println("This is binaryColor: \(resultString)") ...
You can use the "last match or beginning of line" trick in a positive look-behind: String s = "I have been stuck on this problem forever quick brown fox jumps over"; String r = s.replaceAll("(?<=(^|\\G)\\S{0,100}\\s\\S{0,100}\\s\\S{0,100})\\s", "_"); System.out.println(r); The unfortunate consequence of using a look-behind is that you need to provide...
> library(stringr) > mystring <- "\t\t\tFGHGFJKJKJKGDSJS" > x <- "ABCCCBBHHJJJH" > str_replace(mystring,"\\w+",x) [1] "\t\t\tABCCCBBHHJJJH" \w+mean match any character or number or underscore at least once and as many as possible. So each part not a normal char will be replace by your x variable....
The replace(...) method doesn't modify the original String, it returns a new modified String. You need to do something like this: line = line.replace(" alle ", " "); And here is a complete solution: Scanner scannerLines = new Scanner(file); boolean alleFound = false; boolean nrFound = false; boolean minFound =...
string,swift,replace,converter
tmpAddress = tmpAddress.stringByReplacingOccurrencesOfString("-", withString: "+") This replaces '-' with '+' The first parameter is the search string, and the second is the replacement string....
batch-file,replace,folder,filenames
setsyntax for replacing a string: set "var1=my old hat" set "var2=%var1:old=new%" echo %var1% was better than %var2% To get all txt-files: for %%i in (*.txt) do ( echo %%i rem substitution does not work with special %%i variable type set "oldname=%%i" set "newname=!oldname:@=_!" echo rename "!oldname!" "!newname!" ) Why !...
mysql,replace,wildcard,placeholder
Using SUBSTRING_INDEX: UPDATE table1 SET column1 = REPLACE( column1, SUBSTRING_INDEX(column1, '/', 2), 'newsite.com' ) WHERE column1 LIKE 'example.com/%/' This should honour your subfolder structure....
Find what: \\section\{(.+)\} Replace with: <h1>\1</h1> Here, the parentheses (.+) define a group that consists of one or more character, and \1 references the contents of the group....
javascript,regex,string,replace
This will do it (without jQuery): var str = 'hello/page/2/bye'; str = str.replace(/page\/\d+\//,''); The slashes need to be escaped (so they become \/), \d+ is any number, one or more times (so it also matches e.g. page/12/ Demo: http://jsfiddle.net/dd5aohbc/...
jquery,replace,html-lists,href
$("li #ctl01_Auxiliary_Auxiliary_rptWrapper_Auxiliary_rptWrapper_rpt_ctl01_NavigationLink").attr("href", "/Public/Cart") ...
Short answer: no. Long answer: Not really. If you know that the text you're inserting directly replaces the text you're removing you could perform a substitution by seeking to the portion of the file, reading a small block, substituting the new text and replacing that block. This would work for...
php,arrays,regex,string,replace
That's a simple search and replace for your defined markers. $text = 'Hello World [this-echo] Hello Everybody [nobody]'; $markers=array('[this-echo]','[nobody]'); $replacements=array('Today','Anybody'); $text= str_replace($markers,$replacements,$text); Output Hello World Today Hello Everybody Anybody Fiddle...
awk might help in this case: echo "0 1 2 3 4 5" | awk ' { for (i=1; i<=NF; i++) { if ((i-1)%2 == 0) { printf "%d ",$i; } else { print $i } } } ' We split by space and have 6 items. We, then, are...
First of all, the code you wrote here is not working, because when you open an outputStream to the exact file you try read from, it will empty the source file and the statement in.readLine() always returns null. So if this is your real code maybe this is the problem....
public static String filter(String input) { return input.replaceAll("[^A-Za-z0-9' ]", "").replaceAll(" +", " "); } The first replace replaces all characters except alphabetic characters, the single-quote, and spaces. The second replace replaces all instances of one or more spaces, with a single space....
php,string,replace,str-replace,php-5.3
You can't achieve this by str_replace. Use string translation (strtr) which was designed to do so instead: $words = 'word1 word2'; $wordsReplaced = strtr($words, [ 'word1' => 'word2', 'word2' => 'word1' ]); ...
Just use parentheses instead of braces. Regex.Replace(str, @"<(jv:(?:FirstName|MiddleInitial|LastName)>).*?</\1, "<$1</$1"); Braces signify matching any character within them one time. Parentheses match the full string....
Unfortunately VB.NET doesn't support escape sequences but you can use ChrW() to specify code point: s = s.Replace(ChrW(&H201C), """") That's for “, code for ” is &H201D. Note that using code points you're free to search & replace any Unicode character (not just what VB.NET has an escape for -...
The function as.logical will coerce any non-zero number to TRUE and zero to FALSE. Assuming your CSV has been loaded into a data frame called boxdata with the same column names as in the CSV: ifelse(as.logical(boxdata$Receiver_TotalVideoDecoderErrors), 'Error', 'No error') Or you can use broadcasting to do the work for you....
Assuming you can use java.nio.file.Files (since Java 1.7) and your text file isn't too big, I'd read all lines at once and go for the Matcher: Charset charset = Charset.forName("UTF-8"); List<String> lines = Files.readAllLines(file.toPath(), charset); for (String line : lines) { Matcher matcher = regexpPattern.matcher(line); if (matcher.matches()) { // do...
ignore_chars seems to be a perfect fit for this... http://sphinxsearch.com/docs/current.html#conf-ignore-chars...
Try sub('(?<=").*(?=")', y, x, perl=TRUE) #[1] "\t\t<taxon id=\"TOT_A01\"/>" ...
Assuming that your document is well-formed, i.e. <b> opening tags always match with a </b> closing tag, then this may be what you need: sed '[email protected]<[/]\?b>@\n&\[email protected]' path/to/input.txt | awk 'BEGIN {buf=""} /<b>/ {Y=1; buf=""} /<\/b>/ {Y=0; print buf"</b>"} Y {buf = buf$0} ' | tr -s ' ' Output: <b>data1</b>...
How about substring_index()? UPDATE `questions` SET answer = substring_index(substring_index(answer, '"', 2), '"', -1) ...
python,regex,python-2.7,replace,data-cleaning
You should be able to do this with a one RegeEx for all months: import re lines = [ "Yes, I'd say so. Nov 08, 2014 UTC", "Hell yes! Oct 01, 2014 UTC" ] for ln in lines: print re.sub(r'(\w+\s\d{2}, \d{4} UTC)$', r'\t\1', ln) Which will return: Yes, I'd say...
unix,replace,sed,match,condition
There is no need to loop through the file with a bash loop. sed alone can handle it: $ sed -r '/^.{14}R.{12}D/s/(.{52}).{4}/\10000/' file 05196220141228R201412241308D201412200055SA1155WE130800001300SL07Y051 05196220141228R201412241308A201412220350SA0731SU1950LAX C00020202020 05196220141228R201412241308D201412200055SA1155WE130800005300SL07Y051 05196220141228N201412241308A201412240007TU0548WE1107MEL C00000000015 07054820141228N201412220850D201412180300TH1400MO085000040300UL180001 This uses the expression sed '/pattern/s/X/Y/' file:...
XPath cannot modify an XML file. For that you need XSLT or XQuery. If this really is a global change (that is, if you want to change this text regardless of the context where it appears) then I would be very inclined myself to do it using a text editor:...
There have been a lot of responses to this question and for a newbie to regex it is probably a bit overwelming, Overall the best response has been: var str2 = str.replace(/^,/gm, ''); This works by using ^, to check if the first character is a comma and if it...
regex,replace,ms-word,word-vba,pci
Since it seems like you are using the Word wildcard syntax, you probably can use <, which asserts beginning of word, and >, which asserts end of word to prevent the pattern from matching when the text is preceded or succeeded by letters or numbers (which is how it seems...
You will need to put the value that is between the square brackets into a capture group, and substitute that in your replacement. In short, this will do it: Regex.Replace(input, @"Session\[(""\w+"")]\s==", @"Session[$1].ToString() =="); where $1 will insert the contents of your first capture group (determined by parenthesis in the pattern...
It is generally not a good idea to iterate over the output of find since file names can contain the $IFS which would break the loop. Use the -exec option of find instead: find -iname 'PATCH*' -exec sed -i 's#`which cat`#/bin/cat#g' {} \; ...
python,pandas,replace,fill,calculated-columns
Here's a kind of brute-force method. There is probably something more elegant, but you could explicitly loop over the rows like this: df = pd.DataFrame([0, -1, -1, -1, 0 , 0, 0, 1, 0]) df.columns = ['A'] df['B'] = df['A'] # loop here for i in range(1,len(df)): if df.A[i] ==...
Turning on the warnings, you may spot the error yourself: g++ -o main main.cpp -Wall Errors: main.cpp: In function ‘std::string lowercase(std::string)’: main.cpp:9:21: warning: comparison between signed and unsigned integer expressions [-Wsign-compare] for (int i=0; i < ans.size(); i++) ^ main.cpp:14:1: warning: no return statement in function returning non-void [-Wreturn-type] }...
s.replace(/(丁目)(?=\d)/g, "$1 ") This should do it for you.Your earlier regex was not working cos if ?: which makes it non capturing and $1 had nothing in it.See demo. https://regex101.com/r/nS2lT4/10 var re = /(丁目)(?=\d)/g; var str = '北23条東12丁目5-30-405'; var subst = '$1 '; var result = str.replace(re, subst); ...
The loop is killing you. This will remove spaces in an entire column in one shot: Sub SpaceKiller() Worksheets("Sheet1").Columns("A").Replace _ What:=" ", _ Replacement:="", _ SearchOrder:=xlByColumns, _ MatchCase:=True End Sub Adjust the range to suit. If you want to remove double spaces, then: Sub SpaceKiller() Worksheets("Sheet1").Columns("A").Replace _ What:=" ", _...
string,actionscript-3,replace,find
You need to mask "(" and "*" in your pattern: var pattern:RegExp = /\(\*/gi; ...
The easiest way is to isolate sequences like 15、18 and replace all commas in them. text = "3条4~12丁目、15~18条12丁目、2、3条5丁目"; text. replace(/(?:\d+、)+\d+/g, function(match) { return match.replace(/、/g, "&"); }). replace(/条/g, '-jō '). replace(/丁目/g, '-chōme'). replace(/~/g, '-'). replace(/、/g, ', ') // => "3-jō 4-12-chōme, 15-18-jō 12-chōme, 2&3-jō 5-chōme" (Also... Where the heck do you...
javascript,php,regex,string,replace
Something like this? ( °_• ) var txt="or some text or before (.)or(.) anywhere Or"; console.log(txt.replace(/\bor\b/gi,"OR")); // "OR some text OR before (.)OR(.) anywhere OR" ...
sql-server,tsql,replace,case,sql-server-2014
It is obvious that '' means blank (Not null) and it exists between two consecutive letters also. (As it is blank) So Sql can not go on replacing that blank in between every letter with the string you want. That's why you can not replace '' Check this query SELECT...
I think the only thing you're maybe not wrapping your head around is the idea of callbacks and asynchronous methods. You're currently running your console.log statements before the replaceWith occurs. The function() { } block (your "callback") passed as the second parameter to $.get doesn't execute until the AJAX call...
javascript,php,jquery,regex,replace
You can use the following to match: \*\*([^*]*)\*\* And replace with: <strong>$1</strong> $1 - It is the back reference to the first capture group ([^*]*).. basically telling the regex to replace with whatever was captured between first set of parentheses ( ) See DEMO JScode: var str='test1 **test2** test3'; str...
java,string,class,replace,bukkit
Your problem is, most likely, because you expect that your function mutates the original String you passed to the method. Strings are immutable in java, and besides, you cannot modify the original reference you pass to method from that method. To get the result you most likely expect, store returned...
obj.roles[0] is a object {"name":"with whom"}. you cant replace string with object. you need to refer to property "name" in the object obj.roles[0].name Another problem is that var finalXML get a new value every line. you need to add a new value to the variable, not replcae it. var finalXML...
It should be: task writeVersion << { ant.replace( file: 'version.txt', token: 'versionNumber', value: '1.0.0' ) } and: version.number=versionNumber ...
java,string,replace,integer,replaceall
The replace only works on the first execution. After that, value no longer contains -time-. As a solution, you can use a temporary variable to do your replacement: String displayValue = value.replace("-time-", time + ""); ...
javascript,replace,google-spreadsheet
Instead of including DATA1 in the URL, append it with data coming from f.ex a textbox: function loadChart() { var txtbox = document.getElementById('textbox').value; var url = 'https://docs.google.com/spreadsheet/ccc?key=...&sheet=' + txtbox; var query = new google.visualization.Query(url); query.send(handleQueryResponse); } ...
Try it like this: for i=1:nIndices slice = simulatedReturnsEVT1(:,i,:); slice(slice < MinAcceptableVal(i))=MinAcceptableVal(:,i); slice(slice > MaxAcceptableVal(i))=MaxAcceptableVal(:,i); simulatedReturnsEVT1(:,i,:) = slice; end ...