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chosch
Known Participant
February 28, 2024
Answered

Replace() function adding extra parenthesis to string values that have parenthesis.

  • February 28, 2024
  • 4 replies
  • 1518 views

I'm attempting to add text to string values in an array. Some of these strings have parenthesis around them and for some reason the replace() function is adding extra parenthesis around these.

So "apples" becomes "I like apples." but "(oranges)" becomes "(I like (oranges).)"

Why is this happening? Is there a better way to do this?

Here is my code:

var words = new Array();
words = ["apples", "(oranges)", "peaches", "(mangos)"];
for (i = 0; i < words.length; i++) {
    words[i] = words[i].replace(new RegExp(words[i], "g"), "I like " + words[i] + ".");
    alert(words[i]);
}

 

    This topic has been closed for replies.
    Correct answer m1b

    That's not working either. Now I get "()I like ()oranges().()".

    var words = "(oranges)";
words = words.replace(/\(|\)/g, "\($1\)");
words = words.replace(new RegExp(words, "g"), "I like " + words + ".");
alert(words);

    But you're pointing me in the right direction. Guess I need to learn more about regular expressions.


    I'm sure there is a reason that isn't obvious in your simplified example, but why not use this:

     

    var words = ["apples", "(oranges)", "peaches", "(mangos)"];
for (i = 0; i < words.length; i++) {
    words[i] = words[i].replace(words[i], "I like " + words[i] + ".");
    alert(words[i]);
}

     

    I mean, from your example we could do this:

    var words = ["apples", "(oranges)", "peaches", "(mangos)"];
for (i = 0; i < words.length; i++) {
    words[i] = 'I like ' + words[i] + '.';
    $.writeln(words[i]);
}

     

    If that can't be done in your case, can you please improve your example so we can understand what you really need to do?

    - Mark

     

    4 replies

    Marc Autret
    Marc Autret
    Legend
    February 29, 2024

    Hi all,

     

    1. For those seeking a more generic solution, based on localize:

     

    // METHOD 1 - Using localize()
// -----------------------------

const PATTERN = "I like %1."; // ← Specify your pattern using `%1` placeholder.
var words = [ "apples", "(oranges)", "peaches", "(mangos)" ];

const L = $.global.localize;
for( var i=words.length ; i-- ; words[i]=L(PATTERN, words[i]) );

// Display result:
alert( words.join('\r') );

     

    2. For those seeking a faster, generic solution, bypassing the for loop:

     

    // METHOD 2 - No loop.
// -----------------------------

const PATTERN = "I like %1."; // ← Specify your pattern using `%1` placeholder (once).
var words = [ "apples", "(oranges)", "peaches", "(mangos)" ];

const XSEP = '\x01';
var t = PATTERN.split('%1');
words = ( t[0] + words.join(t[1]+XSEP+t[0]) + t[1] ).split(XSEP);

// Display result:
alert( words.join('\r') );

     

    Best,

    Marc

      m1b
      Community Expert
      m1bCommunity Expert
      Community Expert
      March 1, 2024

      Very nice, Marc!

      Peter Kahrel
      Community Expert
      Peter KahrelCommunity Expert
      Community Expert
      February 29, 2024

      Mark is right. You're replacing literals, no need for regular expressions.

      brian_p_dts
      Community Expert
      brian_p_dtsCommunity Expert
      Community Expert
      February 28, 2024

      You are converting "(oranges)" into a regex. Before doing that, you would need to escape the parens if you want to find them: .replace(/(|\)/g,"\$1")

        C
        choschAuthor
        Known Participant
        February 28, 2024

        Ok, that's all well and good. But your code to escape the parens removed them from the string. What if I want to keep the parens? My ultimate goal is for it say "I like (oranges)."

        Here is how I'm using your code:

        var words = "(oranges)";
words = words.replace(/\(|\)/g, "\$1");
words = words.replace(new RegExp(words, "g"), "I like " + words + ".");
alert(words);
        brian_p_dts
        Community Expert
        brian_p_dtsCommunity Expert
        Community Expert
        February 28, 2024

        Got the replace wrong. Should be... ("\($1\)")

        Robert at ID-Tasker
        Robert at ID-Tasker
        Legend
        February 28, 2024

        Isn't it the correct behaviour?

         

        You are using "(oranges)" as a searchValue - and as "()" are used for grouping - RegExp is looking for "oranges" - WITHOUT "()" and replaces word "oranges" with "I like " + "(oranges)" + "." - but as your words[i] is "(oranges)" - it leaves "()" intact and only replaces "oranges".

         

          C
          choschAuthor
          Known Participant
          February 29, 2024

          It's not what I intended, for sure. Your explanation makes sense and tells me I need to learn more about how RegExp works.

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