String replace: multiple replace is not working in JQuery

Question

I have a string and some text (name attr) should be replaced with dynamic text. For example options[team_member][0][name] will be replaced with options[team_member][1][name].

What far i did:

    var current = 1;

    var block = jQuery('#contents').html();
    var replace_str = 'options[team_member]['+(current-1)+']';
    var replace_with = 'options[team_member]['+(current)+']';

    /* replace only first match */
    var rep_block = block.replace(replace_str,replace_with);

    /* replace nothing */
    // var rep_block = block.replace(/replace_str/g,replace_with);

    /* replace nothing */
    // var rep_block = block.replace(/(replace_str)/g,replace_with);

    alert(rep_block);

Please check full code in jsFiddle

I cannot find the way how can i solve this. Please help me. Thanks in advance.

ok. i want to add a html block with some inputs with some dynamic name. How can i make it? Do you like to share your idea please. — itskawsar
– itskawsar, Commented Jul 6, 2013 at 11:05

Hitesh Siddhapura · Accepted Answer · 2013-07-06 11:13:53Z

9

Please replace

var rep_block = block.replace(replace_str,replace_with);

with

var rep_block = block.split(replace_str).join(replace_with);

answered Jul 6, 2013 at 11:13

Hitesh Siddhapura

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Comments

Alex Fitiskin · Accepted Answer · 2013-07-06 19:52:49Z

-1

Use global replace with "g" flag: /reg_ex_here/g or new RegExp(exp, 'g')

var string = 'bla bla bla';
string.replace(/bla/g, 'ok'); // -> ok ok ok
string.replace(new RegExp('bla', 'g'), 'ok'); // -> ok ok ok

In your code:

var replace_str = 'options\\[team_member\\]\\['+(current-1)+'\\]';
rep_block = block.replace(new RegExp(replace_str, 'g'), replace_with);

edited Jul 6, 2013 at 19:52

answered Jul 6, 2013 at 11:06

Alex Fitiskin

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3 Comments

itskawsar Over a year ago

please check jsFiddle first. var rep_block = block.replace(/replace_str/g,replace_with); // replace nothing

Alex Fitiskin Over a year ago

oh, string to replace include brackets... i updated jsfiddle: jsfiddle.net/2QFUw/6 but this is not good solution. Use split + join solution by @Hitesh

ThiefMaster Over a year ago

Please update your answer anyway to use a properly escaped replace_str. Right now the second part is plain wrong.

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