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[RESOLVED] cut a continues string at 32nd position

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Mar 10.2012

I m a js beginer.. i was trying to write a function to cut my string at every 32nd position if there was no spaces.

Explanation:

If the string have spaces dont cut. else if the string is long (no space) then at 32 nd point i want to include a space. again i want to continue and want to include a space at next 33rd postion if there was no spaces till that.

Please anyone help me to write this code..

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JavaScript

5 Comments(s) ↴

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@007JulienMar 10.2012 — Try this code
[CODE]<body>
<label>Before<br><textarea id="bfr" rows="10" cols="32"></textarea></label><br>
<label>After<br><textarea id="ftr" rows="10" cols="32"></textarea></label>

<script type="text/javascript">
var str="Loremipsumdolor sitametconsectetueregetegetetutPhasellus. PharetraQuisquetemporsemconsectetuerSedsapienelitutVivamus Vestibulum. VitaeconvallisfeugiatantecursusProincondimentumauctormetuselitNunc. PellentesqueNamVivamus.";
document.getElementById('bfr').innerHTML=str;
[COLOR="Blue"]var newStr = str.replace(/([w]{31})(?=w)/ig,'$1 ');[/COLOR]
document.getElementById('ftr').innerHTML=newStr;
</script>
</body>[/CODE]We replace any suit of 31 characters followed by another character by this suit (the first sub-pattern $1) followed by a space. With the ?= (a lookahead positive assertion) the second sub-pattern is not matched and count for the possible following pattern (Lookbehind assertions are not supported with javascript).

Edit : With the punctuation, it would be better to work with non space characters
[CODE][COLOR="Blue"]var newStr = str.replace(/([^s]{31})(?=[^s])/ig,'$1 ');[/COLOR][/CODE]
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@johnrockauthorMar 13.2012 — Try this code
[CODE]<body>
<label>Before<br><textarea id="bfr" rows="10" cols="32"></textarea></label><br>
<label>After<br><textarea id="ftr" rows="10" cols="32"></textarea></label>

<script type="text/javascript">
var str="Loremipsumdolor sitametconsectetueregetegetetutPhasellus. PharetraQuisquetemporsemconsectetuerSedsapienelitutVivamus Vestibulum. VitaeconvallisfeugiatantecursusProincondimentumauctormetuselitNunc. PellentesqueNamVivamus.";
document.getElementById('bfr').innerHTML=str;
[COLOR="Blue"]var newStr = str.replace(/([w]{31})(?=w)/ig,'$1 ');[/COLOR]
document.getElementById('ftr').innerHTML=newStr;
</script>
</body>[/CODE]We replace any suit of 31 characters followed by another character by this suit (the first sub-pattern $1) followed by a space. With the ?= (a lookahead positive assertion) the second sub-pattern is not matched and count for the possible following pattern (Lookbehind assertions are not supported with javascript).

Edit : With the punctuation, it would be better to work with non space characters
[CODE][COLOR="Blue"]var newStr = str.replace(/([^s]{31})(?=[^s])/ig,'$1 ');[/COLOR][/CODE][/QUOTE]

Yeah.. this is working fine... But this code seems difficult to understand for me. I dint see this type of code in js before. Can you tell me where i can learn these basics.. esp this part " (/([^s]{31})(?=[^s])/ig,'$1 '); "
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@TheAliveWinnerMar 13.2012 — Learn [B]RegExp[/B]: JS RegExp
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@johnrockauthorMar 13.2012 — Learn [B]RegExp[/B]: JS RegExp[/QUOTE]

thanks bro..!!
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@007JulienMar 13.2012 — You find probably that i, for ignore case, is not useful with non space characters, which could be match with S. Then a shorter regular expression is :

[CODE][COLOR="Blue"]var newStr = str.replace(/(S{31})(?=S)/g,'$1 ');[/COLOR][/CODE]
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