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Remove portion of string unless it contains variable

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  Remove portion of string unless it contains variable Jason C <jwcarlton@gmail.com> - 2012-07-13 16:27 -0700
    Re: Remove portion of string unless it contains variable Jason C <jwcarlton@gmail.com> - 2012-07-13 17:51 -0700
      Re: Remove portion of string unless it contains variable "Evertjan." <exxjxw.hannivoort@inter.nl.net> - 2012-07-14 10:03 +0000
    Re: Remove portion of string unless it contains variable Mike Winter <usenet@michael-winter.me.invalid> - 2012-07-16 01:06 +0100

#15045 — Remove portion of string unless it contains variable

Let's say that I have:

var ignore= "example.com";
var b = "Check out this awesome image!<br><br><img src='http://www.example.com/whatever.jpg' width='300' height='300' border='0'><br><br>Awesome.";

This comes from a message board that allows copy and paste from other sites, so the img tag can change a lot; the only thing for sure is that it will start with <img and have src=... in it.

I'm removing it using:

b = b.replace(/<img.*?src=["']*(.*?)["'}*.*?>/gi, '');

How can I modify this to only replace if $1 does not contain "ignore"?

As an addendum, what if I change ignore to:

var ignore = new Array("example.com", "example_2.com");

and want to exclude img tag that includes any of the array elements in $1?

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#15047

On Friday, July 13, 2012 7:27:09 PM UTC-4, Jason C wrote:
> b = b.replace(/<img.*?src=["']*(.*?)["'}*.*?>/gi, '');

Type; that should have been a square after "', not a curly:

b = b.replace(/<img.*?src=["']*(.*?)["']*.*?>/gi, '');

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#15050

Jason C wrote on 14 jul 2012 in comp.lang.javascript:

> On Friday, July 13, 2012 7:27:09 PM UTC-4, Jason C wrote:
>> b = b.replace(/<img.*?src=["']*(.*?)["'}*.*?>/gi, '');
> 
> Type; that should have been a square after "', not a curly:
> 
> b = b.replace(/<img.*?src=["']*(.*?)["']*.*?>/gi, '');

Try this:

b = b.replace(/<img.*?src=(["'])[^\1]+\1[^>]*>/gi, '');


\1 is replaced by the actual match of (["'])

[^\1]+ quoted url should be not empty

[^>]*> matches anything till and including the next >

-- 
Evertjan.
The Netherlands.
(Please change the x'es to dots in my emailaddress)

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#15062

On 14/07/2012 00:27, Jason C wrote:

> [snip]
>
> b = b.replace(/<img.*?src=["']*(.*?)["'}*.*?>/gi, '');
>
> How can I modify this to only replace if $1 does not contain "ignore"?

Something along the lines of:

   function removeImgElement(string, ignore) {
     return string.replace(/<img.+?src=(["'])([^\1>]*)\1[^>]*>/gi,
       function(substring, qm, attributeValue) {
         return (attributeValue.indexOf(ignore) == -1) ? '' : substring;
       });
   }

> As an addendum, what if I change ignore to:
>
> var ignore = new Array("example.com", "example_2.com");
>
> and want to exclude img tag that includes any of the array elements in $1?

Similarly:

   function removeImgElement(string, ignore) {
     if ((typeof ignore != 'object') || (ignore.constructor != Array))
       ignore = [String(ignore)];

     return string.replace(/<img.+?src=(["'])([^\1>]*)\1[^>]*>/gi,
       function(substring, qm, attributeValue) {
         for (var i = 0; i < ignore.length; i++)
           if (attributeValue.indexOf(ignore[i]) != -1)
             return substring;
         return '';
       });
   }

Hope that helps,
Mike


-- 
Michael Winter
Replace ".invalid" with ".uk" to reply by e-mail.

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