RegEx string “preg_replace”?

I'm not sure I understand what you're asking. Do you mean every line in the file looks like that, and you want to process all of them? If so, this regex would do the trick.

I'm not sure I understand what you're asking. Do you mean every line in the file looks like that, and you want to process all of them? If so, this regex would do the trick: '#^.

*/#' That simply matches everything up to and including the last slash, which is what your regex would do if it weren't for that rogue '$' everyone's talking about. If there are other lines in other formats that you want to leave alone, this regex will probably suit your needs: '#^\. /\w/\w{3}/\w{1,13}/#" Notice how I changed the regex delimiter from '/' to '#' so I don't have to escape the slashes inside.

You can use almost any punctuation character for the delimiters (but of course they both have to be the same).

That's much cleaner, the lines should all be in the same format but I don't want to assume that. I used the second version as it's simpler and cleaner, just needed to change to \w- to account for hyphens as well. Am I right in assuming that \w is alphanumeric characters and underscores?

– phil Sep 8 '09 at 11:15 Yes, \w is the same as A-Za-z0-9_. In some other regex flavors it also matches accented letters plus letters and digits from other writings systems, but PHP's \w is limited to ASCII. – Alan Moore Sep 8 '09 at 12:01.

The only mistake I encouter is the anchor for the string end $ that should be removed. And your expression is also missing the _ character: /^(\. \/0-9a-zA-Z{1}\/0-9a-zA-Z{3}\/0-9a-zA-Z_{1,13}\/)/ A more general pattern would be to just exclude the /: /^(\.

\/^\/{1}\/^\/{3}\/^\/{1,13}\/).

Thanks, works fine now! Nice to know I was only making one tiny mistake! The second example throws out an error however!

Warning: preg_replace() function. Preg-replace: Unknown modifier '' The first one works fine though. Thanks again!

– phil Sep 8 '09 at 10:36 @phil: Fixed it. – Gumbo Sep 8 '09 at 11:09.

You should use PHP's builtin parser for extracting the values out of the csv before matching any patterns.

The values do not have quotation marks surrounding them in the file that this is processing. Purely out of educational interest how would I go about performing the same pattern replacement without using regex? I wouldn't know where to begin I'm afraid.

– phil Sep 8 '09 at 10:50 Sorry, I didn't read your question well enough. I guess you must use regular expressions here, but I would extract the values out of the csv first, and apply the RE afterwards. – soulmerge Sep 8 '09 at 11:18.

The $ means the end of the string. So your pattern would match . /1/024/9780310320241/ and .

/t/fla/8204909_flat/ if they were alone on their line. Remove the $ and it will match the first four parts of your string, replacing them with a space.

$pattern = "/(\. \/0-9a-z{1}\/0-9a-z{3}\/0-9a-z\_+\.(jpg|bmp|jpeg|png))\n/is"; I just saw, that your example string doesn't end with /, so may be you should remove it from your pattern at the end. Also underscore is used in the filename and should be in the character class.

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