（PHP）如何查找以模式开头的单词并替换所有单词？

Question

I have a string. An example might be "Contact /u/someone on reddit, or visit /r/subreddit or /r/subreddit2"

I want to replace any instance of "/r/x" and "/u/x" with " [/r/x](http://reddit.com/r/x) " and " [/u/x](http://reddit.com/u/x) " basically.

So I'm not sure how to 1) find "/r/" and then expand that to the rest of the word (until there's a space), then 2) take that full "/r/x" and replace with my pattern, and most importantly 3) do this for all "/r/" and "/u/" matches in a single go...

The only way I know to do this would be to write a function to walk the string, character by character, until I found "/", then look for "r" and "/" to follow; then keep going until I found a space. That would give me the beginning and ending characters, so I could do a string replacement; then calculate the new end point, and continue walking the string.

This feels... dumb. I have a feeling there's a relatively simple way to do this, and I just don't know how to google to get all the relevant parts.

Answer 1

A simple preg_replace will do what you want .

Try:

$string = preg_replace('#(/(?:u|r)/[a-zA-Z0-9_-]+)#', '[\1](http://reddit.com\1)', $string);

Here is an example: http://ideone.com/dvz2zB

You should see if you can discover what characters are valid in a Reddit name or in a Reddit username and modify the [a-zA-Z0-9_-] charset accordingly.

Answer 2

In my opinion regex would be an overkill for such a simple operation. If you just want to replace instance of "/r/x" with "[r/x](http://reddit.com/r/x)" and "/u/x" with "[/u/x](http://reddit.com/u/x)" you should use str_replace although with preg_replace it'll lessen the code.

str_replace("/r/x","[/r/x](http://reddit.com/r/x)","whatever_string");

use regex for intricate search string and replace. you can also use http://www.jslab.dk/tools.regex.php regular expression generator if you have something complex to capture in the string.

Answer 3

You are looking for a regular expression.

A basic pattern starts out as a fixed string. /u/ or /r/ which would match those exactly. This can be simplified to match one or another with /(?:u|r)/ which would match the same as those two patterns. Next you would want to match everything from that point up to a space. You would use a negative character group [^ ] which will match any character that is not a space, and apply a modifier, * , to match as many characters as possible that match that group. /(?:u|r)/[^ ]*

You can take that pattern further and add a lookbehind, (?<= ) to ensure your match is preceded by a space so you're not matching a partial which results in (?<= )/(?:u|r)/[^ ]* . You wrap all of that to make a capturing group ((?<= )/(?:u|r)/[^ ]*) . This will capture the contents within the parenthesis to allow for a replacement pattern. You can express your chosen replacement using the \\1 reference to the first captured group as [\\1](http://reddit.com\\1) .

In php you would pass the matching pattern, replacement pattern, and subject string to the preg_replace function.