Re: Dividing, and keeping, text from the first space

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On Wed, 2006-08-09 at 15:30 +0900, Dave M G wrote:
> PHP List,
> 
> This regular expression stuff is way tricky.
> 
> Thanks to help from this list, I have an expression that will select the 
> first word of a string, up to the first white space:
> "#^(.*)\s#iU"
> 
> But after some consideration, I realized that I wanted to keep both 
> parts of the original text. The first word, and then everything that 
> came after it, should be divided and stored in separate variables.
> 
> I looked through the php.net manual, and found preg_split().
> 
> At first I thought using my original expression would work, if I 
> included the PREG_SPLIT_DELIM_CAPTURE parameter. It would pull out the 
> first word, but also keep it.
> 
> Turns out that's not true. It removes the first word and the space, and 
> looks for something before and after and only finds something after. 
> Thus, in my current situation, it returns only one variable, which 
> contains everything after the first space.
> 
> I realized what I need to do in this case is not select everything up to 
> the first space, but to find the first space without selecting what 
> comes before it. So the expression I had wasn't suitable for preg_split().
> 
> So then, what is the right expression?
> 
> Looking back over previous discussion and resources about regular 
> expression syntax, I thought I had to say:
> Start at the beginning: ^
> Ignore anything that isn't a space: [^\s]
> Select the space character: \s
> Be case insensitive and not greedy: iU
> 
> Thus, my expression should be (using hash marks, #, as delimiters):
> #^[^\s]|s#iU
> 
> More specifically, my preg_split() syntax is:
> $parts = preg_split("#^[^\s]|s#iU", $word, PREG_SPLIT_DELIM_CAPTURE);
> 
> But it returns a two element array, where the first element is empty, 
> and the second element is the whole original string.
> 
> Where did I go wrong this time?

Use preg_match() and pay special attention to the manual as it refers to
the third parameter :) The expression you need follows:

   "#^([^\\s]*)\\s(.*)$#U"

You don't need the insensitive modifier btw since you aren't actually
matching anything that is applicable.

Cheers,
Rob.
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