PHP Simple String Manipulation

Question

I have a string with multiple lines. For this example, my string will be this:

Name:Jaxo
Description:A person on Stackoverflow
Question:$this->questionName();

How could I get just, say, the 'Description'? I need everything after the description, ie 'A person on Stackoverflow'. I've tried a regex like this, but it doesn't work: /^Description:(.+?)\n/i

Any help is much appreciated!

Thanks

-Jaxo

'description' not is in start of string,you need remove ^ for your regex works fine. — The Mask
– The Mask, Commented Jul 8, 2011 at 21:00

citizen conn · Accepted Answer · 2011-07-08 20:51:42Z

1

This should work for you:

if (preg_match('/Description:(.+)/im', $subject, $regs)) {
    $result = $regs[1];
} else {
    $result = "";
}

Where $result is the Description name.

answered Jul 8, 2011 at 20:51

citizen conn

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1 Comment

Scott Over a year ago

Worked perfectly, thanks!! I shortened it to this, as well: echo $result=preg_match('/Description:(.+)/im',$subject,$r)?$r[1]:null;

uadnal · Accepted Answer · 2011-07-08 20:51:43Z

0

If there is a newline character separating each part of the label you could explode.

$array = explode("\n",$string); // separate params
$desc = explode(":",$array[1]); // separate description

This way you could get any of the parameters.

answered Jul 8, 2011 at 20:51

uadnal

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1 Comment

Donovan Over a year ago

Correct, but if there are semicolons in the description then count($desc) > 2 (just a specification).

The Mask · Accepted Answer · 2011-07-08 20:54:09Z

0

Try this:

$a="Name:Jaxo
Description:A person on Stackoverflow
Question:\$this->questionName();"; 

preg_match("/Description:([^\n]+)/i",$a,$m);
print_r($m);

Output:

Array ( [0] => Description:A person on Stackoverflow [1] => A person on Stackoverflow )

answered Jul 8, 2011 at 20:54

The Mask

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