PHP array with variable in it

Question

I just got some really strange problem.

Check it out:

$yourWord='http://mywebsite.com/combo-points-carmaboy-boys-sixer';
$URLW = "'kid', 'kids', 'boys', 'six'";
$targets = array($URLW);


$pattern = '/\b('.implode('|', $targets).')\b/';

if (preg_match($pattern, $yourWord, $matches)) {
echo 'FOUND!!';
} else {
echo 'NOT FOUND!';
}

In this way it's giving me result NOT FOUND! but when i do that $targets = array('kid', 'kids', 'boys', 'six'); it's working.

So why it's not working on the way i make it in my code sample ?

And how i can use it with variable in the array ?

Thanks

$URLW is a string, it wont get evaluated into an array structure because it looks like one. — Lawrence Cherone
– Lawrence Cherone, Commented Jul 12, 2014 at 20:26
PHP users are so used to evaluate strings to whatever type or construction they want, that they forget how the programming really works... — Ivan Yonkov
– Ivan Yonkov, Commented Jul 12, 2014 at 20:44

FuzzyTree · Accepted Answer · 2014-07-12 20:26:33Z

2

$targets = array("'kid', 'kids', 'boys', 'six'");

is a an array consisting of a single element that's a comma separated string

$targets = array('kid', 'kids', 'boys', 'six');

is an array consisting of 4 string elements

answered Jul 12, 2014 at 20:26

FuzzyTree

32.4k3 gold badges58 silver badges87 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Max · Accepted Answer · 2014-07-12 20:28:17Z

To explain why it's not found, array($URLW) returns an array with ONE element, as it's ONE string containing commas. This makes implode() return false (as it expects the first argument to be an array, and not a string), and your statement completely different than you expected (instead it should convert the value int(0) into a string in the regex-pattern within the word-boundaries, which isn't what you want, and which doesn't exist in your pattern). If you need to use the string with commas, split it first using explode(), then keep that array:

$URLW = explode(", ", "'kid', 'kids', 'boys', 'six'");

If you don't need to use it, then simply declare the array with the regular syntax:

$URLW = Array("kid", "kids", "boys", "six");

ArtisticPhoenix · Accepted Answer · 2014-07-12 20:30:53Z

0

Hi whats this all about?

$URLW = "'kid', 'kids', 'boys', 'six'";
$targets = array($URLW);

Try like this

 $targets = array('kid', 'kids', 'boys', 'six');

Here is your regx before changing it, by the way.

 /\b('kid', 'kids', 'boys', 'six')\b

Here is the regx you wanted,

 /\b(kid|kids|boys|six)\b

And if you need this bit $URLW as a string

 $URLW = "'".implode("', '", $targets )."'";

answered Jul 12, 2014 at 20:30

ArtisticPhoenix

21.7k2 gold badges27 silver badges38 bronze badges

Comments

Gergo Erdosi · Accepted Answer · 2014-07-12 20:31:25Z

0

$URLW is a string, array() won't be able to convert the comma separated items to expected array format. You either need to initialize the array properly:

$targets = ('kid', 'kids', 'boys', 'six');

Or convert the string to an array using something like this:

$targets = array_map(function($target) {
  return trim($target, "' ");
}, explode(',', $URLW));

answered Jul 12, 2014 at 20:31

Gergo Erdosi

42.3k21 gold badges122 silver badges95 bronze badges

Collectives™ on Stack Overflow

PHP array with variable in it

4 Answers 4

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related