PHP array_key_exists key LIKE string, is it possible?

Question

I've done a bunch of searching but can't figure this one out.

I have an array like this:

$array = array(cat => 0, dog => 1);

I have a string like this:

I like cats.

I want to see if the string matches any keys in the array. I try the following but obviously it doesn't work.

array_key_exists("I like cats", $array)

Assuming that I can get any random string at a given time, how can I do something like this?

Pseudo code:

array_key_exists("I like cats", *.$array.*)
//The value for cat is "0"

Note that I want to check if "cat" in any form exists. It can be cats, cathy, even random letter like vbncatnm. I am getting the array from a mysql database and I need to know which ID cat or dog is.

_"I want to see if the string matches any keys in the array"_ Do you mean if any **word** from the string matches any key? — sidyll, Oct 22 '16 at 18:46
@KeithC. I edited your question, adding a comment you left and the "database" tag as it could be relevant. Also, the RDBMS used is unknown, so that could play a role here. If you feel the db stuff is irrelevant, you can roll it back to an earlier revision. — Funk Forty Niner, Oct 22 '16 at 19:03
But in your array you have just the key "cat" and in your string you have "cats", so it won't match exactly by word because the string is plural. Should the key be "cats" instead? — Cave Johnson, Oct 22 '16 at 19:09
@KodosJohnson that's fine. I just want to know if anything in my array matches anything in the string. Even if the string said "I like catastrophes" I'd like to know "cat" was present. — Keith C., Oct 22 '16 at 19:10
aah I see. Can you put that in your question please? It's useful to know. — Cave Johnson, Oct 22 '16 at 19:12
@Fred-ii- I accidentally wrote "car" instead of cat in one of my responses. My mistake. — Keith C., Oct 22 '16 at 19:12

Anthony · Accepted Answer · 2019-01-17T10:20:28.720

7

You can use a regex on keys. So, if any words of your string equal to the key, $found is true. You can save the $key in the variable if you want. preg_match function allows to test a regular expression.

$keys = array_keys($array);
$found = false;
foreach ($keys as $key) {
    //If the key is found in your string, set $found to true
    if (preg_match("/".$key."/", "I like cats")) {
        $found = true;
    }
}

EDIT :

As said in comment, strpos could be better! So using the same code, you can just replace preg_match:

$keys = array_keys($array);
$found = false;
foreach ($keys as $key) {
    //If the key is found in your string, set $found to true
    if (false !== strpos("I like cats", $key)) {
        $found = true;
    }
}

edited Jan 17 '19 at 10:20

answered Oct 22 '16 at 19:03

Anthony

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Thanks Anthony. When I try this I get "Warning: preg_match(): Delimiter must not be alphanumeric or backslash in /var/app/current/reporting/SqsToDbWorker.php on line 56" Not sure why it must NOT be alphanumeric? – Keith C. Oct 22 '16 at 19:25
I'm sorry that was my fault, I fix the issue. In preg_match we have to surround pattern ($key, here) by a delimiter. I added it but you can use a sharp if you want why not. – Anthony Oct 22 '16 at 19:30
I think regular expression are a little too resource-heavy for this kind of thing. You can just use `strpos` – Cave Johnson Oct 22 '16 at 19:33
Yup that fixed it. Thanks. Note, you have a period before the first "/" which doesn't belong. Fix that for anyone who references this in the future. – Keith C. Oct 22 '16 at 19:33
@KodosJohnson that is correct, php documentation suggests the same thing. – Keith C. Oct 22 '16 at 19:34
@KodosJohnson you are right, so I added strpos into my answer. – Anthony Oct 22 '16 at 19:39
Can you do `if (strpos("I like cats", $key) !== false)` instead? The reason is that if `cats` is in the beginning of the string, `strpos` will return 0 since it returns the position number and that is falsey, so it will evaluate to false even if the string is there. – Cave Johnson Oct 22 '16 at 19:42
I just did it, thanks. Indeed strpos returns the position you are right. And if strpos does not find the string it returns a *falsy* value, so check with !== is important as said in documentation. – Anthony Oct 22 '16 at 19:45
@AnthonyB One final note .. it wasn't finding everything while running a test so I did strpos("/I like cats/", $key) which seemed to fix it. I'm not sure if it is good for to put regular expression here but it did fix it. – Keith C. Oct 22 '16 at 19:58
@KeithC. can you give me an exemple ? I'll try to fix it if there is a bug. – Anthony Oct 23 '16 at 11:07

score 2 · Answer 2 · answered Oct 22 '16 at 19:47

2

This should help you achieve what you're trying to do:

$array         = array('cat' => 10, 'dog' => 1);

$findThis      = 'I like cats';

$filteredArray = array_filter($array, function($key) use($string){

    return strpos($string, $key) !== false;

}, ARRAY_FILTER_USE_KEY);

I find that using the array_filter function with a closure/anonymous function to be a much more elegant way than a foreach loop because it maintains one level of indentation.

answered Oct 22 '16 at 19:47

useyourillusiontoo

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I'll give this a shot too as it seems very clean. – Keith C. Oct 22 '16 at 20:00
Worked perfectly for my use case. – Michel Rummens Jul 07 '22 at 16:32

ScaisEdge · Answer 3 · 2016-10-22T19:04:02.083

0

You could do using a preg_match with the value not in array but in search criteria

 if(preg_match('~(cat|dog)~', "I like cats")) {
    echo 'ok';
}

or

$criteria = '~(cat|dog)~';

 if (preg_match($criteria, "I like cats")) {
    echo 'ok';
}

Otherwise you could use a foreach on your array

 foreach($array as $key => $value ) {
     $pos = strpos("I like cats", $key);
     if ($pos > 0) {
      echo $key .  ' '. $value;
     }

 }

edited Oct 22 '16 at 19:04

answered Oct 22 '16 at 18:48

ScaisEdge

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Understood. However, I am getting the array from a database and I need to know which ID car or dog is. – Keith C. Oct 22 '16 at 18:55

PHP array_key_exists key LIKE string, is it possible?

3 Answers3