PHP preg_replace_callback ignore id

Question

I am using the following code to match all letters except i and d.

/{([abcefghj-z]+)}/

How can I change it so that i and d can exist but not in that order (id).

Example strings:

/admin/users/{type}/{id}/edit

Only {type} would get replaced in the above example.

Example 2:

/admin/users/{type}/{id}/{supplied}/edit

In the above example both {type} and {supplied} would get matched.

Is this possible using regex?

UPDATE

The code I am using as per below post is:

$current = 'admin/users/{type}/{id}/changePassword';
$parameters = array('type' => 'administrators', 'id' => '1');

$path = preg_replace_callback('~\{(?!id\})[a-z]+\}~', function ($m) use ($parameters) {
        return $parameters[$m[1]]; 
}, $current);

Wiktor Stribiżew · Accepted Answer · 2015-09-16 12:18:14Z

1

Here is a solution for you: you need to use a capturing group (thus I added round brackets in ([a-z]+)) to be able to get the value you need and I see you need to perform the search in a case-insensitive way (thus, I added ~i):

$current = 'admin/users/{type}/{id}/changePassword';
$parameters = array('type' => 'administrators', 'id' => '1');

$path = preg_replace_callback('~\{(?!id\})([a-z]+)\}~i', function ($m) use ($parameters) {
//                                        ^      ^   ^
        return $parameters[$m[1]]; 
}, $current);

echo $path;

See IDEONE demo

With capturing group, $m[1] will hold "type", and thus you will get the value from the array correctly.

answered Sep 16, 2015 at 12:18

Wiktor Stribiżew

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Comments

Avinash Raj · Accepted Answer · 2015-09-16 10:57:14Z

0

Use a negative lookahead assertion.

\{(?!id\})[a-z]+\}

(?!id\}) asserts the previous token { would be followed by any but not of id}.
If yes, then match one or more lower case letters and then }.

DEMO

Example:

This will replace all the matched substrings with {foo}.

preg_replace('~\{(?!id\})[a-z]+\}~', '{foo}', $str);

DEMO

answered Sep 16, 2015 at 10:57

Avinash Raj

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4 Comments

V4n1ll4 Over a year ago

When I use that regex on my function above I get: preg_replace_callback(): Delimiter must not be alphanumeric or backslash

Avinash Raj Over a year ago

pls use ~ as regex delimiter.

V4n1ll4 Over a year ago

Thanks, but now I get Undefined offset: 1

V4n1ll4 Over a year ago

I've posted my full code at the bottom of my original post.

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