Replace with same named PHP variable

Question

My Example:

$name = "Simon";
$string = "My name is [name].";
echo preg_replace("/\[(.*)]/", ${"$1"}, $string);
// Expected: My name is Simon.
// I get: My name is .
// ${"$1"} should be $name?
exit();

When I do only:

echo preg_replace("/\[(.*)]/", "$1", $string);
// I get: My name is name.
// $1 = name

What am i doing wrong? Why is PHP not using the generated $name var? This is only a example. I would like to work this with any replace:

[foo] --> $foo
[bar] --> $bar
...

Hey Asik, i don't want to use str_replace because i want to do it with much more vars. I want to work this with any replace. `[foo] > $foo`, `[bar] > $bar`, etc.. I don't want to use arrays. — user2987790, Nov 20 '14 at 17:55
Is there a particular reason why you're wanting to use `preg_replace()` etc. and not just `$string = "My name is " .$name;`? This sounds like overkill if you just want to echo out/populate a username. — Funk Forty Niner, Nov 20 '14 at 17:57
I have much more Placeholders, not only [name]. I could put them all in arrays and do str_replace(). It seems for me that the automatic preg_replace() with `$1` would be the more elegant solution. — user2987790, Nov 20 '14 at 18:06
In order for this to work, you will need to change `[name]` to `[$name]` and `${"$1"}` to `"$1"`. Least, from what I could gather. Otherwise, you will need to modify something. If you can live with that, great. — Funk Forty Niner, Nov 20 '14 at 18:15
Fred -ii- this is great man, you can post this as an answer. But why is this `${"$1"}` not working? Do you have any clue where's the problem for PHP? — user2987790, Nov 20 '14 at 18:20

score 4 · Accepted Answer · edited May 23 '17 at 12:05

4

As per OP's wish to post my comment as an answer:

In order for this to work, you will need to change [name] to [$name] and ${"$1"} to "$1"

$name = "Simon";
$string = "My name is [$name].";

echo preg_replace("/\[(.*?)]/", "$1", $string);

PHP needs a variable to go on, so using [name] isn't being populated.

As per another and earlier comment I made, alternatively you could very well do:

$name = "Simon";
$string = "My name is " .$name;

echo $string;

If you have a framework with existing brackets, then that is something you haven't told us, only that you said in comments:
" have much more Placeholders, not only [name]", whether it's part of something bigger, then stick to the accepted method.

As per in comments, you could alternatively use "/\[([^\]]+)\]/" instead of "/\[(.*)]/"
Or "/[(.*?)]/" :)

in regards to using for example: "[$foo]bar[$foo]"

edited May 23 '17 at 12:05

Community

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answered Nov 20 '14 at 18:24

Funk Forty Niner

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What's with the dv's? – Funk Forty Niner Nov 20 '14 at 18:30
If I explained it badly/wrongly, I will adjust my answer. – Funk Forty Niner Nov 20 '14 at 18:34
`preg_replace` to remove squares? seriously? – vp_arth Nov 20 '14 at 18:34
@vp_arth I know, but that's what the OP seems to want to use. There's probably a reason for it, am sure. I asked, but OP didn't respond as to why. As per [**this comment**](http://stackoverflow.com/questions/27046192/replace-with-same-named-php-variable#comment42610960_27046192) – Funk Forty Niner Nov 20 '14 at 18:35
you should ungreed it first. `"[$name]text[$name]"` -> `"$name"` now – vp_arth Nov 20 '14 at 18:36
@vp_arth Oh, seems OP did respond with [**this comment**](http://stackoverflow.com/questions/27046192/replace-with-same-named-php-variable#comment42611254_27046192) – Funk Forty Niner Nov 20 '14 at 18:36
@vp_arth OP has the option. – Funk Forty Niner Nov 20 '14 at 18:37
no, he uses var_names anyway - so php already has this placeholders - `"{$var}"` – vp_arth Nov 20 '14 at 18:37
@vp_arth Exactly and I agree with you. – Funk Forty Niner Nov 20 '14 at 18:38
fix a bug with greedy – vp_arth Nov 20 '14 at 18:38
@vp_arth You could edit your answer with that, I would be glad to upvote it. – Funk Forty Niner Nov 20 '14 at 18:39
Your answer is totally incorrect now... should be `"/\[([^\]]+)\]/"` or `"/\[(.*?)\]/"`. – vp_arth Nov 20 '14 at 18:41
@vp_arth How can it be totally incorrect, I've tested it with `"/\[(.*)]/"` and it worked fine. – Funk Forty Niner Nov 20 '14 at 18:46
try to test with `"[$foo]bar[$foo]"` :) – vp_arth Nov 20 '14 at 18:46
@vp_arth Both methods work, I tested them now. Same result. I could make an edit to that effect, but again, same result. OP is happy with it. :) – Funk Forty Niner Nov 20 '14 at 18:47
`php -r 'echo preg_replace("/\[(.*)]/", "$1", "[foo]aasdf[foo]");'` Result: `foo]aasdf[foo` – vp_arth Nov 20 '14 at 18:50
`OP is happy` is not a SO target. `There is new useful QA` is a target. – vp_arth Nov 20 '14 at 18:57
@vp_arth Would you rather I delete my answer? I would be happy to. – Funk Forty Niner Nov 20 '14 at 18:58
No, just add a question mark to your regular, it's incorrect – vp_arth Nov 20 '14 at 18:58
@vp_arth If you have editing privileges, please feel free. I have to go to work now. – Funk Forty Niner Nov 20 '14 at 18:59

score 1 · Answer 2 · answered Nov 20 '14 at 19:21

1

I suggest you to use something like this:

class Tpl {
  private $tpl;
  public function __construct($tpl) {
     $this->tpl = $tpl;
  }
  public function render($data) {
    $result = $this->tpl;
    foreach ($data as $key => $value) {
      $result = str_replace("%%$key%%", str_replace('%%', '', $value), $result);
    }
    return $result;
  }
}

$greet = new Tpl('Hello, %%name%%');
echo $greet->render(array('name' => 'World'));

But you always must care, that you correctly escape your placeholder formatting. I just remove it :)

answered Nov 20 '14 at 19:21

vp_arth

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Finally i formatted the vars in mysql with placeholders {{FOO}} and used [strtr()](http://stackoverflow.com/questions/10106052/replace-multiple-placeholders-with-php). Thank You All for Your Help! So there is no possibility to store "real" php Vars mixed with Text in mysql? I understand it from the point of security issues but there should be and option to to this. – user2987790 Nov 20 '14 at 20:21

vp_arth · Answer 3 · 2014-11-20T18:45:03.843

0

No need in preg_replace at all.
You use variable_names anyway - so php already has this placeholders - "{$var}":

$name = "Simon";
$string = "My name is {$name}.";

edited Nov 20 '14 at 18:45

answered Nov 20 '14 at 18:28

vp_arth

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That's what [**I wrote**](http://stackoverflow.com/questions/27046192/replace-with-same-named-php-variable#comment42610960_27046192) but the OP seems to want to use that for a particular reason. Why, I have no idea. – Funk Forty Niner Nov 20 '14 at 18:32
in a mysql field i store text mixed with PHP Vars: `bla $foo bla $bar`. If i store this mysql result and echo it i get the same: `bla $foo bla $bar`. if i store text with placeholders `[foo], [bar]` and change them later to $PHPVars the result should be `bla FOOVAR-CONTENT bla BARVAR-CONTENT` (Before i echo the mysql the vars are already set in the script) – user2987790 Nov 20 '14 at 18:48
1

So the OP did have a "special" reason ;) – Funk Forty Niner Nov 20 '14 at 18:52
So whats should i do instead? I want to get the already set values of the vars into my mysql-result ? – user2987790 Nov 20 '14 at 18:52
No reason... no any profit... no difference from `a {$v} b {$v}` to `a [v] b [v]` if you need `$v` variable to make it works... – vp_arth Nov 20 '14 at 18:52
Why your database knows about PHP anything? It's about data, not about implementation. – vp_arth Nov 20 '14 at 18:55
PHP gets a mysql-Result like `$mailbody = "Hello $name";`. `$name` is defined before mysql-request. if i echo `$mailbody` my result is still `"Hello $name"` instead of `"Hello MAILBODYCONTENT"`. When i do it the way with placeholders it works.. – user2987790 Nov 20 '14 at 19:01

Replace with same named PHP variable

3 Answers3