PHP eval syntax

Question

I need help figuring out the syntax for the eval function - or even if eval is the right approach. I have a column in my mysql database which holds the name of a PHP function I need to run. There are also PHP variables that would like to leave as variable until they are passed to the function. Below is what I have so far:

eval($valRec[$key]($key,$value));

$valRec is the array which contains the results of a mysql SELECT. $key is a variable which references the name of the column that contains the function name.

$key and $value are the PHP variables that I need to pass into the function.

In the end, I want to end up with:

functionName($key,$value);

which PHP should run.

Hopefully I explained it clearly - thank you in advance for any help!

never never use eval. There are a fewwwww corner cases where it may be necessary, but I can pretty much guarentee you will never run into them. — Kris, Sep 07 '12 at 03:29

score 4 · Answer 1 · answered Sep 07 '12 at 02:37

4

Eval is probably not what you want. Look at call_user_func and call_user_func_array. They let you call a function whose name is in a variable.

call_user_func($valRec[$key], $key, $value);

answered Sep 07 '12 at 02:37

Barmar

741,623
53
500
612

I didn't end up using eval as Matthew's solution worked - but I will definitely keep call_user_func in mind in case I ever run into a situation where I need something like this again! – Alan Sep 07 '12 at 03:10

Matthew · Accepted Answer · 2012-09-07T02:39:42.637

3

All you need to do is:

$valRec[$key]($key,$value);

Reference: variable functions.

edited Sep 07 '12 at 02:39

answered Sep 07 '12 at 02:27

Matthew

47,584
11
86
98

Could it really be that simple?!? I have a couple tests that I'm going to be running in the next few minutes - will let you know if that works! – Alan Sep 07 '12 at 02:32
`$foo = 'bar'; $foo();` Calls function `bar`. That's all that is happening here, except that the string is contained in an array. Rarely is `eval()` ever needed in PHP. – Matthew Sep 07 '12 at 02:38
I have a few bugs to work out, but this worked. I guess I was just overcomplicating it! Thanks! – Alan Sep 07 '12 at 02:54
(Just be aware that "variable functions" are like "variable variables" and do not represent first-class functions; the variable must still resolve to the *name* of a function.) – Sep 07 '12 at 03:08

score 2 · Answer 3 · answered Sep 07 '12 at 02:46

2

In php, you could do below;

$func = 'strtolower';
$foo = $func($bar);

So in your case, $valRec[$key]($key,$value); will just work.

Check the demo.

Addtion: The reason why your eval not work is because eval need to take a string as parameter, not don't forget ; to end the statement, or it will be syntax error. So you need to do:

eval($valRec[$key].'($key, $value);');

answered Sep 07 '12 at 02:46

xdazz

158,678
38
247
274

Thanks xdazz. Per Matthew's solution above I figured out that I was just overcomplicating this! – Alan Sep 07 '12 at 03:11
@pst Yes, `eval` should not be used, i just mean to point out why `eval` in the op's code not work. – xdazz Sep 07 '12 at 03:13

score 1 · Answer 4 · answered Sep 07 '12 at 02:24

1

I haven't used PHP recently, but the concept of eval is to evaluate a string.

As such, you need to create the string representing the line of code you want run.

In your case that means:

eval($valRec[$key] . '($key, $value);');

You've said it yourself, you want to end up with functionName($key,$value); so you need to make a string that is that, then pass it to eval.

answered Sep 07 '12 at 02:24

Matt Mitchell

40,943
35
118
185

The solution provided by @Matthew above worked - but I really appreciate your help as well!! – Alan Sep 07 '12 at 02:55

score 0 · Answer 5 · edited May 23 '17 at 10:24

0

see this example,

<?php
    $string = 'cup';
    $name = 'coffee';
    $str = 'This is a $string with my $name in it.';
    echo $str. "\n";
    eval("\$str = \"$str\";");
    echo $str. "\n";
?>

this is the output

This is a $string with my $name in it.
This is a cup with my coffee in it.

you need to read this, when is eval evil in php?

edited May 23 '17 at 10:24

Community

1
1

answered Sep 07 '12 at 02:23

John Woo

258,903
69
498
492

John - thanks for the example. This is very similar to the example in the PHP manual - which I read over and tried to apply unsuccessfully. – Alan Sep 07 '12 at 02:27
1

@Alan have a look at [this](http://changelog.ca/log/2005/06/14/the-proper-way-to-use-php-eval) – John Woo Sep 07 '12 at 02:30
Thanks for the link! Per the solution provided by Matthew I didn't end up using eval - but I will make sure to remember this if I end up using it in the future! – Alan Sep 07 '12 at 03:09

PHP eval syntax

5 Answers5