PHP eval syntax

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);

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);');

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.

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?