$var = "['test', 'test2', 'test3']";
how do I create a workable array from this in PHP?
I've tried explode($var, ","); but this didn't seem to work, unless something went wrong with that attempt.
4 Answers
explode($var, ","); is wrong. explode needs the first argument to be the delimiter and the second be the string. Replace [] and then explode -
$var = "['test', 'test2', 'test3']";
$var = str_replace(array('[', ']'), '', $var);
$arr = explode(',', $var);
Comments
[akshay@localhost tmp]$ cat test.php
<?php
$var = "['test', 'test2', 'test3']";
print_r( json_decode(str_replace("'","\"",$var)) );
?>
Output
[akshay@localhost tmp]$ php test.php
Array
(
[0] => test
[1] => test2
[2] => test3
)
Comments
I would say that it looks like a job for json_decode, but its not valid json... There is a way to make it valid however:
How to json_decode invalid JSON with apostrophe instead of quotation mark
Comments
There is an eval() function in PHP which converts string into PHP statements. The string has to be valid PHP statement.
In your case "['test', 'test2', 'test3']"; is not valid statement. You can use something similar to below syntax. Please note that the $x is in single quotes as $x in double quotes will return the value.
$var = "['test', 'test2', 'test3'];";
eval('$x = ' . $var);
print_r($x);
2 Comments
EJTH
Then just pray that this isn't input from a user or 3rd party. I wouldn't use eval for this.
Uday Pawar
Yea that's true; it could be vulnerable to attacks if the string is user input and not sanitized properly.
$myArray = str_getcsv(trim($var, '[]'), ',', "'");