PHPUnit - I get some unecessary exception (undefined variable)

Question

I'm started to use PHPUnit, and I want to connect to my database and get some values from it, but as I have to connect to my database through the file 'database.php'.

'database.php' always had this problem with the undefined variable: $connected. So I read something about with the error-reporting message: error_reporting(E_ALL ^ E_NOTICE);

I did fix it earlier to my other files in my Project, but when I use PHPUnit I get this error message: Undefined variable: connected

My testcase file and it's code:

<?php include("database.php"); ?>
<?php
require_once 'person.php';

class PersonTest extends PHPUnit_Framework_TestCase {

    public function lookupPerson(){

    $result = database::query("select * from Person where forename='Anna'");
    $rows = mysql_numrows($result);
    for($i=0; $i < $rows; $i++){
    $arr[$i] = mysql_result($result, $i, 'forename');
    }
    return $arr;
    }


    public function testLooking(){
    $arr = PersonTest::lookupPerson();
    foreach($arr as $a){
    $this->assertEquals('Anna', $a);
    }
    }
}


?>

So I wonder what can I do?

Regards User

also Alexein

Is this your real test? You don't actually test anything in the Person class, just a method in the test class? — Fenton, May 11 '12 at 11:52

score 1 · Answer 1 · answered May 11 '12 at 11:06

1

At the top of your code add below, this hides all notices... Ofcourse its best to just fix the error :)

error_reporting(0);
ini_set("display_errors", 0);

answered May 11 '12 at 11:06

Hans Wassink

2,529
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score 0 · Answer 2 · edited May 23 '17 at 12:04

0

take a look at this question: test the return value of a method that triggers an error with PHPUnit

answers there show how to prevent PHPUnit from treating notices and warnings as exceptions.

if you can, fix the notice.

edited May 23 '17 at 12:04

Community

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answered May 11 '12 at 11:15

Mariusz Sakowski

3,232
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score 0 · Answer 3 · answered May 11 '12 at 11:19

I did fix it earlier to my other files in my Project, but when I use PHPUnit I get this error message: Undefined variable: connected

From what line in database.php is this notice being thrown? Perhaps you do a if (!$connected) somewhere, while $connected never has been written to. So begin your database.php with $connected = false and you'll lose the notice.

PHPUnit - I get some unecessary exception (undefined variable)

3 Answers3