PHP Mysql While loops overwrites a variable and displaying only one result

Question

I have a PHP script to populate drop down menu with results obtained from database the problem I am having is that only the last result is displayed in drop down menu. I recon it is because the while loop that gets all the results overwrites the variable that stores a string every time is runs.

I have tried to find a solution to fix it but ending up in a dark corner with no solution

Php Code:

$sql2 = "SELECT id, course, location FROM courses WHERE course LIKE '%Forex%' OR  course LIKE '%forex%'";
        $query2 = mysqli_query($link, $sql2);
            $opt = '<select id="Forex" name="Forex">';
            $opt1 = '<option value="">Select Forex Workshop</option>';



                while($result2 = mysqli_fetch_assoc($query2)){
                //I belief the $opt2 variable is overwritten every time the loop runs
                    $opt2 = '<option value="">'.$result2['course'].'</option>';
                    print_r($opt2);
                }
                $opt3 =  '</select>';
                return $opt.$opt1.$opt2.$opt3;  


}

I might be wrong and the issue might be elswhere in the code but when i print_r($result2) all the correct results are there

$opt2 = '<option value=""'.$results2['course'].'</option>'; should be $opt2 .='<option value="">'.... So add a dot before =. Otherwise you will just receive the last value from Database. — StudioArena
– StudioArena, Commented Dec 19, 2013 at 9:52

Pupil · Accepted Answer · 2013-12-19 09:56:45Z

3

Just add a .

$opt2 .= '<option value="">'.$result2['course'].'</option>';
      ^

Your variable is reinitializing, it should be concatenated.

So, the final code should be:

$opt2 = '';
while($result2 = mysqli_fetch_assoc($query2)){
                //I belief the $opt2 variable is overwritten every time the loop runs
                    $opt2 = '<option value="">'.$result2['course'].'</option>';
                    print_r($opt2);
                }
                $opt3 =  '</select>';
                return $opt.$opt1.$opt2.$opt3;

edited Dec 19, 2013 at 9:56

answered Dec 19, 2013 at 9:51

Pupil

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4 Comments

Tomazi Over a year ago

lool simple as thats great I can see all the results now but still getting a Notice Warning " Notice: Undefined variable: opt2 in C:\xampp\htdocs\opes\course.php on line 21"

Pupil Over a year ago

@Tomazi, if this works for you, please mark my answer as solution.

Tomazi Over a year ago

The error is gone now i fixed it by initializing the $opt2 variable $opt2 = ''; trying to accept your answer but is cant as soon i i can i will because you were first. Sometimes it is scary ho simple a solution can be :)

The Blue Dog Over a year ago

Actually he wasn't first, I posted 24 seconds before him. ;)

The Blue Dog · Accepted Answer · 2013-12-19 09:58:09Z

2

Yes, you need to append each value:

$opt2 .= '<option value="">'.$result2['course'].'</option>';

You should also initialise $opt2 before you start the loop.

$opt1 = '<option value="">Select Forex Workshop</option>';
$opt2 = "";

edited Dec 19, 2013 at 9:58

answered Dec 19, 2013 at 9:51

The Blue Dog

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Comments

Qazi · Accepted Answer · 2013-12-19 10:01:34Z

1

You need to concatenate the each value with existing variable in the loop

**$opt2 .= '<option value="">'.$result2['course'].'</option>';**

answered Dec 19, 2013 at 10:01

Qazi

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