CODEWITHSUNDEEP
phpmysql
Tomazi
Tomazi

Reputation: 791

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

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

Upvotes: 3

Views: 442

Answers (3)

Qazi
Qazi

Reputation: 5135

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

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

Upvotes: 1

The Blue Dog
The Blue Dog

Reputation: 2475

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 = "";

Upvotes: 2

Pupil
Pupil

Reputation: 23978

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;

Upvotes: 3

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