PHP Mysql While循环覆盖变量并只显示一个结果

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

Answer 1

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;

Answer 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 = "";

Answer 3

您需要将每个值与循环中的现有变量连接起来

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