Error while storing my db value into dropdown list

Question

I'm having a lot of trouble with storing my values from a database into a dropdown list. The code doesn't bring any errors, but it just freezes...

Code

<div class="input-field col-md-12">

          <?php
              echo "<select name='pcID'>";
              $query = "SELECT benutzername FROM benutzer";
              $result = mysql_query($query);
              if(mysql_num_rows($query)>0){
              while($row = mysql_fetch_array($result)){
              echo "<option value=\"owner1\">" . $row['benutzername'] . "
             </option>";
             }
            }
            echo "</select>";
         ?>
        </div>

      </div>

I found a lot of other solutions for my problem. But no solution worked for me. It just freezes up (on my loading screen). And as soon as I remove the PHP part it unfreezes...

Does anyone know how I can solve my problem?

Edit

yea, like I said. I get no error message in the console at all...

Every time you use [the `mysql_`](http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php) database extension in new code **[this happens](https://media.giphy.com/media/kg9t6wEQKV7u8/giphy.gif)** it is deprecated and has been for years and is gone for ever in PHP7. If you are just learning PHP, spend your energies learning the `PDO` or `mysqli` database extensions and prepared statements. [Start here](http://php.net/manual/en/book.pdo.php) — RiggsFolly, Jan 12 '18 at 13:45
take a second **close** look at this line and fix the glaring error `if(mysql_num_rows($query)>0){` — RiggsFolly, Jan 12 '18 at 13:46
@RiggsFolly yeah, I commented on that and deleted my comment for it. — Funk Forty Niner, Jan 12 '18 at 13:48
@RiggsFolly nothing to be sorry about ;-) I didn't see yours when I deleted it. — Funk Forty Niner, Jan 12 '18 at 13:49
@CallMeLeonardo further clue to RiggsFolly's comment...a SQL query statement is not the same thing as the **result** of that query — ADyson, Jan 12 '18 at 13:50
@RiggsFolly I removed the if statement now. Don't know why I included it tbh... But it wasn't the error. I will have a look at your suggested PDO or mysqli statements — CallMeLeonardo, Jan 12 '18 at 13:53
Why so many downvotes? Pretty hard to learn a language if everyone just goes hard on beginner errors... — CallMeLeonardo, Jan 12 '18 at 13:54
Change `if(mysql_num_rows($query)>0)` to `if(mysql_num_rows($result)>0)` — RiggsFolly, Jan 12 '18 at 13:56
@RiggsFolly Changed it. I see the fault there. But it still doesn't work... — CallMeLeonardo, Jan 12 '18 at 13:58
Add [echo mysql_error();](http://php.net/manual/en/function.mysql-error.php) after the `mysql_query()` — RiggsFolly, Jan 12 '18 at 14:01
"doesn't work" means what? Could be lots of things, we can't do anything with that info. Make sure PHP error reporting is switched on and that you're checking for mysql errors as Riggs suggests, and then tell us what goes wrong. — ADyson, Jan 12 '18 at 14:02
*I get no error message in the console at all* - if you're referring to the browser console, no, you won't get PHP errors in there. PHP runs on the server not the client. — CD001, Jan 12 '18 at 14:02
Thanks for all the help. I solved it by using the PDO as @RiggsFolly suggested. — CallMeLeonardo, Jan 12 '18 at 14:26

CallMeLeonardo · Accepted Answer · 2018-01-12T14:48:44.757

After restructuring my code, I solved it by using the PDO statements.

This is how my solutions looks like now:

<div class="input-field col-md-12">

          <?php
              //query
              $sql = "SELECT benutzername FROM benutzer";
              $statement = $pdo->prepare($sql);
              $statement->execute();
              $users = $statement->fetchAll();

          ?>

          <select>
           <?php foreach ($users as $user): ?>
             <option value="<?= $user['id']; ?>"><?= 
             $user['benutzername'];  ?></option>
           <?php endforeach; ?>
         </select>
        </div>

score 0 · Answer 2 · edited Jan 22 '18 at 04:17

0

You need to connect your db by selecting appropriate database.

$con=mysql_connect("localhost","root","");
mysql_select_db("databasename",$con);

Your option tag in drop-down list just takes the value and no name to display. I'd suggest you to make minimal changes.
```
echo"<option value=".$row["columnname"].">".$row["column2"]."</option>";
```

edited Jan 22 '18 at 04:17

Nisse Engström

4,738
23
27
42

answered Jan 12 '18 at 13:54

user9209261

9
1

1

Turn the tide against teaching/propagating sloppy and dangerous coding practices. If you post an answer without prepared statements [you may want to consider this before posting](http://meta.stackoverflow.com/q/344703/). Additionally [a more valuable answer comes from showing the OP the right method](https://meta.stackoverflow.com/a/290789/1011527). – Jay Blanchard Jan 12 '18 at 14:02

Error while storing my db value into dropdown list

2 Answers2