CODEWITHSUNDEEP
phpmysqlhtml
Alexander Grosse
Alexander Grosse

Reputation: 287

php mysql displaying results as drop down list

I am trying veru hard, to get results from a database to be displayed as a drop down list.

Basically I am fetching the foreign key, its unique code and its name/title. I need to display it as a drop down list, to be entered correctly into a new table

        $sql = "SELECT quali_code, title FROM `qualifications`  ";
    $result = mysql_query($sql, $this->connection);
    $result_array = array();
    while($r=mysql_fetch_array($result))
    {
    $result_array[$r['quali_code']][] = $r;
    }
    return $result_array;

With that I would need to create a drop down list in HTMl/PHP to select the title from, but obviously storing the value. sort of <list name=array[title] value=array[code] >

At the moment I am already stuck at returning my SQl result in an array that works for me, and have no clue how I will then populate a dropdown list with the correct array results.

Sample code above on how i fetch the sql results and here how my array looks like:

Array ( [AAT] => Array ( [0] => Array ( [0] => AAT [quali_code] => AAT [1] => AAT qualification [title] => AAT qualification ) ) [A_lev] => Array ( [0] => Array ( [0] => A_lev [quali_code] => A_lev [1] => A levels [title] => A levels ) ) [HNC] => Array ( [0] => Array ( [0] => HNC [quali_code] => HNC [1] => Fdsc/HNC [title] => Fdsc/HNC ) ) [Lan_ISLT] => Array ( [0] => Array ( [0] => Lan_ISLT [quali_code] => Lan_ISLT [1] => EISLT Qualification [title] => EISLT Qualification ) ) [Lan_qua] => Array ( [0] => Array ( [0] => Lan_qua [quali_code] => Lan_qua [1] => Language qualification [title] => Language qualification ) ) [Nat_Dip] => Array ( [0] => Array ( [0] => Nat_Dip [quali_code] => Nat_Dip [1] => National Diploma [title] => National Diploma ) ) )

Bare in mind I am totally new to PHP. Any help would be appreciated. But this array looks useless to me, and am not sure how to populate a list. The file where I fetch the result / array, is not the location where I populate the array into a list.

Many thanks for any hints on this matter

Upvotes: 1

Views: 3080

Answers (2)

squarephoenix
squarephoenix

Reputation: 1003

Assuming all you're trying to do is echo these rows into a dropdown:

$sql = "SELECT quali_code, title FROM `qualifications` ORDER BY title ";
$result = mysql_query($sql, $this->connection);
while($r=mysql_fetch_array($result))
{
$quali_code = $r['quali_code'];
$result_array[$quali_code][] = $r['title'];
}


echo "<select name='x'>";
foreach($result_array as $q_code => $value)
{
    foreach($value as $title) {
    echo "<option value='" . $q_code . "'>" . $title . "</option>";
    }
}
echo "</select>";

Upvotes: 3

juicebyjustin
juicebyjustin

Reputation: 444

It's been awhile since I've used PHP, however, I remember this is how I printed out columns from a MySQL query. Use $r["columnName"] in the while loop. There may be semantic erros but this should give you a pretty good idea.

 echo "<select>";
 while($r = mysql_fetch_array($result))
 {
    echo "<option value=\".$r['quali_code'].\">.$r['title'].</option>";
 }
 echo "</select>";

I used http://www.tizag.com/phpT/ to teach myself PHP. They have many great tutorials. Here's the link for a MySQL/PHP tutorial on Tizag: http://www.tizag.com/mysqlTutorial/.

Happy coding.

Upvotes: 1

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