php mysql displaying results as drop down list

Question

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

Please stop writing new code with the ancient mysql_* functions. They are no longer maintained and community has begun the deprecation process . Instead you should learn about prepared statements and use either PDO or MySQLi. If you care to learn, here is a quite good PDO-related tutorial. — tereško
– tereško, Commented Apr 20, 2012 at 19:23
thx for that, I will def look into this once i get this project done. Am under big time pressure. What would this one statement then look like, for example ? — Alexander Grosse
– Alexander Grosse, Commented Apr 20, 2012 at 19:26

squarephoenix · Accepted Answer · 2012-04-20 19:55:57Z

3

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>";

edited Apr 20, 2012 at 19:55

answered Apr 20, 2012 at 19:26

squarephoenix

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4 Comments

Alexander Grosse Over a year ago

any idea how I could return a dimensional array before printing them into a list ? My functions to do with DB connections, are in different files tyhan the file where I will use it !

Alexander Grosse Over a year ago

so i first need to create an array I can work with, pass it to a different file, then print them there as a list

Alexander Grosse Over a year ago

apart from the fact that when looking at it, I struggel understanding the loops to print it. Becasue the q_code, value and title variable in the loops is made up, but still it prints the items correctly. I guess the first loop, indicates the first part of teh array, second loop is for second part of array !?

squarephoenix Over a year ago

yeah those var names are just designated to refer to the item the pointer is currently on. foreach($array as $key => $value)

juicebyjustin · Accepted Answer · 2012-04-20 19:34:59Z

1

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.

answered Apr 20, 2012 at 19:34

juicebyjustin

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1 Comment

Alexander Grosse Over a year ago

thx for the answer and the links !! really appreciated, I will def look through the site on occasions !

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