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POHow do I show each item in a multi-select dropdown as already selected if it is in the database?
 

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  1. POHow do I show each item in a multi-select dropdown as already selected if it is in the database?
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    <p>So I have a form that edits a Units already in my database. There is a multi-select dropdown where I can choose Suppliers to the Unit. I need the suppliers that are already assigned to the unit to come up as selected(highlighted) in the dropdown, so the user only has to ctrl+click new Suppliers they want to add to the Unit.</p> <p>units_suppliers is the junction table I have set up that contains paired Unit and Supplier IDs to determine which supplier is assigned to which unit, and vice-versa.</p> <p>This is the code I currently have, but it's not working. I created two different arrays based on different SQL queries and tried to compare whether they have the same Supplier name, and if so the "selected" attribute was added to the tag. But it doesn't seem to be working. Can anyone help me out?</p> <p>EDIT: The queries both seem to produce the right results, so I think the problem is in my nested loops, but I can't figure out what's wrong. Thanks for any advice.</p> <pre><code>&lt;td&gt;&lt;select size=7 multiple="multiple" name="unitsuppliers[]"&gt; &lt;?php $sql = "SELECT suppliers_id, supplier_name FROM suppliers GROUP BY supplier_name ORDER BY supplier_name"; $result = mysql_query($sql,$connection); /* THIS PART FINDS OUT IF THE SUPPLIER SHOULD ALREADY BE HIGHLIGHTED IN THE LIST */ $sql2 = "SELECT suppliers.supplier_name FROM suppliers, units_suppliers WHERE units_suppliers.unit_user_id = "; if(isset($old_details['unit_user_id'])){ $sql2 .= " {$old_details['unit_user_id']} "; } elseif(isset($_POST['old_unit_id'])){ $sql2 .= " {$_POST['old_unit_id']} "; } else { $sql2 .= " 0 "; } $sql2 .= " AND suppliers.suppliers_id = units_suppliers.suppliers_id"; $result2 = mysql_query($sql2,$connection); while ($row = mysql_fetch_array($result)) { echo "&lt;option value=\"{$row['suppliers_id']}\" "; while ($in_junc_table = mysql_fetch_array($result2)) { foreach ($in_junc_table as $array){ if($array['supplier_name'] == $row['supplier_name']){ echo " selected "; } } } echo " &gt;"; echo "{$row['supplier_name']}"; } ?&gt; </code></pre>
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    1. PO
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      1. PTAnswer
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    1. PTQuestion
    1. USJoe Frambach
    1. USSw1sh
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    1. PL
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      1. LTLinked
    1. This table or related slice is empty.
    1. This table or related slice is empty.
    1. PO
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      1. PTAnswer
    1. This table or related slice is empty.
    1. COYou need to sanitize your data, please: http://stackoverflow.com/questions/601300/what-is-sql-injection
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      1. POHow do I show each item in a multi-select dropdown as already selected if it is in the database?
      1. USHalf Crazed
    2. COYou can't just specify `selected`. In most browsers, you still need to specify some truthy value: `selected="selected"`.
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      1. POHow do I show each item in a multi-select dropdown as already selected if it is in the database?
      1. USKyle
    3. COCan't you JOIN the two tables? This should leave you with only the suppliers that are in both tables for conditions you set with the WHERE clause.
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      1. POHow do I show each item in a multi-select dropdown as already selected if it is in the database?
      1. USSumurai8
 

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