Hi All,

I am still relatively new to PHP and MySQL but have been able to get quite a way with info available on the web.

OK ... I want to edit a record in my database using PHP. My main table (the one I want to edit) contains references to other tables. My original entry form dynamically queries the other tables and stores the id in the main table.

So the question now is how do I load an existing record in the edit form, default what was previously entered and show dynamically the other entries in the other table? Hope this makes sense?

e.g. for the main table I query the location table and retrieve the id, description and value fields (used later for calculations). The user selects the appropriate description and the database stores the id.

Thanks in advance for any help!

You're asking about relational db and normalization. In short, your tables are connected via IDs. A table typically has its own ID and other IDs that reference other tables.

In SQL, WHERE clause helps to join tables while querying ..
like WHERE tablename1.id = tablename2.id

If you're still confused, show us few lines of code. We also need know your db structure, to help you, specifically. By the way, MSQL AB (mysql.net) already provides free comprehensive documentation for all the explaination. We all are ourselves reading that.

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Thanks Anshul,

OK I have a table called "ops" and one of many tables called "agerisk". The agerisk table has an id field, description and value fields. I use the id field on the ops table to reference the appropriate row (id) in the "agerisk" table. The user is presented the description field in the combo box on the form. I have included some code from the original form below. I am struggling to understand how I can construct the same lookup to default what was entered by the user when they wish to edit/update the data.

<td>
          <select name=\"agerisk2\">
        <option value=\"\"></option>";

        for($i=0; $i<$ageriskNumber; $i++) {
            $ageriskAgeid = mysql_result($ageriskResult,$i,"agerisk.ageid");
            $ageriskDescription = mysql_result($ageriskResult,$i,"agerisk.Descriptio n");
            print " <option value=\"$ageriskAgeid\">$ageriskDescription</option>";

        }