Question

I’m looking for some help with my INSERT mysql query, I currently have the php loop

<?php 
$sql="SELECT clientID,clientName FROM clients";
$result =mysql_query($sql);
while ($data=mysql_fetch_assoc($result))
{
  ?><option value="
  <?php echo $data['clientID'] ?>" >
  <?php echo $data['clientName'] ?></option>
  <?php 
} ?>

This outputs the list of Clients that I have, but I need to INSERT the chosen client. How do I insert the clientID chosen in a html form and insert it into ‘projects’ ?

projectID ('projects') and clientID ('clients') are related, so I need to insert to clientID into projects to associate them together.

Example:

$strSql="INSERT INTO projects 
   ( projectID , projectName , projectBusiness , projectDescription) VALUES 
   (NULL , '$strprojectName', '$intclientID', '$strProjectDescription')";

The above doesn’t grab the clientID yet..

Answer 1

This is one of the more fundamental aspects of PHP (and HTML to an extent) and so you should get a good grasp of this before going much further.

Wrap the select in a form. You may, or may not have done this already; can’t tell.

<form id="add_client_id" name="add_client_id" method="post" action="script.php">
    <select name="client_id">
        <option value="" selected="selected">-- Select --</option>
        <option value="1">Foo</option>
        <option value="2">Bar</option>
    </select>
    <input type="submit" name="submit" id="submit" value="Submit" />
</form>

Once you submit this form, you can access it from the super global $_POST (this is because the forms method is set to post. Again – fundamentals.

You can now access the value of the submit data in the form of an associative array:

$client_id = mysql_real_escape_string($_POST['client_id']); – where the key comes from the form.

You can now use this variable within your queries and for further processing – after sanitising of course.

Again this is extremely basic; so you should probably do a bit of work looking for basic PHP tutorials.