选自ID填充的行的值

Question

I'm using PHP to populate names into an option list based on values from a database. Once a name is selected I want to populate the values of a row based on whatever name has been selected.

I had no problem populating the option list with the name values, but once I select a name the records of that row will not display in the input boxes.

Here is the code I'm working with:

$query = "SELECT name, id 
FROM artists 
ORDER BY name";

$mysql_stuff = mysql_query($query, $mysql_link);
while($row = mysql_fetch_row($mysql_stuff)) {

 $artists_name = htmlspecialchars(stripslashes($row[0]));
 $artists_id = stripslashes($row[1]);

    // we compare the id of the record returned and the one we have selected
    if ($artists_id) { 
        $selected = $artists_id;
    } else {
        $selected = "";
    }
    print("<option value=\"$artists_id\">$artists_name</option>
    ");
}

print("</select>
<input type=\"Submit\" name=\"submit\" value=\"Update\">
<input type=\"Submit\" name=\"delete\" value=\"Delete\">

<br >
<br />
Edit Biography:
</td>
</tr>
");

if (isset($_POST['submit'])) {
    print("<tr valign=\"top\">
    <td valign=\"top\" width=\"150\">Name</td>
    <td valign=\"top\">Artist Biography</td>
    </tr>

    ");


    print("<tr valign=\"top\" bgcolor=\"$colour\">
    <td valign=\"top\">
    <input type=\"text\" name=\"artists_name[$selected]\" value=\"$artists_name\" size=\"40\" />
    </td>

    <td valign=\"top\">
    <textarea name=\"artists_Biography[$selected]\"  cols=\"40\" rows=\"10\">$artists_Biography</textarea>
    </td>


    </tr>
    ");

}   


print("</table>
</form>
");

Can I please get some assistance with populating the values of the selected name into the input boxes.

Answer 1

So your submitting artist ID to the post. The post would then refresh or change the page. Assuming the action is that page you have the ID in your post. How do you get Artist Name from the ID? Do you run another sql query on the id to find the name? Might need some clarification on the order of your code to help. Thanks

Answer 2

I'm not sure how to interpret this code. Use Ajax. Use Jquery on change() method to take the value of the select and send it via ajax() to a php script that gets the info from the DB and returns what you want. Then use the Success part of the .ajax() method to put that data into you input on the form.

$(function(){
    $("select#artist").change(function() 
    {
        var artist = $("select#artist").val();
    $.ajax({
    type: "POST",
    url: "ajax-find-artist.php",
    data: "artistID=" + artist, 
    cache: false,
    success: function(html){
    var newElems = html;
    $("input#artistRestult").html(newElems);
    }
    });
    });
});