Displaying info from one table - insert into another

[harkly]

I have 2 tables, images & movements, they are connected with the medID field.

I am pulling info from the images table and have created an update page. I want to be able to pull the medID field from movements, there are a lot and I will have a drop down box, select that info an update it to the images table.

Can someone help me with displaying that info and updating to another field?

Here's my line for the update.php - there are a lot more, all work

echo "<tr><td colspan=4><h5>Movements Associated to --</b></td></tr>  \n";

I want to have this so I don't make a mistake and add the wrong medId in the box

      <br><br><br><h5>Movement Choices<br><br>
<?php
    $result = mysql_query("SELECT * FROM movement order by movName") or die(mysql_error());
    $num_rows=mysql_num_rows($result);
      if ($num_rows != "0")
        {
                echo "<form method='post' action='updateArtist.php'>";
                echo "<select name='movement'>";
            while ($r=mysql_fetch_array($result))
              {	
                $movName=$r["movName"];
                echo "<option value='$movid_1' selected>$movName</option>";
              }
                echo "</select>";
                echo "<input type='hidden'>";
                echo "</form>";
        }
 ?>

[djjjozsi]

$movid_1 is undefined in your code, means all the listed movement will passed to the update page as 0 ...

embed the primary key $r["moveid"] where moveid is the primary key's name.

And in your code selected has no sense. You should built a select form which . If you know the image's one modevementID, make an if condition inside the while cycle.

 <label for="movement">movement</label>
  <select name="movement" id="movement">
   <?php while ( $r = mysql_fetsch_object( $results ) ) {
    if ( $r->moveid == $movid_1 )
        $sel = " selected=\"selected\"";
    else
        $sel = "";

echo "<option value=\"{$r->moveid}\" $sel>{$r->movName}</option>\n";
} 

?>
  </select>