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POCannot get IF $_GET[video] = edit, display a form
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  1. POCannot get IF $_GET[video] = edit, display a form
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    copied!<p>I cannot seem to figure this out. I'm trying to get row[video] value from results, and if there is a entry in the DB for video and it's not null, display a link that says Edit Video Link with a href= of edit_media.php?video=edit (which is this page). If the user clicks this link, the page reloads with the URL including video=edit and that triggers the variable videoEdit to be TRUE which should display a form input box that displays the current value in the database.</p> <p>However, It doesn't seem to be working, for some reason I'm seeing the form input box even before I click Edit Video link, (url does not have video=edit). and when I click "Modify Video Link" and the url does have the video=edit, I see a textbox with the words Video in it. I'm not sure why. I have the correct MYSQL Query before this code, but I didn't include it here. How can I get this working?</p> <pre><code> // if the user clicked edit video link if ($_GET['video'] = EDIT) { $videoEdit = TRUE } while ($row = mysql_fetch_array($result, mysql_assoc)) { $video = $row['video']; } if ($videoEdit = TRUE) { echo '&lt;form action="edit_media.php" method="post"&gt; &lt;p&gt;Video Link: &lt;input type="text" name="serial" size="25" maxlength="100" value='; if (isset($video)) echo $video; ' /&gt;&lt;/p&gt;&lt;p&gt;&lt;input type="submit" name="submit" value="Submit Video" /&gt;&lt;/p&gt; &lt;input type="hidden" name="submitted" value="TRUE" /&gt; &lt;/form&gt;'; } </code></pre> <p>And it goes on to say:</p> <pre><code>if ($video != "") { echo '&lt;iframe width="420" height="315" src=' . $video . ' frameborder="0" allowfullscreen&gt;&lt;/iframe&gt; | &lt;a href=edit_media.php?video=EDIT&gt;Modify Video Link&lt;/a&gt;'; } else { echo "Video Not Present | &lt;a href=edit_media.php?video=EDIT&gt;Add a Video&lt;/a&gt;"; } </code></pre> <p>Can you help me in making this script work>?</p>
 

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