updating data in databse?

Porkie · September 2, 2009

Hey,

As you can see in my code, i have a section whicxh loads the latest feeds from my database. However if i click edit to the side of one of my feeds i want it to load that feed into my table to allow me to edit my data. How do i do this? can someone please explain?

Regards

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>RSS Feed </title>
<style type="text/css">
<!--
.style1 {
font-family: Arial, Helvetica, sans-serif;
font-size: 14px;
margin: 0px 0px 0px 0px;
}
.style2 {
font-family: Arial, Helvetica, sans-serif;
font-size: 18px;
font-weight: bold;
color: #CC00CC;
margin: 0px 0px 0px 0px;
}

#feed_table {
  font-family: Arial, Helvetica, sans-serif;
  color: #FFFF00;
  position: absolute;
  margin: 100px 0px 0px 600px;
}
.style_search {
font-family: Arial, Helvetica, sans-serif;
font-size: 20px;
margin: 0px 0px 0px 600px;
}
-->
</style>
</head>
<?php
require('core.inc.php');
echo page_header('Register');
echo $error;

?>
  
<?php
    
$con = mysql_connect("");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("", $con);
$q ="SELECT * FROM  ORDER BY date DESC ";
$sql5 = mysql_query($q);
echo '<table id="feed_table" width="400" border="1">';
echo '<tr><td>TOP 5 FEEDS</td><td colspan="2"><a href="rss_show.php">View All</td></tr>';
while ($row = mysql_fetch_assoc($sql5)) {
echo '<tr>';
    echo '<td>';
    echo '<img style="width:16px;height:16px;" src="img/rss_image.jpg"/> '.$row['title'];
    echo '</td>';
echo '<td>';
echo '<a href="rss_edit.php">Edit </a>';
echo '</td>';
echo '<td>';
echo '<a href="rss_delete.php">Delete </a>';
echo '</td>';
    echo '</tr>';
  	}
echo '</table>';
echo '</P>';
?>

<form method="post" enctype="multipart/form-data">
<br />
<form id="addfeed" name="addfeed" method="post" action="rss_edit.php">
  <table width="400" border="1">
    <tr>
      <th width="105" scope="col" align="left">RSS Feed </th>
      <th width="198" scope="col"> </th>
    </tr>
    <tr>
      <td width="150pt">Title : </td>
      <td><label>
        <input name="rsstitle" type="text" id="rsstitle" size="33" />
      </label></td>
    </tr>
    <tr>
      <td width="150pt">Link : </td>
      <td><label>
      <input name="rsslink" type="text" id="rsslink" size="33" value="http://" />
      </label></td>
    </tr>
    <tr>
      <td width="150pt">Description : </td>
      <td>
  <label>
        <textarea name="rssdescription" cols="30" rows="5" id="rssdescription"></textarea>
      </label>
  </td>
    </tr>
    <tr>
      <td width="150pt">Author : </td>
      <td><label>
        <input name="rssauthor" type="text" id="rssauthor" size="33" />
      </label></td>
    </tr>
    <tr>
      <td width="150pt">Date : </td>
      <td><label>
        <select name="doby"><?php for ($c = 1900; $c <= 2009; ++$c) { echo '<option value="'.$c.'">'.$c.'</option>'; } ?></select>
	<select name="dobm"><?php for ($c = 1; $c <= 12; ++$c) { echo '<option value="'.$c.'">'.$c.'</option>'; } ?></select>
	<select name="dobd"><?php for ($c = 1; $c <= 31; ++$c) { echo '<option value="'.$c.'">'.$c.'</option>'; } ?></select>
      </label></td>
    </tr>
<tr>
	<td width="150pt">Image Upload  :</td>
	<td><input type="file" name="avatar" /></td>
</tr>
<tr>
	<td width="150pt">Category  :</td>
	<td>
		<select name="type">
			<option value="ALL">ALL</option>
			<option value="Unsigned Music">Unsigned Music</option>
			<option value="Song Writing">Song Writing</option>
	</td>
</tr>
    <tr>
      <td> </td>
      <td> </td>
    </tr>
    <tr>
      <td> </td>
      <td><label>
        <input type="submit" name="Submit" value="Submit" />
        <input type="reset" name="Reset" value="Reset" />
      </label></td>
    </tr>
  </table>
  

</form>


<?php
echo page_footer();

?>

tommyda · September 2, 2009

It would be much easier if you didnt hide your table names.

<?php
//Check if the form was submitted if yes update the database

if (isset($_POST['save']))

{
//GET Feed ID ED: yoursite.com/editfeed.php?ID=33

$feed_id = $_GET['ID'];

mysql_query("UPDATE feeds SET content='$_POST[content]' WHERE id = '$feed_id'")or die(mysql_error());
echo 'Saved';
};

//Get feed from database

$getfeed = mysql_query("SELECT * FROM feeds WHERE id = '$feed_id'")or die(mysql_error());
$feed = mysql_fetch_array($getfeed);

//Display it in the edit feed form

<form method="post">
<input type="text" name="content" value="<?php echo $feed['content'];?>"/>
<input type="submit" name="save" value="save"/>
</form>

That should work

Sign In

updating data in databse?

Recommended Posts

Porkie

Link to comment

Share on other sites

tommyda

Link to comment

Share on other sites

Archived

Browse

Activity

Important Information