Edit/Delete From DB

Will_742

Ok so I have a news system and i want to be able to edit and delete previous posts...

The problem i get when i go to delete is that once i click submit and i have chosen the post i want to delete, the page refreshes and nothing happens, and it doesnt delete.

This is the delete code:

<body bgcolor=1F1F1F>
<font face="verdana" size=1 color="424242">
<form method="get" action="<? echo $PHP_SELF; ?>">
<?
mysql_pconnect("localhost","cotjah","****"); //connect to the mySQL
mysql_select_db("cotjah_news"); //select the db
if(!$submit)//if submit has not already been pressed
{
    $result = mysql_query("select * from news order by id"); //select all the news
    while($r=mysql_fetch_array($result)) //run the while loop
    { 
        $title=$r["title"];//take out the title
        $id=$r["id"];//take out the id
		echo "<div align=\"left\">";
        echo "<INPUT TYPE=\"RADIO\" NAME=\"id\" VALUE=\"$id\">";
        echo $id;
		echo "&nbsp;";
		echo "|";
		echo "&nbsp;";
        echo $title;
		echo "<br>";
    }
	echo "<br>";
	echo "<input type=\"submit\" value=\"Delete!\" name=\"Submit\"></form>";
	} else {
    $sql = "DELETE FROM news WHERE id=$id";
    $result = mysql_query($sql);
	echo "<div align=\'center\'>";
    echo "News deleted!";
}
?>

Also, in my edit page, i have it so you click which post you want to edit and click submit and there is a form where you can edit the news.. but i get the error : "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home2/cotjah/public_html/form/edit.php on line 62" once i clicked submit and no information shows up in the form.

Here is the code :

<form action="<? echo $PHP_SELF ?>" method="post">
<?
mysql_pconnect("localhost","cotjah","*****"); //enter name and password
mysql_select_db("cotjah_news"); //select the database
if(!$cmd)
{
	$result = mysql_query("select * from news"); //replace news with your table name
	while($r=mysql_fetch_array($result))
	{	
		$title=$r["title"];
		$message=$r["newstext"];
?>
<INPUT TYPE="RADIO" NAME="id" VALUE="<?php echo $id;?>"><? echo $id;?> <? echo $title ?><br>
<? } ?>
<input type="submit" name="cmd" value="edit"></form>
<? } ?>
<?
if($cmd=="edit")
{
	if (!$submit)
	{
 		$sql = "SELECT * FROM news WHERE id=$id";
		//replace news with your table name above
 		$result = mysql_query($sql);        

 		$myrow = mysql_fetch_array($result);
?>
<input type=hidden name="id" value="<?php echo $myrow["id"] ?>">
Title:<INPUT TYPE="TEXT" NAME="title" VALUE="<?php echo $myrow["title"] ?>" SIZE=30><br>
Message:<TEXTAREA NAME="message" ROWS=10 COLS=30><? echo $myrow["message"] ?></TEXTAREA><br>
<input type="hidden" name="cmd" value="edit">
<input type="Submit" name="submit" value="Post Comment">
</form>
<? } ?>
<?
	if($submit)
	{
 		$sql = "UPDATE news SET title=\'$title\',message=\'$message\',who=\'$who\' WHERE id=$id";
		//replace news with your table name above
 		$result = mysql_query($sql);
 		echo "Thank you! Information updated.";
	}
}
?>

So, does anyone know whats up? Any help greatly appreciated.

Thanks,
Will 🙂

noimad1

To start, lets maybe try some error checking on your database queries.

Insert this after each of your $result = Lines:


if($result == false){
		echo mysql_errno().":".mysql_error()."<BR>";
		exit;
		}

That way we can make sure you aren't getting any database connection errors.

If not, then we can go on to the code...

On your second page of script, I'm not sure if this can create a problem or not, but it looks like you have one to many spaces after Set in the "UPDATE news SET title"

Will_742

Still get the same error(s)

noimad1

still a blank page on the first one?

Hmm....maybe your if condition has a problem...

Can you try to comment out all of your if code.

So basically it just says:

mysql_connect("localhost","cotjah","****"); //connect to the mySQL
mysql_select_db("cotjah_news"); 
$sql = "DELETE FROM news WHERE id=$id";
    $result = mysql_query($sql);
    echo "<div align='center'>";
    echo "News deleted!";

that way we can break the code down into sections and find out where the error is occuring...

Also, are you sure you put the error checking after all of your database queries? That should have given a more specific error than the "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL "

Also on your second page, I would clean this code up a bit:

if(!$cmd)
{
    $result = mysql_query("select * from news"); //replace news with your table name
    while($r=mysql_fetch_array($result))
    {    

        $title=$r["title"];
        $message=$r["newstext"];
?>
<INPUT TYPE="RADIO" NAME="id" VALUE="<?php echo $id;?>"><? echo $id;?> <? echo $title ?><br>
<? } ?>
<input type="submit" name="cmd" value="edit"></form>
<? } ?

change to:

if(!$cmd)
{
    $result = mysql_query("select * from news"); //replace news with your table name
    while($r=mysql_fetch_array($result))
    {    

        $title=$r["title"];
        $message=$r["newstext"];
echo "<INPUT TYPE=\"RADIO\" NAME=\"id\" VALUE=\"$id;\">$id $title <br>";
}

echo "<input type=\"submit\" name=\"cmd\" value="edit"></form>";
}

And I would do that for all of your code.

I don't think you want to start and if/while statement, end your php code with the ?> then start it again later.

That could be causing the error on the second page.

Try to clean that code up so it is one large php block instead of php/html/php/html etc...