Php problem

[razorsese]

The problem is whit the var $threadid .

It dosent work here(it dosen't display the data from database):  $dbh = "SELECT *FROM comments WHERE threadid = '".$threadid."' ";

But if I assign a value before the that code the html page show correctly.

 

 

include("config.php");

$name = $_POST['name'];
$comment = $_POST['comment'];
$threadid = $_POST['threadid'];

if($name & $comment)
{
	$dbh="INSERT INTO comments (name,comment,threadid) VALUES ('$name','$comment','$threadid') ";
	mysql_query($dbh);


}



  
    
    $dbh = "SELECT *FROM comments WHERE threadid = '".$threadid."' ";
    $req = mysql_query($dbh);
    while($row=mysql_fetch_array($req))
    {
         echo "<li>";
        echo "<br>"."<b>".$row['name']."</b></br>"."<br>".$row['comment']."</br>";
	echo "</li>";
    }

 

$('.submit').click(function(){
		  location.reload();
		var name = $("#name").val();
            var comment = $("#comment").val();


	var threadid  = $("#v").val();

		var dat = 'name='+name+'&comment='+comment+'&threadid='+threadid;

            $.ajax({
			type:"post",
		    url:"comment.php",
			data:dat,
			success:function(){ 
			console.log("dat"); 
			 }
	      });
              
		 return false;

	 });

[smerny]

try putting a space between * and FROM

 

also, you can do this to see the error in the query...

 

$req = mysql_query($dbh) or die("MySQL Error: ". mysql_error());

[litebearer]

Also seriously need to validate/sanitize your post values BEFORE you attempt to use them

[razorsese]

Still the same problem and mysql_error()) don't report anything

[smerny]

what exactly is it doing?

 

are you getting empty list items? or nothing at all that is within the mysql loop?

 

if you aren't even getting the empty list items sent to the browser, then it would appear that nothing in the comments database matches the threadid you are supplying?

[razorsese]

Well when i submit a comment in the html page the PHP code insert correctly in the database but fails at SELECT * FROM

And also if i manually assign a value to $threadid right before the second query for example 2 in the html page the value from database appear correctly:

 

Without assigned value (Not displaying correctly): http://s17.postimage.org/c1ldbk6gf/example1.png

With assigned value(Displaying correctly):http://s17.postimage.org/8knwmbwrz/example2.png

[Vel]

  Quote

if($name & $comment)

 

Should be if($name && $comment)

 

And as others have pointed out, validate/sanitize your code before putting them anywhere near an SQL statement.

[razorsese]

After I inspected whit FireBug in console log the values appear correctly but for some reason don't appear in the html page

[Vel]

I just noticed your Ajax statment, you're not telling the HTML to update. You need to adjust you success statement to:

success: function(data) {
console.log("dat");
$('.classofdivtobeupdated').html(data);
}

[razorsese]

Yep.I saw that before entering the forum

[Vel]

  Quote

Yep.I saw that before entering the forum

If you've fixed that, and any other potential mistakes, then you should post your latest code here for us.

 

If you add alert(data); to the success function what do you get?