What is wrong with this ?

nimzrocks · February 15, 2010

Can you please tell me what is wrong with this comment script ? it is not showing anything. Just a blank page

This is comment.php

<!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=utf-8" />
<title>Untitled Document</title>
</head>

<body>

<?php

$id = $_REQUEST["id"]; 
include('write_comment.php'); 
$result = mysql_query ("SELECT * FROM commenttable WHERE postid='$id'
ORDER BY id desc") or exit (mysql_error()); 

while($row = mysql_fetch_array($result)) 

{ 

   echo  $row['name']; 
   echo  $row['mail']; 
   echo  $row['comment'];
   echo  $row['timestamp']; 

} 

?> 



<form action="index.php?id=<? echo $id;?>" method="post"/>
Name:<input name="name" type="text" ><br /> 
Mail:<input name="mail" type="text" ><br /> 
Kommentar: <textarea name="comment" cols="50" rows="6"></textarea><br /> 
<input name="submit" type="submit" class="submit" value="Send"/> 



</form >






</body>
</html>

This is write_comment.php

<!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=utf-8" />
<title>Untitled Document</title>
</head>

<body>




<?php

$con = mysql_connect("localhost","username","password");
if (!$con)
 {    
die('Could not connect: ' . mysql_error()); 
 } 
mysql_select_db("databasename", $con); 
if(isset($_POST['submit'])) { 

$name = $_REQUEST['name']; 
$mail = $_REQUEST['mail']; 
$commen = $_REQUEST['comment']; 
$postid = $_REQUEST['id']; 

if( (!empty($name)) && (!empty($mail)) && (!empty($comment)) ) { 
$sql = "INSERT INTO  prototype (name, mail, comment, postid)  

VALUES('$name', '$mail', '$comment', '$postid')"; 

mysql_query($sql) or die(mysql_error()); 

}

} 

?> 

</body>
</html>

Please someone help me.......

Thank you

Deoctor · February 15, 2010

echo the value of $id and see whether u are getting any results into that..

first check this documentation regarding the $_request

http://www.alternateinterior.com/2007/01/what-is-_request.html

bugcoder · February 15, 2010

cant get exactly what can be the issue but will like to know that what you will get after adding following line on both pages.

error_reporting(E_ALL);

nblackwood · February 15, 2010

try this code:

<?php

require_once('write-comment.php');

$id = $_REQUEST["id"];

$result = mysql_query ("SELECT * FROM commenttable WHERE postid='$id'

ORDER BY id desc") or exit (mysql_error());

while($row = mysql_fetch_array($result));

{

echo $row['name'];

echo $row['mail'];

echo $row['comment'];

echo $row['timestamp'];

}

?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<head>

<title>Untitled Document</title>

</head>

<body>

Name:

<br />

Mail:

<br />

Kommentar:

<br />

</form>

</body>

</html>

</body>

</html>

nblackwood · February 15, 2010

I forgot one thing...

make this change to the code i provided:

nimzrocks · February 15, 2010

Thank you nblackwood but I'm new to php. I do not see a code to connect to my sql data base can you tell me how to put that code also ?

Thank you

bugcoder · February 15, 2010

$con = mysql_connect("localhost","username","password");

instead of it give the username and password of your localhost database username password

by default it should be like

$con = mysql_connect("localhost","root","");

nimzrocks · February 15, 2010

My problem is I cant see anything when I go to comment.php I do not see my form there is anyone has a script ? Please give me. I'm really sick of this now. I typed a lot to get help every where but no one is helping me properly It is two weeks now still could not make this

http://www.phpfreaks.com/forums/index.php/topic,287535.msg1363149.html#msg1363149

you can see someone was helping me but he did not finish it Is anyone can complete it ?

please help me

wildteen88 · February 15, 2010

My problem is I cant see anything when I go to comment.php I do not see my form there

If you're getting a blank screen then it is more than likely there is an error in your code. In order to see the error you need to enable a PHP setting call display_errors and set error_reporting to E_ALL within your php.ini. If your host doesn't allow access to the php.ini then add the following two lines at the top of your php script (on the line after the opening PHP tag <?php)

error_reporting(E_ALL);
ini_set('display_errors', 1);

Post the errors you're getting here.

nimzrocks · February 15, 2010

Thank you very much. These are my errors

Warning: require_once(write-comment.php) [function.require-once]: failed to open stream: No such file or directory in /home/vol11/0fees.net/fees0_4383963/mywebsite.com/htdocs/script1/comment.php on line 15

Fatal error: require_once() [function.require]: Failed opening required 'write-comment.php' (include_path='.') in /home/vol11/0fees.net/fees0_4383963/mywebsite.com/htdocs/script1/comment.php on line 15

and I found out a what is the error so I just renamed write_comment.php to write-comment.php

and now I'm getting this error now

Notice: Undefined index: id in /home/vol11/0fees.net/fees0_4383963/mywebsite.com/htdocs/script1/comment.php on line 17

No database selected

What is the problem ?

wildteen88 · February 15, 2010

The main problem is this error

No database selected

For some reason your code is not connecting to your database. Try the following code, I have merged comment.php with write-comment.php. Place the following code in comment.php

<?php

$con = mysql_connect("localhost","username","password");
if (!$con)
    die('Could not connect: ' . mysql_error());

mysql_select_db("databasename", $con);

if(isset($_POST['submit']))
{
    $name = mysql_real_escape_sting($_POST['name']);
    $mail = mysql_real_escape_sting($_POST['mail']);
    $commen = mysql_real_escape_sting($_POST['comment']);
    $postid = (int) $_POST['id'];

    if( (!empty($name)) && (!empty($mail)) && (!empty($comment)) )
    {
        $sql = "INSERT INTO  prototype (name, mail, comment, postid) VALUES('$name', '$mail', '$comment', '$postid')";
        mysql_query($sql) or die('Unable to insert comment: '. mysql_error());
    }
}

?>

<!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=utf-8" />
<title>Untitled Document</title>
</head>

<body>

<?php

if(isset($_REQUEST["id"]) && is_numeric($_REQUEST["id"]))
{
    $id = (int) $_REQUEST["id"];
    $result = mysql_query ("SELECT * FROM commenttable WHERE postid='$id' ORDER BY id desc") or die ('Unable to grab comments: '.mysql_error());

    while($row = mysql_fetch_array($result));
    {
       echo  $row['name'];
       echo  $row['mail'];
       echo  $row['comment'];
       echo  $row['timestamp'];
    }
}

?>

<form id="form1" name="form1" method="post" action="">
  Name:
  <input name="name" type="text" />
  <br />
Mail:
<input name="mail" type="text" />
<br />
Kommentar:
<textarea type="text" name="comment" cols="50" rows="6"></textarea>
<br />
<input name="submit" type="submit" class="submit" value="Send"/>
</form>

</body>
</html>

nimzrocks · February 15, 2010

ok now I can see the form and I can fill it but when i clickes on submit I'm getting this error now.

Fatal error: Call to undefined function mysql_real_escape_sting() in /home/vol11/0fees.net/fees0_4383963/mywebsite.com/htdocs/script1/comment2.php on line 14

what can be the error now ?

wildteen88 · February 15, 2010

Ops. All instances of mysql_real_escape_sting should be mysql_real_escape_string

I mistype the function name.

nimzrocks · February 15, 2010

ok thank you I corrected other two sting to string but now I'm getting this error

Notice: Undefined index: id in /home/vol11/0fees.net/fees0_4383963/mywebsite.com/htdocs/script1/comment2.php on line 17

what is this time ?

nimzrocks · February 15, 2010

Seems got stuck again

wildteen88 · February 15, 2010

Change

<form id="form1" name="form1" method="post" action="">

To

<form id="form1" name="form1" method="post" action="comment2.php?id=<?php echo $id; ?>">

Now change

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

to

if(isset($_POST['submit'], $_GET['id']) && is_numeric($_GET['id']))

Also change

$postid = (int) $_POST['id'];

to

$postid = (int) $_GET['id'];

nimzrocks · February 16, 2010

No errors. I'm getting a new form back and data base is not getting any data

This is is script

<?php

error_reporting(E_ALL);
ini_set('display_errors', 1);

$con = mysql_connect("...","....","...");
if (!$con)
    die('Could not connect: ' . mysql_error());

mysql_select_db("fees0_4383963_test", $con);

if(isset($_POST['submit'], $_GET['id']) && is_numeric($_GET['id']))
{
    $name = mysql_real_escape_string($_POST['name']);
    $mail = mysql_real_escape_string($_POST['mail']);
    $commen = mysql_real_escape_string($_POST['comment']);
   $postid = (int) $_GET['id'];

    if( (!empty($name)) && (!empty($mail)) && (!empty($comment)) )
    {
        $sql = "INSERT INTO  prototype (name, mail, comment, postid) VALUES('$name', '$mail', '$comment', '$postid')";
        mysql_query($sql) or die('Unable to insert comment: '. mysql_error());
    }
}

?>

<!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=utf-8" />
<title>Untitled Document</title>
</head>

<body>

<?php

if(isset($_REQUEST["id"]) && is_numeric($_REQUEST["id"]))
{
    $id = (int) $_REQUEST["id"];
    $result = mysql_query ("SELECT * FROM prototype WHERE postid='$id' ORDER BY id desc") or die ('Unable to grab comments: '.mysql_error());

    while($row = mysql_fetch_array($result));
    {
       echo  $row['name'];
       echo  $row['mail'];
       echo  $row['comment'];
       echo  $row['timestamp'];
    }
}

?>

<form id="form1" name="form1" method="post" action="comment2.php?id=<?php echo $id; ?>">
  Name:
  <input name="name" type="text" />
  <br />
Mail:
<input name="mail" type="text" />
<br />
Kommentar:
<textarea type="text" name="comment" cols="50" rows="6"></textarea>
<br />
<input name="submit" type="submit" class="submit" value="Send"/>
</form>

</body>
</html>

This is table

CREATE TABLE `databasename`.`prototype` (

`name` VARCHAR( 50 ) NOT NULL ,

`mail` VARCHAR( 50 ) NOT NULL ,

`comment` VARCHAR( 500 ) NOT NULL ,

`postid` INT NOT NULL AUTO_INCREMENT PRIMARY KEY

) ENGINE = MYISAM

What should I do now ? Please help

jl5501 · February 16, 2010

$commen = mysql_real_escape_string($_POST['comment']);

is that meant to be $comment ?

nimzrocks · February 16, 2010

Yes you are right it is comment but nothing happened even i corrected it

any help ?

gizmola · February 16, 2010

This is kind of a mess. Let's start with your database. What is the table structure that these comments are going to be related to? Please don't say postid, because you have defined that column to be the primary key, and auto_increment.

freakstyle · February 17, 2010

The error is saying that you are not passing the variable "id" along with the form.

Looking back through your posts it is unclear as to why you would want a user to set the id of the post, if that's the use you intended it to have, then you simply need to add a new text input field with the name "id" and allow users to enter the id for that post.

Typically the post id would be set incrementally in your database table. I don't see your table schema anywhere so we can't help with that side of it.

The script below will output the contents of the posted data from your form, in case you aren't sure how to do that.

echo '<pre>';
print_r($_POST); 
echo '</pre>';

Good luck.

gizmola · February 18, 2010

There's another thread by the OP that covers the same ground. I'm locking this one so nobody else wastes their time.

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