PHP / MYSQL Wont post after adding Field

cchiffers

Hello,

I am having problems submitting data from a form.

Bassically, the form works fine, UNTILL I add a field at the end of the table in my MySQL database called rq_report. When I add the field, the data from the form fails to get added to the database.

Here is the code I currently use to submit the form :

<?PHP

$ip = getenv ( 'REMOTE_ADDR' );

$domain = GetHostByName($REMOTE_ADDR);

$done = "";

if ( $_POST['submit'] )
{

if (!$POST['name'])
{
$done = "Please enter your habboname...";
}
elseif (!$POST['artist'])
{
$done = "Please enter an artist name...";
}
elseif (!$_POST['title'])
{
$done = "Please enter a song title...";
}
else
{
$date = date("Y-m-d H:i:s");

  $db = mysql_connect("*****", "*****", "*****");
  mysql_select_db("nsfm_nsfmsite", $db);

  mysql_query("INSERT INTO tm_requests VALUES ('', '{$_POST['name']}', '{$_POST['artist']}', '{$_POST['title']}', '{$_POST['messege']}', '{$ip}', '{$date}')", $db);

  $_POST['name'] = "";
  $_POST['artist'] = "";
  $_POST['title'] = "";
  $_POST['messege'] = "";

  header("Location: thanks.htm");

}
}

print <<<EOF
<html>

<head>
<meta http-equiv="Content-Language" content="en-us">
<meta http-equiv="Content-Type" content="text/html; charset=windows-1252">
<title>NamistaiFM</title>
<link rel="stylesheet" href="blue.css">
<style type="text/css">

</style>
</head>
<form name="subscribe" action="request.php" method="post">
<body background="images/notes.gif">
<p align="center"><img src = "images/jockeys.gif" width="140" height="107"></p>
<div align="center"><font size="2" face="tahoma">{$done}</font>
<table width="270" height="79" border="0" cellpadding="0" cellspacing="0">
<tr>
<td>
<table width="273" height="125" border="0" cellpadding="0" cellspacing="1">
<tr>
<td width="123" height="22"><strong><font size="2" face="Tahoma">Name:</font></strong></td>
<td width="147"><input name="name" type="text" id="name" value="{$POST['name']}"></td>
</tr>
<tr>
<td height="22"><strong><font size="2" face="Tahoma">Song
Artist:</font></strong></td>
<td><input name="artist" type="text" id="artist" value="{$POST['artist']}"></td>
</tr>
<tr>
<td height="22"><strong><font size="2" face="Tahoma">Song
Title:</font></strong></td>
<td><input name="title" type="text" id="title" maxlength="210" value="{$POST['title']}"></td>
</tr>
<tr>
<td height="54"><strong><font size="2" face="Tahoma">Message:</font></strong></td>
<td><textarea name="messege" cols="16" rows="5" id="messege">{$POST['message']}</textarea></td>
</tr>
</table></td>
</tr>
<tr>
<td> </td>
</tr>
</table>
<input type="hidden" name="ip" valume="ip">

<input type="submit" name="submit" value="Send It!" />
<br><br>
<font face="tahoma" size ="1" color ="#FF0000"><strong>YOUR I.P : </strong></font><font face="tahoma" size ="1"><strong><span class="style1">$ip</span></strong></font><font face="tahoma" size ="1" color ="#FF0000"><strong> Will
be sent with your request<br></br>
PLEASE READ BEFORE SUBMITTING YOUR REQUEST:</strong></font><br>
<font size="1" face="Tahoma">Sending in the same request more than once will guarantee that your request will NOT be played on the air. Please do not flood our request system. Also, note that your IP is recorded every time you send in a request. Flooding, swearing, and/or abusing the request system could result in an IP ban. Please be respectful.</font>
</center>
</div>
</body>
</form>
</html>
EOF;
?>

I have a field in the mysql table called rq_id, that is autoincriminent, and is the first field in the table.

The system works fine, until I had the rq_report field, then it fails to post the data to the database.

I have no idea why this happens, bassically I wanted the rq_report field so that I could set the standard value of it as 1, then when someone reports a post to an administrator, the value of the rq_report changes to 2, then I wanted code on my page that displays the data from the database to beable to read the rq_report value for each post, if It was 1, do nothing and if it was 2, flag the messege table in red.

First things first though!

bastien

If you were to add some error handling to see if the sql generated an an error [by doing this]
[php

$sql = "INSERT INTO tm_requests VALUES ('', '{$POST['name']}', '{$POST['artist']}', '{$POST['title']}', '{$POST['messege']}', '{$ip}', '{$date}')":
mysql_query($sql, $db) or die ("Can't complete query [$sql] because ".mysql_error());
[/code]

you would have seen this error message

Syntax eror, column count does not match table

(or something close to that)

Solutions
1. Add error handling to your code. Makes this easier to debug.
2. Decide how to handle error
a. define the columns in the sql statement

insert into table(col1, col2... coln) values...

b. add the values for the last column to the current sql statement. If its to be empty or null simply append 'null'

$sql = "INSERT INTO tm_requests VALUES ('', '{$_POST['name']}', '{$_POST['artist']}', '{$_POST['title']}', '{$_POST['messege']}', '{$ip}', '{$date}',null)":

cchiffers

Thanks alot!

That works perfectly!

heres the new code :

mysql_query ("INSERT INTO tm_requests (rq_id, rq_name, rq_artist, rq_sname, rq_mess, rq_ip, rq_date, rq_report) VALUES (null, '{$POST['name']}', '{$POST['artist']}', '{$POST['title']}', '{$POST['messege']}', '{$ip}', '{$date}', 1)", $db) or die ("Can't complete query [$sql] because ".mysql_error());

I set it to the rq_report value to 1,

What im planning on doing now, is on the request view page, have a little script that says, if rq_id =1, do nothing, then if rq_id = 2, make table row red,

im also gonna have it so when people check the boxes next to the requests on the page, and click report, it copys them over to a new database (that works already)and set the rq_report value to = 2,

Heres some real basic code that I started off with :

$do = mysql_query("INSERT INTO tm_reports SELECT * FROM tm_requests WHERE rq_id='$value') (INSERT INTO tm_requests (rq_report) VALUES ('2') FROM tm_requests WHERE rq_id='$value') (INSERT INTO tm_reports (rq_report) VALUES ('2') FROM tm_requests WHERE rq_id='$value'") or die ("Can't complete query [$sql] because ".mysql_error());
}

When I run that I get this :

Can't complete query [] because You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ') (INSERT INTO tm_requests (rq_report) VALUES ('2') FROM tm_requests WHERE rq_id' at line 1

I'm kinda learning as I go 😛 I wana learn lots and lots!

Thanks for ALL your help!

bastien

sql syntax is wrong...

if you are doing an update to existing data

the syntax is

update tm_requests set rq_report =('2' WHERE rq_id = $id

if a straight insert statement

INSERT INTO tm_requests (rq_report) VALUES ('2')

cchiffers

hmmm,

I tried this code :

mysql_query("update tm_requests set rq_report =('2' SELECT * FROM tm_requests WHERE rq_id = '$id' AND tm_report=('1')") or die ("Can't complete query [$sql] because ".mysql_error());

And I get This Error :

Can't complete query [] because Duplicate entry '1025' for key 1