Jump to content

having trouble inserting data into 2 tables


yobo

Recommended Posts

hey all,

 

i am having problems inserting data into 2 tables at the same time after submitting my form here is my form code

 

<?php

session_start();

if (!isset($_COOKIE['loggedin']))

die("You are not logged in!");

?>

<head>

<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />

<title>Add a Hack</title>

</head>

<?php

 

 

$userid=$_GET['userid'];

$_SESSION['userid']=$userid;

 

include("config.php");

 

// connect to the mysql server

$link = mysql_connect($server, $db_user, $db_pass)

or die ("Could not connect to mysql because ".mysql_error());

 

// select the database

mysql_select_db($database)

or die ("Could not select database because ".mysql_error());

 

$type = @mysql_query('SELECT typeid, name FROM type');

if(!$type) {

exit('<p>unable to obtain category list from the ' .

'databas.</p>');

}

 

?>

<body>

<h2> Submit a New Hack!</h2>

<form action="hackadded.php" method="post" enctype="multipart/form-data">

<input type="hidden" name="userid" value="<?php echo $userid;?>" /><br/>

<label> Hack Catagory: <select name="tid" size="1">

<option selected value="">select one</option>

<option value="">---------</option><br/>

<?php

while ($hacktype = mysql_fetch_array($type)) {

$tid = $hacktype['typeid'];

$aname = htmlspecialchars($hacktype['name']);

echo "<option value='$tid'>$aname</option>\n";

}

?>

</select><br/>

<label> Hack Name: <input type="text" name="hackname" /><br/></label>

<label>Upload Hack: <input type="file" name="upload" /><br/></label>

<label>Hack Description: <input type="text" name="desc" maxlength="255" /><br/></label>

<label>Hack Version: <input type="text" name="version" /><br/></label>

<input type="submit" value="SUBMIT HACK"

</form>

<?php

echo " <a href=members.php?userid=$userid>Back To Profile<br>";

?>

</body>

</html>

 

and this is the code that handles the form submission.

 

<?php

include("config.php");

 

// connect to the mysql server

$link = mysql_connect($server, $db_user, $db_pass)

or die ("Could not connect to mysql because ".mysql_error());

 

// select the database

mysql_select_db($database)

or die ("Could not select database because ".mysql_error());

 

//bail out if the file isnt really an upload

if (!is_uploaded_file($_FILES['upload']['tmp_name'])) {

exit('there was no file uploaded!');

}

$hackname = $_POST['hackname'];

$uploadfile = $_FILES['upload']['tmp_name'];

$uploadname = $_FILES['upload']['name'];

$uploadtype = $_FILES['upload']['type'];

$uploaddesc = $_POST['desc'];

$hackversion = $_POST['version'];

$userid = $_POST['userid'];

$tid = $_POST['tid'];

 

if ($tid == '') {

exit('<p>you must select a catagory.</p>');

}

 

//open file for binary reading

$tempfile = fopen($uploadfile, 'rb');

 

//read the entire file into memory using php's

//filesize function to get the file size

$filedata = fread($tempfile, filesize($uploadfile));

 

//prepaer for database insert by adding backslashes

//before speaciail chractors

$filedata = addslashes($filedata);

 

//close connection

fclose($tempfile);

 

//create the sql query

$sql = "INSERT  INTO hacks SET

filedata = '$filedata',

mimetype = '$uploadtype',

hackname = '$hackname',

description = '$uploaddesc',

version = '$hackversion',

username = '$userid'";

 

$sql2 = "INSERT INTO hacktype SET

hacksid = '$hid',

typeid = '$tid'";

 

 

//perform the insert

$ok = @mysql_query($sql);

if (!$ok) {

exit('database error: ' . mysql_error());

}else{

echo"hack added";

}

$hid = mysql_insert_id();

 

 

 

now the data insertes into the hacks table but not the hacktype table what i need is the hacks id and the selected catagory id to be inserted into the hacktype table

Archived

This topic is now archived and is closed to further replies.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.