Form submit

longstand · #1 Nov 24th, 2007, 11:48

Below is a form with PHP code, can anyone tell me why the data from the form once filled is not being submitted into my database?

Created a form like this plenty of times, but when i make it larger it appears not to work! Its real basic PHP code, and i have probebly missed somthink being dozy!

Thanks!

PHP: Select all


<!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>
<form enctype="multipart/form-data" action="add.php" method="POST"> 
Name: <input type="text" name="first_name"><br> 
Surname: <input type="text" name = "last_name"><br> 
Address: <input type="text" name = "address"><br>
City: <input type="text" name = "city"><br>
Area: <input type="text" name = "area"><br>
Email: <input type="text" name = "email"><br>
Phone: <input type="text" name = "phone"><br>
User_name: <input type="text" name = "user_name"><br>
Password: <input type="text" name = "password"><br>
Work_location: <input type="text" name = "work_location"><br>
Agency_name: <input type="text" name = "agency_name"><br>
Agency_number: <input type="text" name = "agency_number"><br>
Skills: <input type="text" name = "skills"><br>
about: <input type="text" name = "about_me"><br>
Myspace_address: <input type="text" name = "myspace_address"><br>
Hotmail_yahoo_address: <input type="text" name = "hotmail_yahoo_address"><br>
Gender: <input type="text" name = "gender"><br>
Age: <input type="text" name = "age"><br>
Height_one: <input type="text" name = "height_one"><br>
Height_two: <input type="text" name = "height_two"><br>
Build: <input type="text" name = "build"><br>
Eye_colour: <input type="text" name = "eye_colour"><br>
Hair_colour: <input type="text" name = "hair_colour"><br>
Desired_work: <input type="text" name = "desired_work"><br>
 
 
 
 
Photo: <input type="file" name="image_one"><br>
Photo: <input type="file" name="image_two"><br>
Photo: <input type="file" name="image_three"><br>
Photo: <input type="file" name="image_four"><br> 
<input type="submit" value="Add">
</form>
</body>
<?php 
//This is the directory where images will be saved 
$target = "images/"; 
$target = $target . basename( $_FILES['photo']['image_one']);
$target = $target . basename( $_FILES['photo']['image_two']);
$target = $target . basename( $_FILES['photo']['image_three']);
$target = $target . basename( $_FILES['photo']['image_four']);
 
//This gets all the other information from the form 
$Name=$_POST['first_name']; 
$Surname=$_POST['last_name']; 
$Address=$_POST['address'];
$City=$_POST['city'];
$Area=$_POST['area'];
$Email=$_POST['email'];
$Phone=$_POST['phone'];
$User_name=$_POST['user_name'];
$Password=$_POST['password'];
$Work_location=$_POST['work_location'];
$Agency_name=$_POST['agency_name'];
$Agency_number=$_POST['agency_number'];
$Skills=$_POST['skills'];
$About=$_POST['about_me'];
$Myspace_address=$_POST['myspace_address'];
$Hotmail_yahoo_address=$_POST['hotmail_yahoo_address'];
$Gender=$_POST['gender'];
$Age=$_POST['age'];
$Height_one=$_POST['height_one'];
$Height_two=$_POST['height_two'];
$Build=$_POST['build'];
$Eye_colour=$_POST['eye_colour'];
$Hair_colour=$_POST['hair_colour'];
$Desired_work=$_POST['desired_work'];
 
 
 
$pic=($_FILES['photo']['image_one']);
$pic=($_FILES['photo']['image_two']);
$pic=($_FILES['photo']['image_three']);
$pic=($_FILES['photo']['image_four']); 
 
mysql_connect("localhost", "root", "longstand1") or die(mysql_error()) ; 
mysql_select_db("dbsignup") or die(mysql_error()) ; 

mysql_query("INSERT INTO `customer` VALUES ('$first_name', '$last_name', '$address', '$city', '$area', '$email', '$phone', '$user_name', '$password', '$work_location', '$agency_name', '$agency_number', '$skills', '$about_me', '$myspace_address', '$hotmail_yahoo_address', '$gender', '$age', '$height_one', '$height_two', '$build', '$eye_colour', '$hair_colour','$desired_work')") ; 
 
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target)) 
{ 
 
echo "The file ". basename( $_FILES['uploadedfile']['image_one']). "has been uploaded, and your information has been added to the directory"; 
} 
else { 

echo "Sorry Andy There Was A Problem Uploading The File."; 
} 
?> 
</html>

Marc · #2 Nov 24th, 2007, 11:54

Do you not need to list your column names in your Insert statement?

This is one of mine...

Code: Select all

   $query = mysql_query("INSERT INTO author_details (id, auth_name, website_url, auth_email, current_status, team_contact, final_response, cont_again, comments) VALUES ( '', '{$authname}', '{$authurl}', '{$authemail}', '{$currentstat}', '{$teamname}', '{$finalresponse}', '{$contactagain}', '{$comments}')") or die(mysql_error());

You may also want to add 'or die(mysql_error())' to see if it is an error that is causing it not to insert.

Hope that helps.

longstand · #3 Nov 24th, 2007, 14:11

Hello Marc, how are you today?

I have just tried what you said & altered the code to suit, its still not working!

I am using Mysql administrator to check if the data has gone into teh database, and it is creating new lines but leaving the coloumns blank, any ideas

Cheers,

c010depunkk · #4 Nov 24th, 2007, 15:27

These variable names:

PHP: Select all


$Name=$_POST['first_name']; 
$Surname=$_POST['last_name']; 
$Address=$_POST['address'];
$City=$_POST['city'];
$Area=$_POST['area'];
$Email=$_POST['email'];
$Phone=$_POST['phone'];
$User_name=$_POST['user_name'];
$Password=$_POST['password'];
$Work_location=$_POST['work_location'];
$Agency_name=$_POST['agency_name'];
$Agency_number=$_POST['agency_number'];
$Skills=$_POST['skills'];
$About=$_POST['about_me'];
$Myspace_address=$_POST['myspace_address'];
$Hotmail_yahoo_address=$_POST['hotmail_yahoo_address'];
$Gender=$_POST['gender'];
$Age=$_POST['age'];
$Height_one=$_POST['height_one'];
$Height_two=$_POST['height_two'];
$Build=$_POST['build'];
$Eye_colour=$_POST['eye_colour'];
$Hair_colour=$_POST['hair_colour'];
$Desired_work=$_POST['desired_work'];

and these here:

PHP: Select all


mysql_query("INSERT INTO `customer` VALUES ('$first_name', '$last_name', '$address', '$city', '$area', '$email', '$phone', '$user_name', '$password', '$work_location', '$agency_name', '$agency_number', '$skills', '$about_me', '$myspace_address', '$hotmail_yahoo_address', '$gender', '$age', '$height_one', '$height_two', '$build', '$eye_colour', '$hair_colour','$desired_work')") ;

are different. Variable names are case-sensitive. And some of the variable are just completely different.... The query is working perfectly, it's inserting empty values into the database because the variables are all empty.....

longstand · #5 Nov 24th, 2007, 15:34

Damn i have been told this before by Alex Geek and i have made the same mistake twice, am getting better but still making dosy "A_s mistakes all the time

Ha ha its the only way you learn i suppose,

Thanks for your help mate!

By the way i added you to me MSN, is the end of your address supposed to say .de or .com ?

Take Care!

longstand · #6 Nov 24th, 2007, 15:53

Ok c010depunkk was right by fixing the varables all the data from the form has been posted into the database fine the only problem i am having now is the images are still not being sent to the destiantion directory "images"

And the name of the images is still not being posted into the databse with the rest of the form data!

So basically am back to wear i started, i had the same problem when writing the form in an array in me other post thread,

Please dont give up on me now, ave been trying to fix this problem for a week,

Can anyone help

Thanks

ScottR · #7 Nov 25th, 2007, 01:14

It looks like your overwriting the $target variable. Each time you use the assignment operator with $target it rewrites the data of that variable. Maybe try setting your directory, and then adding the filenames to it.

Code: Select all

//This is the directory where images will be saved 
$target = "images/"; 
$image1 = $target . basename( $_FILES['photo']['image_one']);
$image2 = $target . basename( $_FILES['photo']['image_two']);
$image3 = $target . basename( $_FILES['photo']['image_three']);
$image4 = $target . basename( $_FILES['photo']['image_four']);

You may also have the state the directory differently depending on where it resides in your document structure. I.E. ../images/ to move back one level in the document tree.

Make sense??

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