[SOLVED] Mysql Not Submitting Anything

[imdead]

Hey Guys, Im Working on a script for a client although i'm stuck

i cant find the error

i even tried echo'ing out the INSERT values to see if they were the problem but there not

Please Help

Thanks

 

 

Signup

						<form action="tregister.php" method="post">
					<p>First Name: <input type="text" name="first" size="10"></p>
					<p>Username: <input type="text" name="username" size="10"></p>
					<p>Password: <input type="password" name="password" size="10"></p>
					<p>Second Name: <input type="text" name="second" size="10"></p>
					<p>Address: <input type="text" name="address" size="20"></p>
					<p>Email: <input type="text" name="email" size="20"></p>
					<p>What Is Your Current Job? <input type="text" name="job" size="15"></p>
					<p>Current Salary? <input type="text" name="salary" size="10"></p>
					<p>What Would You Like To Do? <input type="text" name="like" size="10"></p>
					<p>Salary Expectations? <input type="text" name="salarye" size="10"></p>
					<p>Do You Have Your Own Transport? <input type="text" name="transport" size="10"></p>
					<p>Contact Number: <input type="text" name="contact" size="15"></p>
					<p><input type="submit" value="Submit" name="submit"></p>
					</form>

 

Takesignup

$id = mysql_insert_id();
$first = $_POST["first"];
$username = $_POST["username"];
$password = $_POST["password"];
$second = $_POST["second"];
$address = $_POST["address"];
$email = $_POST["email"];
$job = $_POST["job"];
$salary = $_POST["salary"];
$like = $_POST["like"];
$salarye = $_POST["salarye"];
$transport = $_POST["transport"];
$contact = $_POST["contact"];
//insert the values
$result = @mysql_query("INSERT INTO users (id, first, username, password, second, address, email, job, salary, like, salarye, transport, contact) VALUES ($id, $first, $username, $password, $second, $address, $email, $job, $salary, $like, $salarye, $transport, $contact)");
    if (mysql_affected_rows() == 1)
	print "You have successfully registered and may now login";
else
	print mysql_error();

[MadTechie]

change

$result = @mysql_query("INSERT INTO users (id, first, username, password, second, address, email, job, salary, like, salarye, transport, contact) VALUES ($id, $first, $username, $password, $second, $address, $email, $job, $salary, $like, $salarye, $transport, $contact)");

 

to

$result = mysql_query("INSERT INTO users (id, first, username, password, second, address, email, job, salary, like, salarye, transport, contact) VALUES ($id, '$first', '$username', '$password', '$second', '$address', '$email', '$job', '$salary', '$like', '$salarye', '$transport', '$contact')") or die(mysql_error());

 

Should reveal the error (or more info atleast)

 

EDIT: also added the quotes

[dptr1988]

Have your tried echoing the query and checking if it's correct? Example:

<?php
$query = "INSERT INTO users (id, first, username, password, second, address, email, job, salary, like, salarye, transport, contact) VALUES ($id, $first, $username, $password, $second, $address, $email, $job, $salary, $like, $salarye, $transport, $contact)";
echo $query;
$result =mysql_query($query);
?>

 

Also, why are you suppressing errors on the mysql_query() function? What error messages are you getting? Do you even have a mysql connection?

 

[imdead]

Ok thanks i now get the error

 

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 'like, salarye, transport, contact) VALUES (0, Kevin, kevski, PASSSSWORDD, LASTNAME,' at line 1

[dptr1988]

the word 'like' is a mysql keyword

 

You need to quote that column name in backticks(`)

[MadTechie]

also you need to quotes your values!

[imdead]

Ok thanks for such quick help guys

 

i've changed the code to

$result = mysql_query("INSERT INTO users (id, first, username, password, second, address, email, job, salary, like, salarye, transport, contact) VALUES (`$id`, `$first`, `$username`, `$password`, `$second`, `$address`, `$email`, `$job`, `$salary`, `$like`, `$salarye`, `$transport`, `$contact`)") or die(mysql_error());
    if (mysql_affected_rows() == 1)
	print "You have successfully registered and may now login";
else
	print mysql_error();

 

And now im recieving this error:

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 'like, salarye, transport, contact) VALUES (`0`, `Kevin`, `kevski`, `PASSsWORD`, ' at line 1

[dptr1988]

the field name 'like' is still not quoted.

 

It should be like this

<?php
$result = mysql_query("INSERT INTO users (id, first, username, password, second, address, email, job, salary, `like`, salarye, transport, contact) VALUES (`$id`, `$first`, `$username`, `$password`, `$second`, `$address`, `$email`, `$job`, `$salary`, `$like`, `$salarye`, `$transport`, `$contact`)") or die(mysql_error());
?>

 

 

It would be even better if you could rename the field 'like'  to some other name. It is not a good idea to use mysql reserved words for field names

[imdead]

ah ok i changed it to the code above and now the error is

 

 Unknown column '0' in 'field list'

[dptr1988]

Have you tried looking at the echoed query as described in this message: http://www.phpfreaks.com/forums/index.php/topic,194387.msg875311.html#msg875311

 

 

 

 

[MadTechie]

the field name 'like' is still not quoted.

 

It should be like this

<?php
$result = mysql_query("INSERT INTO users (id, first, username, password, second, address, email, job, salary, `like`, salarye, transport, contact) VALUES (`$id`, `$first`, `$username`, `$password`, `$second`, `$address`, `$email`, `$job`, `$salary`, `$like`, `$salarye`, `$transport`, `$contact`)") or die(mysql_error());
?>

 

No it should be like this

 

$result = mysql_query("INSERT INTO `users` (`id`, `first`, `username`, `password`, `second`, `address`, `email`, `job`, `salary`, `like`, `salarye`, `transport`, `contact`) VALUES ('$id', '$first', '$username', '$password', '$second', '$address', '$email', '$job', '$salary', '$like', '$salarye', '$transport', '$contact')") or die(mysql_error());

[imdead]

Fixed it

 

Thanks alot guys!