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shakac

Active Member
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2011
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hi all friends
i code in php for a multipage registration form like
first page :first name, last name
2nd page:parental details
.
.
last will submit all data to mysql database
the code is here by page but after last page that will submit data to database got a problem.
the problem is
Code:
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 '='shalah',                 email_addres=  'uddin',                 membership_' at line 2

here code for all page

page1:
PHP:
<form method="post" action="page2.php">
First Name:<input type="text" name="first_name" /><br>
Last Name:<input type="text" name="last_name"  /><br>
<input type="submit" name="submit" />
</form>

page2:
PHP:
<?php
$fname=$_POST['first_name'];
$lname=$_POST['last_name'];
?>
<html>
<head></head>
<body>
<form method="post" action="page3.php">
<input type="hidden" name="first_name" value="<?php echo $fname;?>" />
<input type="hidden" name="last_name" value="<?php echo $lname;?>" />
Education:<input type="text" name="education" /><br>
School:<input type="text" name="school" /><br>
<input type="submit" name="submit" />
</form>
</body>
</html>

page3:
PHP:
<?php
$fname=$_POST['first_name'];
$lname=$_POST['last_name'];
$education=$_POST['education'];
$school=$_POST['school'];
?>
<html>
<head></head>
<body>
<form method="post" action="page4.php">
<input type="hidden" name="first_name" value="<?php echo $fname;?>" />
<input type="hidden" name="last_name" value="<?php echo $lname;?>" />
<input type="hidden" name="education" value="<?php echo $education;?>" />
<input type="hidden" name="school" value="<?php echo $school;?>" />
Experience:<input type="text" name="experience" /><br>
<input type="submit" name="submit" />
</form>
</body>
</html>

page3:
PHP:
<?php
$fname=$_POST['first_name'];
$lname=$_POST['last_name'];
$education=$_POST['education'];
$school=$_POST['school'];
$expe=$_POST['experiance'];
//let's create the query
$insert_query = "insert into subscriptions (
                name='$fname',
                email_addres=  '$lname',
                membership_type= '$education',
                terms_and_conditions='$school',
                name_on_card='$expe')";
//let's run the query
$connect=mysql_connect("localhost","root","");
if(!$connect)
{
die('Could not connect: ' . mysql_error());
}
    mysql_select_db("reg",$connect);
if (!mysql_query($insert_query,$connect))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";

mysql_close($con);
?>
please anybody help me to fix this.
thanks in advance
 
5 comments
The problem is with your mysql query syntax

PHP:
$insert_query = "insert into subscriptions (
                name='$fname',
                email_addres=  '$lname',
                membership_type= '$education',
                terms_and_conditions='$school',
                name_on_card='$expe')";
It should be like this:

PHP:
$insert_query = "INSERT INTO subscriptions (name, email_addres, membership_type, terms_and_conditions, name_on_card)
                VALUES ($fname, $lname, $education, $school, $expe) ";
It should fix your problem

Just replace page3 with this:

PHP:
<?php
$fname=$_POST['first_name'];
$lname=$_POST['last_name'];
$education=$_POST['education'];
$school=$_POST['school'];
$expe=$_POST['experiance'];
//let's create the query
$insert_query = "INSERT INTO subscriptions (name, email_addres, membership_type, terms_and_conditions, name_on_card)
                VALUES ($fname, $lname, $education, $school, $expe) ";
//let's run the query
$connect=mysql_connect("localhost","root","");
if(!$connect)
{
die('Could not connect: ' . mysql_error());
}
    mysql_select_db("reg",$connect);
if (!mysql_query($insert_query,$connect))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";

mysql_close($con);
?>

Regards
 
depending on ur Table data type add those....

PHP:
$insert_query = "INSERT INTO subscriptions(name, email_addres, membership_type, terms_and_conditions, name_on_card) VALUES ('$fname', '$lname', '$education', '$school', '$expe')";

single quotes are needed for string / VarChaR type field !
 
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