Hi all, I need some help I have a mysql tables and want to take some values from a form and place them into the database. This is the code I have so far I am pretty new to php so sorry for any school boi errors. :P

<?php
include("application_top.php");
//validate the form
if (!empty($_REQUEST['Email'])){
$Email = $_REQUEST['Email'];	
} else {
$Email = NULL;
echo '<p class="error"> You forgot to enter a Email!</p>';
}
$Movies = $_REQUEST['Movies'];
$Score = $_REQUEST['Score'];
$Score1 = $_REQUEST['Score1'];
$Score2 = $_REQUEST['Score2'];
$Score3 = $_REQUEST['Score3'];

INSERT INTO Survey VALUES ('$Email','$Movies','$Score','$Score1','$Score2','$Score');

?>

any help would be amazing thanks all

Instead of that last line of code, you need something like this.

PHP Code:

$connection = mysql_connect('localhost','username','password');
if($connection){
if(mysql_select_db('database_name', $connection)){
//keep people from doing SQL injection attacks.
$Email = addslashes($Email);
$Movies = addslashes($Movies);
$Score = addslashes($Score);
$Score1 = addslashes($Score1);
$Score2 = addslashes($Score2);
$Score3 = addslashes($Score3);

$query = "INSERT INTO Survey VALUES ('$Email','$Movies','$Score','$Score1','$Score2','$Score');";
if(mysql_query($query, $connection)){
echo 'success';
}else{
echo 'failure';
}
}else{
echo 'failure';
}else{
echo 'failure';
}
}else{
echo 'failure';
} 


you can avoid all the failure issues if you put that in a function and return on success, but meh.

EDIT: oh, and be sure to replace 'username', 'password', 'database_name', and possibly 'localhost' with correct values (ask your webhost if you're not sure).

Hi done that but when i go to sql and say select * from Survey it does not show any results

The thing is I am conecting to a server thought putty as localhost dont know if that has anythink to do with it

if you're connection with putty as localhost, but your server is not on your own computer, then of course you won't get any results.

You should use phpmyadmin to view the results online.

Try running the code I gave you (inside your own script) and see what it prints out. if it prints success, there should be values. if it prints failure, there is a problem.

trying your code I get this error

Parse error: parse error in /users/start07/jpr23/public_html/Handelform.php on line 31

and line 31 is?

30echo 'failure';
31}else{
32echo 'failure';
33}

I thougth there was a brace missing when i added one it moved the error to 35

sorry, I think I had one too many elses in there. try this.

PHP Code:

$connection = mysql_connect('localhost','username','password');
if($connection){
    if(mysql_select_db('database_name', $connection)){
        //keep people from doing SQL injection attacks.
        $Email = addslashes($Email);
        $Movies = addslashes($Movies);
        $Score = addslashes($Score);
        $Score1 = addslashes($Score1);
        $Score2 = addslashes($Score2);
        $Score3 = addslashes($Score3);
        
        $query = "INSERT INTO Survey VALUES ('$Email','$Movies','$Score','$Score1','$Score2','$Score');";
        if(mysql_query($query, $connection)){
            echo 'success';
        } else{
            echo 'failure';
        }
    } else{
        echo 'failure';
    }
    mysql_close($connection);
} else{
    echo 'failure';
} 


Warning: mysql_select_db(): supplied argument is not a valid MySQL-Link resource in /users/start07/jpr23/public_html/Handelform.php on line 14
failure

14if(mysql_select_db('m08jpr23', $connetion)){

dang it, typo. should be $connection, not $connetion

it comes up with failure does that mean problem with my connection ??

Oh! also, I forgot to close the connection. It will be closed by default at the end of the script, but it's usually a good idea to close it when you're done with it.

You should now see the function call toward the end of the second piece of code I posted. (4th to last line)

how do I close the Database is it mysql_close($connection); and all i get now is failure

yes, it's the mysql_close call.

try adding this to see what went wrong.

PHP Code:

echo mysql_error(); 


adding it to where !

the end of the code

this is what I have so far

<?php
$connection = mysql_connect('host','username','password')or die(mysql_error());
if($connection){
if(mysql_select_db('m08jpr23', $connection)){

$Email = addslashes($Email);
$Movies = addslashes($Movies);
$Score = addslashes($Score);
$Score1 = addslashes($Score1);
$Score2 = addslashes($Score2);
$Score3 = addslashes($Score3);

$query = "INSERT INTO Survey VALUES ('$Email','$Movies','$Score','$Score1','$Score2','$Score')";
if(mysql_query($query, $connection)){
 echo 'success';
        } else{
            echo 'failure';
        }
    } else{
        echo 'failure';
    }
    mysql_close($connection);
} else{
    echo 'failure';
}  
echo mysql_error();  
?>

and I just get failure

Personally, I'd write it that way:

PHP Code:

<?php
$connection = mysql_connect('host','username','password')or die(mysql_error());
if($connection===FALSE){
    echo mysql_error();
    die();
}
if(mysql_select_db('m08jpr23', $connection)===FALSE){
    echo mysql_error();
    die();
}

$Email = addslashes($Email);
$Movies = addslashes($Movies);
$Score = addslashes($Score);
$Score1 = addslashes($Score1);
$Score2 = addslashes($Score2);
$Score3 = addslashes($Score3);

$query = "INSERT INTO Survey VALUES ('$Email','$Movies','$Score','$Score1','$Score2','$Score')";
$ret=mysql_query($query, $connection);

if($ret===FALSE){
    echo 'failure\n<br/>';
    echo mysql_error();
}
else{
    echo 'success';
}
mysql_close($connection);
?>

Hi all, tried all that still it wont load the information into the database could it be because of the contents I am trying to load into the database?

here is all my code
form code

<form name="Register" method="post" action="Handelform.php">
              <label>
              <fieldset>
              <legend> Please Enter your information all fields are required:</legend>
              
              <div align="left">
                <p>Email
                  <input type="text" name="Email" id="Email" size="25" maxlength="60">                
                  </p>
                <p>
                  <label>
                  <select name="Movies" size="1" id="Movies">
                    <option>Indian Jones TOD</option>
                    <option>Bad Boys 2</option>
                    <option>Pulp Fiction</option>
                  </select>
                  </label>
                  </p>
                <p>Action:
                  <label>
                  <select name="Score" size="1" id="Score">
                    <option>1</option>
                    <option>2</option>
                    <option>3</option>
                    <option>4</option>
                    <option>5</option>
                    <option>6</option>
                    <option>7</option>
                    <option>8</option>
                    <option>9</option>
                    <option>10</option>
                  </select>
                  </label>
                  </p>
                <p>Visual Effects:
                  <select name="Score1" size="1" id="Score1">
                    <option>1</option>
                    <option>2</option>
                    <option>3</option>
                    <option>4</option>
                    <option>5</option>
                    <option>6</option>
                    <option>7</option>
                    <option>8</option>
                    <option>9</option>
                    <option>10</option>
                  </select>
                </p>
                <p>Story Line:
                  <select name="Score2" size="1" id="Score2">
                    <option>1</option>
                    <option>2</option>
                    <option>3</option>
                    <option>4</option>
                    <option>5</option>
                    <option>6</option>
                    <option>7</option>
                    <option>8</option>
                    <option>9</option>
                    <option>10</option>
                  </select>
                </p>
                <p>Overall Score:
                  <select name="Score3" size="1" id="Score3">
                    <option>1</option>
                    <option>2</option>
                    <option>3</option>
                    <option>4</option>
                    <option>5</option>
                    <option>6</option>
                    <option>7</option>
                    <option>8</option>
                    <option>9</option>
                    <option>10</option>
                  </select>
                  <br>
                  </p>
              </div>
              <div align="left"></div>
<div align="left"><label><br>
                            <br>
                            <input type="submit" name="Submit" id="Submit" value="Submit">
                            </label>
                            <br>
                            </div>
                          </label>
                          </fieldset>
            </form>            </td>

and my php code that handel this

<?php
$db = mysql_connect("host","username","password");
$result = mysql_select_db("m08jpr23", $db);

$Movies = $_REQUEST['Movies'];
$Score = $_REQUEST['Score'];
$Score1 = $_REQUEST['Score1'];
$Score2 = $_REQUEST['Score2'];
$Score3 = $_REQUEST['Score3'];

$query = "INSERT INTO Survey VALUES ('$Email','$Movies','$Score','$Score1','$Score2','$Score')";
?>

and finally my code for my SQL

create table Survey( 
SurveyId INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
Email VARCHAR(30),
Movies VARCHAR(25),
Score INT (5),
Score1 INT (5),
Score2 INT (5),
Score3 INT (5)
);

any help would be amazing thanks in advance

still it wont load the information into the database

Could you be more descriptive ?
Have you got an error?
Does it gives you "success" and you don't see nothing in the db ?
You don't have errors, but the datas are not in the db ?
If you try to add an

PHP Code:

echo $query; 


just after the mysql_query(), what is it's value ?