forms connecting and updating to the database

Question

I can't update the data in my database, but it shows that the data is key in successfully,also i have several errors, but i don't how to to correct them, as i'm new to php..please help me out.

My firstform.php :

<?php
session_start();
include 'connect.php';
$query = mysql_query("SELECT *FROM firstform WHERE $id='$_SESSION[manager_name]'");

while($row = mysql_fetch_assoc($query))
{
    $manager =$row['manager_name'];
    $evaluator =$row['evaluator_name'];
    $outlet =$row['outlet_location'];
    $date =$row['date'];
    $day =$row['day'];
    $timein =$row['time_in'];
    $timeout =$row['time_out'];
    $overall=$row['overall'];
    $quality=$row['quality'];
    $service=$row['service'];
    $cleanliness=$row['cleanliness'];
}
?>

This is my updatefirstform.php :

<?php
//error_reporting(0);
include'connect.php';
session_start();

$manager =$_POST['manager_name'];
$evaluator =$_POST['evaluator_name'];
$outlet =$_POST['outlet_location'];
$date =$_POST['date'];
$day =$_POST['day'];
$timein =$_POST['time_in'];
$timeout =$_POST['time_out'];
$overall=$_POST['overall'];
$quality=$_POST['quality'];
$service=$_POST['service'];
$cleanliness=$_POST['cleanliness'];

mysql_query("UPDATE firstform SET id ='$id' manager_name='$manager', evaluator_name ='$evaluator',outlet_location='$outlet',date='$date',day='$day',time_in ='$timein',time_out='$timeout',overall='$overall',quality='$quality',service='$service',cleanliness='$cleanliness' WHERE $id ='$_SESSION[manager_name]'");

echo ("<SCRIPT LANGUAGE ='JavaScript'>
         window.alert ('Data are Sucessfully Updated!')
         window.location.href ='secondform.php'
         </SCRIPT>");

?>

The errors:

Errors :Notice: Undefined variable: manager in C:\xampp\htdocs\marrybrown_clean\firstform.php on line 4

Notice: Undefined index: manager_name in C:\xampp\htdocs\marrybrown_clean\firstform.php on line 4

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\marrybrown_clean\firstform.php on line 6

Answer 1

i have edited my firstform.php :

<?php
session_start();
include 'connect.php';

$query = mysql_query("SELECT * FROM firstform  $id ='$_SESSION[user_id]'");
while($row = mysqli_fetch_assoc($query))
{

    $id=$row['user_id'];
    $manager =$row['manager_name'];
    $evaluator =$row['evaluator_name'];
    $outlet =$row['outlet_location'];
    $date =$row['date'];
    $day =$row['day'];
    $timein =$row['time_in'];
    $timeout =$row['time_out'];
    $overall=$row['overall'];
    $quality=$row['quality'];
    $service=$row['service'];
    $cleanliness=$row['cleanliness'];
}

?>

and also my updatefirstform.php :

<?php
//error_reporting(0);
include'connect.php';
session_start();
if(isset($_POST['submit'])) {


    $id=mysqli_real_escape_string($_POST['user_id']);
    $manager =mysqli_real_escape_string($_POST['manager_name']);
    $evaluator =mysqli_real_escape_string($_POST['evaluator_name']);
    $outlet =mysqli_real_escape_string($_POST['outlet_location']);
    $date =mysqli_real_escape_string($_POST['date']);
    $day =mysqli_real_escape_string($_POST['day']);
    $timein =mysqli_real_escape_string($_POST['time_in']);
    $timeout =mysqli_real_escape_string($_POST['time_out']);
    $overall=mysqli_real_escape_string($_POST['overall']);
    $quality=mysqli_real_escape_string($_POST['quality']);
    $service=mysqli_real_escape_string($_POST['service']);
    $cleanliness=mysqli_real_escape_string($_POST['cleanliness']);

mysqli_query("UPDATE firstform SET manager_name='$manager', evaluator_name ='$evaluator',outlet_location='$outlet',date='$date',day='$day',time_in ='$timein',time_out='$timeout',overall='$overall',quality='$quality',service='$service',cleanliness='$cleanliness' WHERE $id ='$_SESSION[user_id]'");
}
if (isset($_GET['submit'])) {

echo ("<SCRIPT LANGUAGE ='JavaScript'>
         window.alert ('Data are Sucessfully Updated!')
         window.location.href ='secondform.php'
         </SCRIPT>");
}
?>

<?php
mysql_close($connection);
?>

and i'm getting this error again, how to fix it?


    Notice: Undefined variable: id in C:\xampp\htdocs\marrybrown_clean\firstform.php on line 5

    Notice: Undefined index: user_id in C:\xampp\htdocs\marrybrown_clean\firstform.php on line 5

    Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\marrybrown_clean\firstform.php on line 6

also each my data in my form shows undefined variable like :

    <br /><b>Notice</b>:  Undefined variable: manager in <b>C:\xampp\htdocs\marrybrown_clean\firstform.php</b> on line <b>139</b><br />

i think my form is okay, but why is it showing error? if i take the id away, how to start my session by manager name only?

    This would be my form :

    <h2><font size ="5">Pre Evaluation Checklist</font></h2>
                        <form action="updatefirstform.php" method="post">
                            <div class="form_settings">
                                <p><input class ="input" type ="hidden" name="user_id" value ="<?php echo $id; ?>" /> </p>
                                <p><span>Manager on Duty  :</span><input type="manager" id="manager_name" name="manager_name" value="<?php echo $manager;?>" /></p>
                                <p><span>Evaluator's Name :</span><input type="evaluator_name" name="evaluator_name" value="<?php echo $evaluator;?>" /></p>
                                <p><span>Outlet Location  :</span><input type="outlet_location" name="outlet_location" value="<?php echo $outlet;?>"/></p>
                                <p><span>Date             :</span><input type="date" name="date" value="<?php echo $date;?>"  /></p>
                                <p><span>Day              :</span><input type="day" name="day" value="<?php echo $day;?>"  /></p>
                                <p><span>Time In          :</span><input type="timein" name="time_in" value="<?php echo $timein;?>" /></p>
                                <p><span>Time Out         :</span><input type="timeout" name="time_out" value="<?php echo $timeout;?>" /></p>
                                <table width="100%" height="79" border-spacing:0;>
                        <tr><th width="15%">Overall </th><th width="15%">Quality</th><th width="15%">Service </th><th width="15%"> Cleanliness </th></tr>
                        <tr><td><input type="overall" name="overall" value ="<?php echo $overall;?>"/></td><td><input type="overall" name="quality" value="<?php echo$quality;?>"/> </td><td><input type="service" name="service" value ="<?php echo $service;?>"/></td><td><input type="cleanliness" name="cleanliness" value="<?php echo $cleanliness;?>"/></td></tr>

                        </table>

Answer 2

You forgot , after setting the ID. And why $id instead of id in your update condition? What is the value of your $id by the way? It should look like this:

mysql_query("UPDATE firstform SET id ='$id',
                                  manager_name='$manager',
                                  evaluator_name ='$evaluator',
                                  outlet_location='$outlet',
                                  date='$date',
                                  day='$day',
                                  time_in ='$timein',
                                  time_out='$timeout',
                                  overall='$overall',
                                  quality='$quality',
                                  service='$service',
                                  cleanliness='$cleanliness'
                              WHERE id = '$_SESSION[manager_name]'");

Same with your SELECT statement:

$query = mysql_query("SELECT * FROM firstform WHERE id = '".$_SESSION['manager_name']."'");

Use also *_real_escape_string before binding a variable to your query.

$manager = mysql_real_escape_string($_POST['manager_name']);
//         ^^ do this for the rest of your passed-on variables

You should also consider using mysqli_* rather than the deprecated mysql_*.

Answer 3

At first, manager_name post data cannot from your update form, please check it.

After that use it

mysql_query("UPDATE firstform SET id ='$id', manager_name='$manager', evaluator_name ='$evaluator', outlet_location='$outlet', date='$date', day='$day', time_in ='$timein', time_out='$timeout', overall='$overall', quality='$quality', service='$service', cleanliness='$cleanliness' WHERE $id ='$_SESSION[manager_name]'");

Answer 4

mysql_query("SELECT * FROM firstform WHERE $id='".$_SESSION["manager_name"]."'");

Same problem in the Update