[SOLVED] Updating data in a database - Without Session

taz321 · January 2, 2008

Hi

As you can see from the previous post that my update was'nt working because i was using sessions, but this time i dont want to use a session, so below is my code for updating a record in the database which doesnt work, i get a error message saying

Unable to perform queryupdate employee set title ='', firstname ='', surname ='', username ='', password ='' where employeeID =

My code is -

<?php

require "connect.php";

$employeeID = $_GET['employeeID'];

$title = $_GET['title'];

$firstname = $_GET['firstname'];

$surname = $_GET['surname'];

$username = $_GET['username'];

$password = $_GET['password'];

$query = "update employee set title ='".$title."', firstname ='".$firstname."', surname ='".$surname."', username ='".$username."', password ='".$password."' where employeeID = ".$employeeID;

$result = mysql_query($query, $connection) or die ("Unable to perform query$query");

header("Location: displayEmployeeAdmin.php");

exit();

?>

Iv been doing for days but still not working.

Any help would be grateful

Thanks

revraz · January 2, 2008

Lets see the form that sends this info over.

taz321 · January 2, 2008

Its quite big

<?php
session_start();
if (isset($_SESSION['username']) == false){
	header("Location:Adminlogin.php");
	exit();
}
?>

<?php
    require "connect.php";
$employeeID = $_GET['employeeID'];
$query = "select * from employee where employeeID ='".$_GET['employeeID']."'";
$result = mysql_query($query, $connection) or die ("Unable to perform query<br>$query");
$row= mysql_fetch_array($result);
?>

<html>
<head>
<title>Jupiter Development Support</title>
<link href="Images/mystyle.css" rel="stylesheet" type="text/css"> 
<style type="text/css">
<!--
#Layer2 {position:absolute;
width:174px;
height:48px;
z-index:2;
left: 16px;
top: 143px;
}
#Layer1 {position:absolute;
width:200px;
height:76px;
z-index:1;
left: 254px;
top: 84px;
}
.style1 {font-family: "Copperplate Gothic Bold"}
.style3 {font-family: "Copperplate Gothic Bold"; font-size: x-large; }
#Layer3 {
position:absolute;
width:386px;
height:40px;
z-index:3;
left: 418px;
top: 226px;
}
#Layer10 {position:absolute;
width:200px;
height:115px;
z-index:10;
left: 804px;
top: 318px;
}
.style4 {	font-family: "Copperplate Gothic Bold";
font-size: medium;
font-weight: bold;
}
#Layer4 {
position:absolute;
width:155px;
height:42px;
z-index:11;
left: 59px;
top: 367px;
}
#Layer5 {
position:absolute;
width:157px;
height:41px;
z-index:12;
left: 58px;
top: 421px;
}
#Layer6 {
position:absolute;
width:159px;
height:41px;
z-index:13;
left: 58px;
top: 311px;
}
-->
</style>
</head>

<body>
<div id="header">
<h1 align="center"> </h1>
<h1 align="center"> </h1>
<h1 align="center"> </h1>
<div id="Layer1"><img src="Images/JDS.jpg" width="704" height="79"></div>
<h1 align="center"> </h1>
<h1 align="center"> </h1>
<div id="Layer2"><img src="Images/logo.jpg" width="173" height="43"></div>
<h1 align="center"> </h1>

<form action="EditUser.php" method="get">
<table width="1225" border="0">
  <tr>
    <td width="286" height="39"> </td>
    <td width="137"> </td>
    <td width="172"> </td>
    <td width="117"> </td>
    <td width="491"> </td>
  </tr>
  <tr>
    <td height="39"> </td>
    <td> </td>
    <td> </td>
    <td> </td>
    <td> </td>
  </tr>
  <tr>
    <td height="40"><div id="Layer5"><a href="deletepersonScreen.php"><img src="Images/deleteUser.jpg" width="158" height="41" border="0"></a></div>
      <div id="Layer6"><a href="AdminMainPage.php"><img src="Images/AdminHomepagebutton.jpg" width="158" height="42" border="0"></a></div></td>
    <td height="40"><span class="style1">Employee ID </span></td>

    <td><input name="title2" type="text" readonly="true" value="<?php echo $row['employeeID'];?>"></td>
    <td> </td>
    <td><div id="Layer10"><img src="Images/Symbol.jpg" width="235" height="231"></div></td>
  </tr>
  <tr>
    <td height="40"> </td>
    <td height="40"><span class="style1">Title</span></td>
    <td><input name="title" type="text" value="<?php echo $row['title'];?>"></td>
    <td><div id="Layer3"><span class="style3">Update User Details</span></div></td>
    <td> </td>
  </tr>
  <tr>
    <td height="40"> </td>
    <td height="40"><span class="style1">FirstName</span></td>
    <td><input name="firstname" type="text" value="<?php echo $row['firstname'];?>"></td>
    <td> </td>
    <td> </td>
  </tr>
  <tr>
    <td height="40"> </td>
    <td height="40"><span class="style1">SurName</span></td>
    <td><input name="surname" type="text" value="<?php echo $row['surname'];?>"></td>
    <td> </td>
    <td> </td>
  </tr>
  <tr>
    <td height="40"><div id="Layer4"><a href="CreateUser.php"><img src="Images/createUserButton.jpg" width="158" height="42" border="0"></a></div></td>
    <td height="40"><span class="style1">UserName</span></td>
    <td><input name="username" type="text" value="<?php echo $row['username'];?>"></td>
    <td> </td>
    <td> </td>
  </tr>
  <tr>
    <td height="40"> </td>
    <td height="40"><span class="style1">Password</span></td>
    <td><input name="password" type="text" value="<?php echo $row['password'];?>"></td>
    <td>
      <label></label>
      <div align="center">
        <input name="Save" type="submit" class="style4" id="Update" value="Update">
      </div>       </td>
    <td> </td>
  </tr>
  <tr>
    <td height="52"> </td>
    <td> </td>
    <td> </td>
    <td><label>
    </label></td>
    <td> </td>
  </tr>
  <tr>
    <td height="52"> </td>
    <td> </td>
    <td> </td>
    <td> </td>
    <td> </td>
  </tr>
</table>

<p align="center"> </p>
</body>
</form>
</html>

wildteen88 · January 2, 2008

To find out why you query isn't running add mysql_error to your or die clause:

$result = mysql_query($query, $connection) or die ("Unable to perform query: $query<p>".mysql_error());

I have a feeling that it is because you are have column named as a reserved word (password). Rerun your code again and it should output an error from mysql why your query is failing

taz321 · January 2, 2008

i have re run the system and the following error message is displayed

Unable to perform query: update employee set title ='', firstname ='', surname ='', username ='', password ='' where employeeID =

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 '' at line 1

taz321 · January 2, 2008

have column named as a reserved word (password). Rerun your code again and it should output an error from mysql why your query is failing

I dont quite understand the above comment, please elaborate.

revraz · January 2, 2008

You set employeeid to title2 here

<td height="40"><span class="style1">Employee ID </span></td>

   <td><input name="title2" type="text" readonly="true" value="<?php echo $row['employeeID'];?>"></td>

But you assign $employeeID here to

$employeeID = $_GET['employeeID'];

I would double check all your variable names as well.

What wildteen is saying is that your column name Password should be changed.

taz321 · January 2, 2008

Ahh sorry, stupid me, i have now amended the 'title2' :-S

When you say change the column name, do you mean -

<td height="40"><span class="style1">Password</span></td>

Change the "input name" to something else ?

Sorry it may be a dumb question.

revraz · January 2, 2008

No, in MySQL, change the fieldname.

But your error appears to come before that, verify your tablename is in fact employee

taz321 · January 2, 2008

Thanks it appears to have worked :-)

But im abit confused about something, i have two PHP pages -

1) The actual SQL thats runs the form

2) The form itself

Now when i display the form in a browser, it all looks fine and the update works fine.

But when i open the actual SQL script, it shows me an error message saying

Unable to perform query: update employee set title ='', firstname ='', surname ='', username ='', password ='' where employeeID =

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 '' at line 1

I know the SQL won't display in the browser but as the SQL is now working for me, why is it still showing me an error message ?

redarrow · January 2, 2008

I guess session id not working echo employee session............

revraz · January 2, 2008

This error is coming from somewhere else. Look at this line:

$result = mysql_query($query, $connection) or die ("Unable to perform query$query");

Notice your error, Unable to perform query$query. This would produce an error like

Unable to perform queryupdate employee set title blah blah

But your error is

Unable to perform query: update employee set title blah blah

Notice the colon and space? This is coming from a error like:

Unable to perform query: $query

So I don't think we're looking at the right page somewhere.

taz321 · January 2, 2008

well this is the update script

<?php

require "connect.php";

$employeeID = $_GET['employeeID'];

$title = $_GET['title'];

$firstname = $_GET['firstname'];

$surname = $_GET['surname'];

$username = $_GET['username'];

$password1 = $_GET['password'];

$query = "update employee set title ='".$title."', firstname ='".$firstname."', surname ='".$surname."', username ='".$username."', password ='".$password1."' where employeeID = ".$employeeID;

$result = mysql_query($query, $connection) or die ("Unable to perform query: $query<p>".mysql_error());

header("Location: displayEmployeeAdmin.php");

exit();

?>

And this is the SQL in the form

<?php

require "connect.php";

$employeeID = $_GET['employeeID'];

$query = "select * from employee where employeeID = ".$employeeID;

$result = mysql_query($query, $connection) or die ("Unable to perform query: $query<p>".mysql_error());

$row= mysql_fetch_array($result);

?>

Both the $results in the two scripts are te same so, im not quite understanding what you mean by we might not be looking at the right page.

revraz · January 2, 2008

Ok, then you changed the error reporting since the first post.

It really looks like it's running twice for some reason. The first time it works fine, but then it runs again and all your variables are empty.

taz321 · January 2, 2008

Because its showing an error message, would this affect my system being able to update data in the employee table, as it seems to be working fine at the momment ?

revraz · January 2, 2008

It would be in your best interest to find out why and fix it.

taz321 · January 2, 2008

Hmmm i wouldnt know where to begin, as it the code is for the two pages

The actual script

AND the form

Are you saying there could be an error outside of these two pages ?

revraz · January 2, 2008

I would start adding echo's in the current scripts so you can see what's going on and when. Then remove them after you see where it's coming from.

taz321 · January 2, 2008

Sorry may be a dumb question.

I know what echo's do, but where would i start adding echo's and not quite what u mean there.

taz321 · January 2, 2008

Thanks for your help guys.

Thanks to you i can rest in peice now :-)

revraz · January 2, 2008

So what was it?

taz321 · January 4, 2008

I didnt resolve the bit when you asked me to echo it, but the update works and the error message remains as im not sure what to actually echo.

redarrow · January 4, 2008

Ok mate i added the print_r function to print your current query, i also change the query names and they where both the same, remember if query names are the same on the same php page there might be unexspected problams always have seprate query names ....

if print_r dosent work replace with echo,

good luck..........

<?php
   require "connect.php";
   $employeeID = $_GET['employeeID'];
   $title = $_GET['title'];
   $firstname = $_GET['firstname'];
   $surname = $_GET['surname'];
   $username = $_GET['username'];
   $password1 = $_GET['password'];
   $query1 = "update employee set title ='".$title."', firstname ='".$firstname."', surname ='".$surname."', username ='".$username."',  password ='".$password1."' where   employeeID = ".$employeeID;

PRINT_R($query1);// debug mode. 

  $result1 = mysql_query($query1, $connection) or die ("Unable to perform query: $query1<p>".mysql_error());
   header("Location: displayEmployeeAdmin.php");
   exit();
?>


And this is the SQL in the form

<?php
    require "connect.php";
   $employeeID = $_GET['employeeID'];
   $query2 = "select * from employee where employeeID = ".$employeeID;

PRINT_R($query2);//debug mode.
   
$result2 = mysql_query($query2, $connection) or die ("Unable to perform query: $query2<p>".mysql_error());
   $row= mysql_fetch_array($result2);
?>

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[SOLVED] Updating data in a database - Without Session

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