Using Session()

[nishmgopal]

Hi guys

 

I am trying to pass a variable across pages and I believe using sessions is probably the best way.

 

The code on the page where the variable is based is shown below.  the variable 'job' is from a form.

 

<?php session_start();
  
$_SESSION['job'] = $_POST['job'];
  
  ?>

 

to read the variable on other pages I am using:

 

<?php session_start();
   
    $getjob = $_SESSION['job'];
   
     ?>

 

And I am trying to use it in the following SQL query:

 

$query1 = "SELECT * FROM Job_Spec WHERE Project_Name ='" . mysql_real_escape_string($_POST['job']) . "'";

 

When I execute this query I get a blank screen with no errors!

 

Please help

[JonnoTheDev]

You are using the POST array and not the SESSION

$query1 = "SELECT * FROM Job_Spec WHERE Project_Name ='" . mysql_real_escape_string($_POST['job']) . "'";

 

$query1 = "SELECT * FROM Job_Spec WHERE Project_Name ='" . mysql_real_escape_string($_SESSION['job']) . "'";

[nishmgopal]

I have changed it to what you suggested but again, it just returns a blank screen. No errors.

 

[JonnoTheDev]

print the query to the screen to make sure it looks ok. Die if the query produces an error:

 

// does the query look ok - if ok then comment out and run the query below
print $query1;

// run the query
$result = mysql_query($query1) or die(mysql_error());

 

[nishmgopal]

I ran this query:

 

<?php 
session_start(); 

$getjob = $_SESSION['job'];

?>
<?php
    
$host="co-project";
$user="conmg";
$password="********";

$conn = mysql_connect($host, $user, $password) or die ('Error connecting to mysql');
$dbname="conmg";

$dbname=@mysql_select_db($dbname, $conn)
       or die("Couldn't select database");
	   
//First Table Queries


$query1 = "SELECT * FROM Job_Spec WHERE Project_Name ='" . mysql_real_escape_string($_SESSION['job']) . "'";
// does the query look ok - if ok then comment out and run the query below
print $query1;

// run the query
//$result = mysql_query($query1) or die(mysql_error());


?>

 

And I got a blank screen.  Then I ran this:

 

<?php 
session_start(); 

$getjob = $_SESSION['job'];

?>
<?php
    
$host="co-project";
$user="conmg";
$password="rey38pgb";

$conn = mysql_connect($host, $user, $password) or die ('Error connecting to mysql');
$dbname="conmg";

$dbname=@mysql_select_db($dbname, $conn)
       or die("Couldn't select database");
	   
//First Table Queries


$query1 = "SELECT * FROM Job_Spec WHERE Project_Name ='" . mysql_real_escape_string($_SESSION['job']) . "'";
// does the query look ok - if ok then comment out and run the query below
//print $query1;

// run the query
$result = mysql_query($query1) or die(mysql_error());


?>

 

And again I got a blank screen

[JonnoTheDev]

Firstly you do not need to open and close php tags like you are.

?>
<?php

 

The fact that you are getting a blank screen suggests there is an error but you have error reporting disabled. You can change this in your php ini file or add the following to the top of your your script.

I would suggest placing this in a common include (included on all pages of your site):

 

ini_set('display_errors', 1); 
ini_set('error_reporting', E_ALL & ~E_NOTICE); 

 

Also the @ on your functions supresses errors. You should get rid of this.

@mysql_select_db($dbname, $conn)

mysql_select_db($dbname, $conn)

 

 

 

[nishmgopal]

But all my other errors are shown, whenever I've had them.

[JonnoTheDev]

Does this print the query to the screen? You said you got a blank screen.

print $query1;

 

If you get nothing issuing this command then there must be a script error. If you do see the query then all is OK. I cannot see what you are expecting to be on the screen as your code is cut off.

 

 

[nishmgopal]

When i print the query, its prints to the screen.

 

the full code is:

 

<? 
session_start(); 

$getjob = $_SESSION['job'];

?>
$conn = mysql_connect($host, $user, $password) or die ('Error connecting to mysql');
$dbname="conmg";
$dbname=@mysql_select_db($dbname, $conn)
       or die("Couldn't select database");

//////


$query1 = "SELECT * FROM Job_Spec WHERE Project_Name ='$getjob'";


  $result1 = mysql_query($query1)
       or die ("Couldn't execute query.");

$colcnt = 0;


   while ($row = mysql_fetch_array($result1,MYSQL_ASSOC)) {
  {
     if (!$colcnt)
     {     
          $colcnt = 7;    
    }

   $colcnt--;{
   
      $Name1 = $row['Project_Name'];
      $Java1 = $row['Java'];
  $Leadership1 = $row['Leadership'];
  $Communication1 = $row['Communication'];
  $Teamwork1 = $row ['Team_Work'];
  $Problem_Solving1 = $row['Problem_Solving'];
  }
}

echo 
   "$Java1";

}
}
?>

[JonnoTheDev]

This is bad syntax. You braces have no apparent function.

$colcnt--;{

 

Also

echo "$Java1";

Does not need to be contained with "

echo $Java1

 

Check your code carefully. Is your query returning the correct results?

 

 

[nishmgopal]

yes my query is returning proper results when i use a row name such as 'Project_Name='Manager' but if I try to use the variable $getjob, it returns a blank screen!

[JonnoTheDev]

Do you see the correct query with the correct job id within the query when you print to the screen?

print $query1;

[nishmgopal]

When I print the query:

 

$query1 = "SELECT * FROM Job_Spec WHERE Project_Name ='Manager'";

 

I get the correct results

 

When I print the query:

 

$query1 = "SELECT * FROM Job_Spec WHERE Project_Name ='$getjob'";

 

SELECT * FROM Job_Spec WHERE Project_Name =''

 

 

[JonnoTheDev]

there is your problem. your value is not set in the session. check your code that sets this value.