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Storing Variables In Session()
Posted: Thu Mar 12, 2009 3:53 pm
by nishmgopal
Hi Guys please refer to my code below:
Code: Select all
$query = "SELECT * FROM Job_ID";
$result = mysql_query($query);
echo "<h1 class='title2'>Upcoming Project Roles</h1>
<p>From the menu below please select the Project Role:</p>
<form id='form2' method='post' action='new_go.php'>
<p>
<label>
<select name='job' id='job'>";
[color=#BF0000] $Job_ID = $row['Job_ID'];[/color] - [color=#40FF00]HOW CAN I STORE THIS IN SESSION?[/color]
$Project_Name= $row['Job_Name'];
echo "<option value=\"$Project_Name\" \"$Job_ID\">$Job_ID - $Project_Name</option>";
}
echo "</select></label></p>
<input type='submit' name='notjob' id='button' value='Go' />
</label>
</p>
</form>
";
Re: Storing Variables In Session()
Posted: Fri Mar 13, 2009 6:54 am
by nishmgopal
can anyone please help?
Re: Storing Variables In Session()
Posted: Fri Mar 13, 2009 7:24 am
by mattpointblank
Underneath it (before any output is displayed):
Code: Select all
session_start();
$_SESSION['Job_ID'] = $Job_ID;
You can now use this variable ($_SESSION['Job_ID']) on any page on your site, as long as you call session_start() beforehand. Really though, a quick search on Google or the PHP manual site would have told you this.
Re: Storing Variables In Session()
Posted: Fri Mar 13, 2009 8:01 am
by nishmgopal
I have already tried this and it didnt work. the top of my page where the variable is coming from looks like:
Code: Select all
<?php
session_start();
$_SESSION['Job_ID'] = $Job_ID;
?>
and on everyother page where I want to use this variable looks like this:
Code: Select all
<?php
session_start();
$_SESSION['Job_ID'];
?>
and to use this variable:
Code: Select all
<div class="title2"> <? echo $_SESSION['Job_ID'] ?> Specification:</div>
This doesnt work. What am i doing wrong?
Re: Storing Variables In Session()
Posted: Fri Mar 13, 2009 8:17 am
by nishmgopal
I have figured out the problem! Got it to work
But now I am having trouble callin the variable $Project_Name, this is what my code looks like:
Code: Select all
$Job_ID = $row['Job_ID'];
session_start();
$_SESSION['Job_ID'] = $Job_ID;
$Project_Name= $row['Job_Name'];
$_SESSION['Job_Name'] = $Project_Name;
I can call the $Job_ID by using $_SESSION['Job_ID'] = $Job_ID, but when I try to use $_SESSION['Job_Name'] = $Project_Name it doesnt work.
Re: Storing Variables In Session()
Posted: Fri Mar 13, 2009 8:38 am
by susrisha
post any errors that you are getting and add an echo before assigning to session to see whats the value that you are getting in $Project_Name
Re: Storing Variables In Session()
Posted: Fri Mar 13, 2009 8:50 am
by nishmgopal
I think I should explain the whole system again, as even my $Job_ID is not working as I would want.
I have page 1 where a drop down menu is generated by my sql statement, the code looks like:
Code: Select all
$query = "SELECT * FROM Job_ID";
$result = mysql_query($query);
echo "<h1 class='title2'>Upcoming Project Roles</h1>
<p>From the menu below please select the Project Role:</p>
<form id='form2' method='post' action='new_go.php'>
<p>
<label>
<select name='job' id='job'>";
while ($row = mysql_fetch_array($result,MYSQL_ASSOC)) {
$Job_ID = $row['Job_ID'];
session_start();
$_SESSION['Job_ID'] = $Job_ID;
$Project_Name= $row['Job_Name'];
echo "<option value=\"$Project_Name\" \"$Job_ID\">$Job_ID - $Project_Name</option>";
}
The on Submit, the action goes to page2, where I want to display all the records from the job table where the Job ID is passed from Page 1 ($Job_ID). At first I thought this was working, but when I select different records from the drop down menu and hit submit, the same value keeps showing. the code for page 2:
Code: Select all
<?php
session_start();
$_SESSION['Job_ID'];
?>
$query1 = "SELECT * FROM Job_ID WHERE Job_ID ='$_SESSION[Job_ID]'";
$result1 = mysql_query($query1)
or die ("Couldn't execute query.");
while($row = mysql_fetch_assoc($result1)){
$array[] = $row; }
array2table($array,200);
Re: Storing Variables In Session()
Posted: Fri Mar 13, 2009 12:06 pm
by nishmgopal
please help!