Variable not passed

kamy99

This problem is so easy , I feel embarassed to post it, but since I hit a brick wall here it oes:

Simple idea, list items, click on an item go to details.

Code to list items:

<?
db_connect();
$getInfo=mysql_query("SELECT * FROM orders_txt ");
while ($i=mysql_fetch_array($getInfo))
{
?>
<a href="order_detail.php?Orders=<?echo $i[0]; ?>"><?echo $i[4]; ?></a>

<?}?>

It works, products get lited correctly.

Code to display details; (order-detail.php):

<?
db_connect();

$showInfo=mysql_query("SELECT * FROM orders_txt WHERE OrderID= '$Orders' ");

while ( $i=mysql_fetch_array($showInfo) )
{ ?>
 <?echo $i[4]?> 
<?echo $i[3]?>

<?}?>

I get a blank page!

I substituted OrderID= '$Orders' with OrderID= '3' (and other numbers) and it DOES work!!!!!!

Please help me..................

bernouli

use $HTTP_GET_VARS['var'] or $HTP_POST_VARS['vars']
depends on which method do you use while sending the form.
If you use get then use GET_VARS.

<?
db_connect();

$showInfo=mysql_query("SELECT * FROM orders_txt WHERE OrderID= '$HTTP_POST_VARS['Orders']' ");

kamy99

But I am not using a form . I didn't think I had to.

bernouli

when you dont use a form, where do you get the var?
i dont see any var with name Order in the codes

kamy99

I pass the variable through URl

<a href="order_detail.php?Orders=<?echo $i[0]; ?>"><?echo $i[4]; ?></a> 🙂

I'm looking @ my other codes and they're all the same and all work fine... 🙁
I don't know what's wrong with this one.

bernouli

I pass the variable through URl

<a href="order_detail.php?Orders=<?echo $i[0]; ?>"><?echo $i[4]; ?></a>

aha

but the file ist simple not the same one 😃

Code to display details; (order-detail.php):

kamy99

actually the files are named correctly.
The orders-detail.php was a typo on my end!

bernouli

are you sure, there is some value in Orders?

try to echo $Orders out in order_detail.php , and see what is it

Weedpacket

Like bernouli suggests, use $HTTP_GET_VARS['Orders']. That will contain whatever value was given to Orders in the querystring. ($HTTP_GET_VARS is so-called because they're variables supplied in an http GET request, which is what the browser sends to the server when you follow a link).

prasit

Originally posted by kamy99

while ( $i=mysql_fetch_array($showInfo) )
{ ?>
 <?echo $i[4]?> 
<?echo $i[3]?>

<?}?>
Please help me..................

it's my first day trying to learn PHP, so i visit this forum. However I think your need to put

<?echo $i[4]; ?>

i mean semi-colon after $i[4] and $i[3].

Is it the correct answer?

prasit

sorry, i was wrong.

the correct answer would be $HTTP_GET_VARS['Orders'].

it seems like you can just use $Orders in the ealier PHP version, but for the new PHP you must use $HTTP_GET_VARS['Orders']..

prasit

i got a PHP application and I have to convert the all cgi variables 🙁

Is there a way to use just $Orders instead of $HTTP_GET_VARS['Orders']. Otherwise I have to modify the whole application codes and it's a big job.

thank you.

intenz

Try...

<? 
db_connect();
$sql = mysql_query("SELECT * FROM orders_txt WHERE OrderID = '$_GET[Orders]'") or die ('Sql prob: ' . $sql . '<p>' . mysql_errno() . ': ' . mysql_error());
while($i = mysql_fetch_array($sql))
{
?>
     <b><?=$i[3]?></b><br>
     <b><?=$i[4]?></b><br>

<?
} 
?>

citfor

At this point it sounds like it's time for old school debugging. Echo out each variable after a procedure. If that doesn't display the problem, try copying the SQL statements generated and entering them into the DB SQL interpreter.

One of the very few down sides to PHP (Please no flaming me) is that it is so fault tolerant. You may have had the same error in a dozen rewrites, but he DB isn't displaying the error to you. Hope this helps. It always does when I hit the wall.

~Living below, looking above.

adamgeorge

Hi,

firstly, with your sql statement, if all you require is the orderID and orderName, you should only select them in your query:

$getInfo=mysql_query("SELECT OrderID, orderName FROM orders_txt ");

Next, you should check that it's actually printing out the orderID:

<? while ($i=mysql_fetch_array($getInfo)) 
{ 
?> 
<a href="order_detail.php?Orders=<?echo $i['OrderID']; ?>">
<?echo $i['orderName'] . $i['OrderID']; ?></a><br> 
<?}?>

Ok, if that's all good, then you need to check your query on the next page. Try printing it out to see if the $Orders var got passed correctly. (Again, you should only select the fields you require from the database. It will make your code easier to read and debug - I've just put in a fake field here. Use whichever ones you meant to use.)

<? 
db_connect(); 
$sql = "SELECT orderDate, orderName FROM orders_txt WHERE OrderID= '".$Orders."' ";

echo $sql;

$showInfo=mysql_query($sql); 

while ( $i=mysql_fetch_array($showInfo) ) 
{ ?> 
<b> <?echo $i['orderName']?><br> 
<?echo $i['orderDate']?></b> 

<?}?>

One thing you should be carefull with in php is enclosing variables in single quotes ' ' . they can be taken litterally instead of as variables. So, that's why I split the string up when I do that.

ok. it should print out the query. check that the query is correct., and that $Orders has been passed. It should have been. Copy and paste the query into a mySql query engine or whatever you have and check that the query actually works if you have to. If the query is correct, and the $Orders var is getting passed, then I have no idea what is going on.

That's all I can think of! This is just the basic debugging process you should go through when you get stuck like this.

hope it helps.
-Adam 🙂

adamgeorge

Also, I might like to add, how come your orderID is a string? or isn't it?

I assume you would use some sort of unique number for storing an ID, but in your select you are enclosing it in quotes... I don't know if this has anything to do with you not getting any results, but it would be worth having a look at.

eg, you wrote:

$showInfo=mysql_query("SELECT * FROM orders_txt WHERE OrderID= '$Orders' ");

but you should probably write this, if OrderID is a integer:

$sql = "SELECT * FROM orders_txt WHERE OrderID= $Orders";
$showInfo=mysql_query($sql);

Like I say... don't know if this is an issue or not... prolly mysql would have chucked a wobbly at you if it was - but it may be worth changing if nothing else works, and just because it's not very good coding practice to enclose your numbers in quotes.

cya
-Adam 🙂

adamgeorge

Oh.. I guess you could be using a GUID (Globally Unique Identifier) though. Er... my mistake. If you are you should rename the field so that it reflects that. It can be confusing.

Although I didn't think mySql had GUID's ....

anyways... I'll be quite now!

cya
-Adam 🙂

prasit

Originally posted by adamgeorge

but you should probably write this, if OrderID is a integer:
$sql = "SELECT * FROM orders_txt WHERE OrderID= $Orders";
$showInfo=mysql_query($sql);
Like I say... don't know if this is an issue or not... prolly mysql
-Adam 🙂 [/B]

mysql is find with or without quote for an integer, but you need quote for characters