carrying $ID to detail page

mondragon

I have a publications database that lists the search results on one page and then I would like the user to be able to click on a specific record that will then take them to a details page for that record.

On the results page I have the follwoing code:
echo "<a href = details.php?ID=$ID>Report Number</a>";

On the details page I have:
@ $db = mysql_pconnect("servername", "username", "password" );

if (!$db)
{
echo "Error: Could not connect to the database. Please try again later.";
exit;
}

mysql_select_db("dbname");
$query = "SELECT * FROM tablename WHERE ID=$ID";
$result=mysql_query($query);

while ($row=mysql_fetch_array($result))
{
$ID=$row['ID'];
$rptid=$row['rptid'];
$title=$row['title'];

echo $ID;
echo $rptid;
echo $title;

}

mysql_close();

I have also tried:
$query="SELECT * FROM tablename WHERE ID=\"$ID\"";
and a couple other derivations.

Whatever I try, I still can't get the details page to display the information. If I replace
$query = "SELECT FROM tablename WHERE ID=$ID";
with
$query = "SELECT FROM tablename WHERE ID=2";

The details page will display the information for the record with the primary key of 2. So my guess is that the details page is not be handed the $ID from the results page.

I am obviously missing something (other than the appropriate amount of brain cells to figure this out). Anybody have any ideas?? Thanks!!

Anon

Before you even connect to the database, try echoing the $ID variable and see what you get. This will tell you if the variable is being carried over and what it is set to.

mondragon

I placed the echo statement as shown below and it did not print. Any idea why it's not getting passed? Also I am getting an error statement on the "while" line of code. Is this because no information is being passed or do I have other issues?

Thanks for your help!!

echo $ID;
@ $db = mysql_pconnect("servername", "username", "password" );

if (!$db)
{
echo "Error: Could not connect to the database. Please try again later.";
exit;
}

mysql_select_db("databasename");
$query = "SELECT * FROM tablename WHERE ID=$ID";
$result=mysql_query($query);

while ($row=mysql_fetch_array($result))
{
$ID=$row['ID'];
$rptid=$row['rptid'];
$title=$row['title'];



}


echo "$ID";
if($rptid !=""){
echo "$rptid";
}

Anon

Stupid question: When you click on the link, does the URL look like this?
http://whatever.com/my_page.php?ID=2

(I'm just wanting to be sure that it doesn't evaluates $ID.)

Secondly, instead of:
echo $ID;
Try this:
echo $HTTP_GET_VARS["ID"];

See if it gives you a value.

mondragon

When you place your cursor on the link the URL shows:
http://whatever.com/my_page.php?ID=

It does not have a value for the ID. Should there be?

I also tried your suggestion of using
echo $HTTP_GET_VARS["id'];
I incorporated this as shown below.

Am I correct in believing that the $ID is not being passed to the details page or is there another problem. Also, I am still getting an error message in the browser window that points to the "while statement"...is this because the $ID is not being passed??

Thanks again for everyones help!

echo $HTTP_GET_VARS["ID"];
@ $db = mysql_pconnect("servername", "username", "password" );

if (!$db)
{
echo "Error: Could not connect to the database. Please try again later.";
exit;
}

mysql_select_db("dbname");
$query = "SELECT * FROM tablename WHERE ID=$ID";
$result=mysql_query($query);

while ($row=mysql_fetch_array($result))
{
$ID=$row['ID'];
$rptid=$row['rptid'];
$title=$row['title'];



}


echo "$ID";
if($rptid !=""){
echo "$rptid";
}

Anon

Looks to me like your link is not being created correctly. How are you getting the $ID variable ?

Anon

Yes, there should be a value in the URL. Right now, your URL is telling PHP that ID="".

Also, the while is giving you errors because $result is invalid, because the query is invalid, because $ID="". Make sense?

Two suggestions:
#1 - Manually add a valid id there. See if your script works as expected. That is:
http://whatever.com/my_page.php?ID=2
Based on what you've already said, that should work.

#2 - The URL problem is probably in the code that you use to generate the link URL. Take a close look at that. The bug is in there somewhere.

mondragon

David,

I'm not exactly sure what you mean by getting the $ID variable but hopefully the following answers your question. I am attempting to pass the $ID variable in the link (as shown below), however, I am not sure if I have passed it correctly to the details page. I have tried echoing the $ID in several places on the 2 pages and it doesn't seem to get passed to the details page (also shown below).

Thanks for your help!

//Results page code:
$row = mysql_fetch_array($result);
echo ($row["ID"]);
echo "<br>";
//the code above results in the ID number being displayed on the results page & makes me assume the ID is getting this far
echo "<a href = details.php?ID=$ID>Report Number</a>";
echo htmlspecialchars ( stripslashes($row["rptid"]));

//Details page

echo $HTTP_GET_VARS["ID"];
//tried to get it to display $ID with above statement but it didn't show the ID
@ $db = mysql_pconnect("servername", "username", "password" );

if (!$db)
{
echo "Error: Could not connect to the database. Please try again later.";
exit;
}

mysql_select_db("dbname");

$query="SELECT * FROM tablename WHERE ID=$ID";
//in the above query, if I change $ID to a number...say '4' a result will display
$result=mysql_query($query);

while ($row=mysql_fetch_array($result))
{
$ID=$row['ID'];
$rptid=$row['rptid'];
$title=$row['title'];



}


echo "$ID";
if($rptid !=""){
echo "$rptid";
}

Any suggestions/ideas???

Thanks!

mondragon

You are correct!! When I insert a value into the link (e.g. change http://whatever.co/my_page.php?ID=
to
http://whatever.co/my_page.php?ID=2
the details page works fine.

I am attempting to pass the variable from the results page to the details page in the following manner:
echo "<a href = details.php?ID=$ID>Report Number</a>";

Any suggestions on what's missing/wrong??

Thanks!!

noeffred

try this, you're missing some '' around the filename and don't use space between html tags and their values, this confuses most browsers.

echo "<a href='details.php?ID=$ID'>Report Number</a>";

hope this helps!

mondragon

Thanks but still no dice!! The URL is still lacking the ID value when the cursor is on it (e.g. http://whatever.com/details.php?ID= ) instead of http://whatever.com/details.php?ID=2 (or some other number).

Any ideas on another way to pass the $ID value or ways to overcome this problem that has plagued me for the past 2 days?

Thanks again!

steffen

Try breaking up the echo with concatenation:

echo "<a href=details.php?ID=".$ID.">Report Number</a>";

Anon

Jeff,

For the link to the details page, try this:

echo "<a href = details.php?ID=$row["ID"]>Report Number</a>";

That should work.

James

mondragon

I tried breaking it apart as suggested:
echo "<a href=details.php?ID=".$ID.">Report Number</a>";

Still no luck...

I also tried
echo "<a href='details.php?ID=$ID'>Report Number</a>";

Is there another way to pass the variable or does anybody have any more suggestions on what's wrong with the current method I'm using??

Thanks!!

Anon

Change
echo "<a href = details.php?ID=$ID>Report Number</a>";

to
echo "<a href='details.php?ID=$ID'>Report Number</a>";

You are calling $ID twice, once from the URL and once in the while loop. Try renaming $ID to something else eg $newID in the while loop and see if that works.

while ($row=mysql_fetch_array($result))
{
$newID=$row['ID'];
$rptid=$row['rptid'];
$title=$row['title'];

echo $newID;
echo $rptid;
echo $title;

}
mysql_close();

Good luck

Anon

Where is the variable $ID supposed to get its value from on the page where the link is generated? Obviously, that's the bit that's not working, because $ID's not getting its intended value.

What does it say round about where that page's script says "$ID=..."? The error may be around there. Failing that, it may be in the vicinity of wherever the code that gives $ID its value gets its value.

Hunt the problem down by working backwards.

mondragon

This problem has been plaguing me for 2 very long days and I have tried a variety of techniques to pass this variable. I am VERY open to other suggestions...If you passed the variable like this:
echo "<a href='details.php?ID=$ID'>Report Number</a>";

how would you get its value on the details page?

I believe that there is no variable being passed because the link shows:
http://mypage.com/details.php?ID=

Thus it appears to me that nothing is being sent to the details page.

PLEASE let me know if you have any ideas. Thanks!!

Where is the variable $ID supposed to get its value from on the page
where the link is generated? Obviously, that's the bit that's not
working, because $ID's not getting its intended value.

                  What does it say round about where that page's script says "$ID=..."?
                  The error may be around there. Failing that, it may be in the vicinity
                  of wherever the code that gives $ID its value gets _its_ value. 

                  Hunt the problem down by working backwards.

Anon

I guess that is the only thing left. Can you include the code from the original page that populates the $ID variable? If so, I would bet that it is a problem with assigning a value to the variable.

stevenrhiggins

It seems to me that you have not name ID in your results page:

you should include

$ID=$row["ID"] prior to the link.

mondragon

YIPEE!! You are right!! That was exactly the problem! I needed to include $ID=$row["ID"] prior to the link on the results page.

THANK YOU VERY MUCH!!