Variable from URL

Anon

Maybe this is obvious... but I seem to be missing the boat.

I want to call a url (like bio.php?id=1). Then on the next page I want to display data that matches hte id variable from the URL as the ID field on my data source.

I'm trying to do it with this code:

<body bgcolor="#FFFFFF" text="#000000">
<div align="center">
<?
mysql_connect("localhost","user","pass");
mysql_select_db("db");

$SQL = "SELECT * FROM employees WHERE id = $id";

$result = mysql_query($SQL);

?>
<table><tr>
<td><a href="bio.php?id=<? echo $row["ID"]; ?>"><? echo $row["first"]; ?></a></td>
<td><? echo $row["last"]; ?></td>
<td><? echo $row["extension"]; ?></td>
<td><? echo $row["tnumber"]; ?></td>
<td><a href=mailto:<? echo $row["email"]; ?>><? echo $row["email"]; ?></a></td>
<td><? echo $row["department"]; ?></td>
</tr></table>
</div>
</body>

Anon

This code, of course, hasn't been tested, but try making changes along these lines:

<body bgcolor="#FFFFFF" text="#000000">
<div align="center">
<?php

// change the following variables
$server = "localhost";
$user = "type_username_here";
$pass = "type_password_here";
$database = "type_database_here";

$connection = mysql_connect($server, $user, $pass);

mysql_select_db($database, $connection);

$SQL = "SELECT * FROM employees WHERE id = '$id'";

$result = mysql_query($SQL, $connection);

$row = mysql_fetch_array($result);

Anon

Still nothing... It couldn't be the way that I'm linking could it?? Is there something special I need to do on my code on the links page??

You can see what I'm trying to do at:

http://209.235.192.178/cuzone/staff/stafflist_email.php

Anon

I think this line has problems:

$SQL = "SELECT * FROM employees WHERE id = $id";

if ID is defined as a string, try:

$SQL = "SELECT * FROM employees WHERE id = '" . $id . "'";

if ID is defined as an int or long, etc:

$SQL = "SELECT * FROM employees WHERE id = " . $id;

Anon

The id is defined as an int. So I changed the $SQL line... But now I get this:

"Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource."

for the

$row = mysql_fetch_array($result);

line...

Ideas?

mrskhris

What version of PHP do you use? Anything above 4.2 has variable tracing turned off by default. If it's the case, you have to use $_GET["id"] to fetch your id variable from URL.

Anon

just for test purposes, try hardcoding it like this and see if you still get an error:

$SQL = "SELECT * FROM employees WHERE id = 1";

Be sure you have a record in your table with ID = 1.

Anon

I have PHP 4.2.2. So where do I use the $_GET["id"]??

Anon

here is some code that you can look at that may hold the answer for you. It looks like you need to fetch the values once you have comleted the query.

<?php
require("har_globals.php");
require("har_common.php");

$rc = FALSE;

// Connect to the database
if (!($connection=ConnectDB($hostName, $userName, $paswrd, $databaseName))) {
$rc = TRUE;}

// Get photos passed from main selection
if (!$rc) {
$sql = "SELECT photos.*, register.reg_firstname, register.reg_lastname FROM photos LEFT JOIN register ON ph_loginname = reg_loginname WHERE ph_filepath = '" . $id . "' AND ph_photoid > '' order by ph_filepath, ph_shorttitle;";
$sql_result=mysql_query($sql, $connection);}

// Check for system error
if (($rc==FALSE) && ($sql_result==0)) {
$errmsg = "Sorry, an error was encountered. The Webmaster has been notified.";
$message = "000_photothumb: Error selecting from database." . $newline . $sql . $newline;
$message .= sprintf("error:%d %s", mysql_errno($connection), mysql_error($connection));
NotifyWebmaster($message);
$rc = TRUE;}

if (!$rc) {
echo ("<TABLE BORDER=0 CELLSPACING=0 CELLPADDING=5 WIDTH=480><TR>");
$numpics=0;

  while ($row = mysql_fetch_array($sql_result)) {
     $filepath = "../assets/photos/" . $row["ph_filepath"] . "/" . $row["ph_thumbnail"] . ".jpg" ;
     $title = $row["ph_shorttitle"];
     $owner = "&#169; " . $row["reg_firstname"] . " " . $row["reg_lastname"];
     $filenm = $row["ph_filename"];
     $photoid = $row["ph_photoid"];
     $numpics++;

     echo ("<TD VALIGN=TOP WIDTH=154><P><TABLE WIDTH=154 BORDER=0 CELLSPACING=0 CELLPADDING=0><TR><TD>");
     echo ("<A HREF=\"../a1000/000_photolarge.php?filenm=$filenm&photoid=$photoid\"><IMG BORDER=0 SRC=\"$filepath\" ALT=\"$title\"></A>");
     echo ("</TD></TR></TABLE>");
     echo ("<B><FONT SIZE=-1>$title<BR></FONT></B>");
     echo ("<FONT SIZE=-2>$owner</FONT></TD>");

     if ($numpics==3) {
        echo ("</TR><TR><TD>&nbsp;</TD></TR><TR>");
        $numpics=0;
     }
  }
  echo ("</TR></TABLE>");

}

Anon

Hard coding it works. Maybe I do need to use the $_GET["id"]??

Anon

Take a look at this function in my prior post. I think you need to fetch the results after the query:

mysql_fetch_array($sql_result)

Anon

be sure to change $sql_result to $result for your coding.

mrskhris

Dan,

Did you even read her posting? shauna has PHP4.2.2 - does it tell you something?

mrskhris

Instead of $id. It's new security feature (which I really dislike, but that's how it is). Check this page:
http://www.php.net/manual/en/language.variables.predefined.php

Anon

I still seem to be having problems. Thanks so much for all your help (both of you).

Right now my code looks like:

...

$SQL = "SELECT * FROM employees WHERE id = "$_GET["id"]";

$result = mysql_query($SQL, $connection);

while ($row = mysql_fetch_array($result)) {

echo ("<td><? echo $row["first"]; ?></td>");

...

And I'm getting an error that says "parse error, unexpected T_VARIABLE."

Anon

Try putting a line at the beginning that says:

$id = $_GET["id"];

Then you can just use $id, like this:

$SQL = "SELECT * FROM employees WHERE id = '$id'";

You're getting the error because there are quotes within the quotes on this line:

$SQL = "SELECT * FROM employees WHERE id = "$_GET["id"]";

To do it that way, you would have to use single quotes within the string, like this:

$SQL = "SELECT * FROM employees WHERE id = '$_GET['id'];

$SQL = "SELECT * FROM employees WHERE id = '$_GET[id]';

Make sense?

Lacey

Anon

Yay!! That did it!!! Thank you thank you all for the help!!!!

I've learned a lot today!!

Anon

Good to hear!

Anon

It tells me something about you...

mrskhris

I'm sorry if it looked harsh. I understand you will to help, but you spilt so much information at once... and it was mostly useless and confusing.