MySQL and PHP... errors..

I got this error message:
"Warning: 0 is not a MySQL result index"

I use the following code:

$connection = mysql_connect("localhost", "username", "password");
mysql_select_db("qdesign", $connection);


$query = "SELECT * FROM guestbook WHERE id=$display ";
$result = mysql_query($query, $connection);


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

print("<b>n");
print($row["subject"] . "<br>n");
print("</b>n");
print($row["name"] . "<br>n");
print($row["email"] . "<br>n");
print($row["url"] . "<br><br>n");
print("<table border=0><tr><td width=300>nn");
print($row["msg"] . "<br><br><br>n");
print("</td></tr></table>nn");
print("<a href="guestbook.php3?what=view">Go back to the guestbook</a>");

}


what's wrong...

When I enter the page I have a query with me.
Like this

www.server.com/page.php3?display=1

I want to view the same id in the database
as in the display variable.

Thanks.

After your query add the following line:

print mysql_error();

to see what's wrong. The 0 result index means that mysql returned an error.

It's not readily apparent from your code what the error might be.

test . . .please ignore

you query is flawed

$query = "SELECT * FROM myTable WHERE myField = 'ThisItem'";
ThisItem must be enclosed in quotes...

- Regards

The query is not flawed if id is a column which uses a numeric data type (*int, which is probably what it is). Only strings must be enclosed in quotes.

I'm having trouble with these lines:
$dept_name = mysql_query ("SELECT department_name FROM dept_info WHERE dept_id = '$row["dept_id"]'");
print $dept_name;

It seems to be printing the array index, and not a result from the mysql_query. The Query works if I type it in the command line with an explicit value substituted for the $row["dept_id"]

This line :
print $row["dept_id"];
is printing the correct value to HTML.

Here's the code:

$result = mysql_query ("SELECT * FROM jobs");

print ("<table>");
if ($row = mysql_fetch_array($result)) {
do {
print ("<tr><td bgcolor=#cccccc>");
print $row["position_name"];
print ("</td><td>");
print ("<b>");
print $row["dept_id"];
print ("</b>");
$dept_name = mysql_query ("SELECT department_name FROM dept_info WHERE dept_id = '$row["dept_id"]'");
print $dept_name;
print ("</td></tr>"); }
while($row = mysql_fetch_array($result)); }
else {print "Sorry, no records were found!";}

print ("</table>");



[This message has been edited by brett_webb (edited February 02, 2000).]

Hi,

There are 'conflicting' double-quotes in the string, you can try:

$selectquery = "SELECT department_name FROM dept_info WHERE dept_id = " . $row["dept_id"];
$dept_name = mysql_query($selectquery);

Peter