Multiple queries in one script

[Bob Sanderson]

I am relatively new to PHP and MySQL. This is the first time I've tried
to use multiple queries in a single script.

I have the following PHP script which gets a Job Number from a search
form and generates a web page which displays the record for that job:

$username="root";
$password="";
$database="foobar";

$Searchterm = $_GET['JobNumber'];

mysql_connect('localhost',$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

$query="select * from jobs where JobNumber like $Searchterm";

$result=mysql_query($query);
$num_results=mysql_num_rows($result);
$JobNumber=mysql_result($result,$i,"JobNumber");
$NSN=mysql_result($result,$i,"NSN");

echo various fields here

This part works fine.

----------------------------------

I now want to list other jobs which use the same NSN on the same page
under that display. This is the script I'm using:

$query2="select * from jobs where NSN like $NSN";
$result=mysql_query($query2);
$num_results=mysql_num_rows($result);
$JobNumber=mysql_result($result,$i,"JobNumber");

echo query2; (returns the correct query, including the NSN)
echo $num_results (returns nothing)

$i=0;
while ($i < $num) {

echo various fields here (returns nothing)

$i++;
}

Can anyone tell me what I'm missing?

[Erwin Moller]

Bob Sanderson wrote:

Hi Bob,

I am relatively new to PHP and MySQL. This is the first time I've tried
to use multiple queries in a single script.

I have the following PHP script which gets a Job Number from a search
form and generates a web page which displays the record for that job:

$username="root";
$password="";
$database="foobar";

$Searchterm = $_GET['JobNumber'];

mysql_connect('localhost',$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

Ok so far.
$query="select * from jobs where JobNumber like $Searchterm";
This is very dangerous.
NEVER EVER thrust input originating from a form that is filled in by some
user.
You are wide open to the SQL-Injection attack this way.

If you have magic_quotes on, you are a lot safer, but please be sure what
you are doing...

$result=mysql_query($query);
$num_results=mysql_num_rows($result);
$JobNumber=mysql_result($result,$i,"JobNumber");
$NSN=mysql_result($result,$i,"NSN");
What is $i here?
$i defines the row to be retrieved, but you didn't give it any value.

echo various fields here

This part works fine.
good. :-)

Suprisingly because you didn't define $i.....

----------------------------------

I now want to list other jobs which use the same NSN on the same page
under that display. This is the script I'm using:

$query2="select * from jobs where NSN like $NSN";
$result=mysql_query($query2);
$num_results=mysql_num_rows($result);
$JobNumber=mysql_result($result,$i,"JobNumber");
What is $i here?

echo query2; (returns the correct query, including the NSN)
That should be:
echo $query2;

You forgot the $
echo $num_results (returns nothing)
should return something if you fix the previous code. :-)

$i=0;
while ($i < $num) {
What is $num?
Do you mean $num_results???

echo various fields here (returns nothing)

$i++;
}

Can anyone tell me what I'm missing?

Fix the various mistakes. :-)
Good luck!

Regards,
Erwin Moller