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Posted by Erwin Moller on 10/13/47 11:39
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
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