errors on MySQL

mentalfloss

Warning: Unable to jump to row 0 on MySQL result index 2 in c:\inetpub\wwwroot\partordering\partshow.php on line 12

It works fine, if I have a part# 1234 in the field.

But if I put a part# 12-34, I get this Warning.

Why is this?

Thanks
Terry

[deleted]

most likely: you are seeking inside a resultset, but the query you ran did not return any results.

Always check the mysql_num_row() to see if there are any results before you start processing them.

mentalfloss

I am a newbie, so if I sound stupid 🙂

mysql_num_row() so run this like a error check, return a error if 0 rows?

why would the dash character, cause this?

I know databases get a little funny with meta characters.

How can I get by this if I need to us a - in my number?

[deleted]

"mysql_num_row() so run this like a error check, return a error if 0 rows?"

mysql_num_rows() is not an error check, it just returns the number of records that your query returned.
If you run a query that finds no records, this function will return zero. Wether that is an error depends on your application.

"But if I put a part# 12-34, I get this Warning"

Most likely because there are no records where the partnumber is '12-34'.
Again, wether that is an error is up to you.

mentalfloss

Yeah its an error to me.

the record is exactly the same as the one without the dash in it.

example table:

I hope this comes out ok 🙂

The only thing different is the Part_num

and it returns that error...I can't figure out why the dash would cause that.

mentalfloss

By the way I read your rants "Why should I not put images in a database?"....So, I am changing my code, so it won't do that. 🙂

My goal is to use this as a parts list with pictures. I thought loading the image on a database was a cool idea. I changed my mine.

Thanks,
Terry

[deleted]

Did you put quotes areound "12-34" in the query? otherwise PHP might interpret it as 12-34= -22, and look for records containing -22

mentalfloss

Your right about the quotes. Here is what I was doing wrong.

I had something like:

$myquery = "SELECT field1 from table where field2 = $variable";

I need to put '$variable' round the variable.

Thanks For The Help