How do I get this code to work!

programming
1

I cant seem to get this right. I have created if statements to diagnose where I keep on having the query error, and just like i thought, it was in the INSERT INTO code. Here is the code, can someone look at this for me and see if The code is in anyway wrong, (misspelling, punctuation and etc.?)

//error here.
   $query = "INSERT INTO register (id, company, att, racernum, phone, names, date) VALUES ('', '$company', '$att', '$racernum', '$phone', '$names', '$date')";

   mysql_query($query) or die('Error, query failed');


exit();
2

why are you putting a blank value into the id field?
if its an auto-increment primary key, then just leave that alone and it will automatically add it itself.

3

also, try two things:

  1. print out all your variables that you are inserting before the query…this way u can make sure that ur variables have no problems

  2. print out the query (ie. print $query;)
    this will show you wat you are trying to do so u can identify the problem in the query

4

try:

$query = "INSERT INTO register (company, att, racernum, phone, names, date) VALUES ('$company', '$att', '$racernum', '$phone', '$names', '$date')";
mysql_query($query) or die(mysql_error());

adding to what DHDesign said, if your id field is auto_increment, you should not be inserting any id value. it will automatically populate.

make sure your die’s are mysql_error() when testing so you get a better error message.

5

ok. Well, I seem to still be getting an error, but the detailed error did give me a different suggestion as to what the problem is, but I have no idea what the error means. I get this as the error.

“Duplicate entry ‘0’ for key 1”

6

Okay, I gotta be a retard, I just forgot to put the row as “auto increment”

hello.

well, thanks to everyone who helped, there were some tips that will be very useful.

7

no problem. for those that didn’t understand, the
"Duplicate entry ‘0’ for key 1" error message was because the field was set as a primary key, but skyler was unintentionally trying to insert a duplicate value. to fix the problem, just set as auto increment and make sure the starting auto increment number does not already exist in the table.