[SOLVED] Small PHP/SQL Problem

[grahamb314]

Hi all,

 

I want to check to see if "Show" (From the previous form) is already in the database I have. If it is, then echo something, if it isnt, then add it

 

 

I think i`m nearly there !

<?php
require_once 'mysql_connect.php';
$result = mysqli_query($mysqli, "SELECT show FROM shows WHERE show= '".$_POST["Show"]."'");

if (mysqli_num_rows($result) == 1) {
echo "Show name already exists";
}
	else {$sql = mysqli_query($mysqli, "INSERT INTO `grahamb_test`.`shows` (`id`, `show`) VALUES (NULL,'".$_POST["Show"]."')");
	}
?>

 

Thanks!

 

Graham

 

[ratcateme]

whats the problem??

 

Scott.

[grahamb314]

Oh yea, sorry.

The error:

 

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in /URL HERE on line 5

[ratcateme]

it looks like you first query is failing try changing it to

$result = mysqli_query($mysqli, "SELECT show FROM shows WHERE show= '".$_POST["Show"]."'") or die(mysqli_error($mysqli));

and see what the output is

 

Scott.

[grahamb314]

Thanks,

Now we have:

 

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'show FROM shows WHERE show= ''' at line 1

[grahamb314]

Its the POST thing thats a bit odd. - The quotes or soemthing? -

 

Any thoughts?

 

$result = mysqli_query($mysqli, "SELECT show FROM shows WHERE show= '".$_POST["Show"]."'") or die(mysqli_error($mysqli));

[ratcateme]

try changing it to

$query = "SELECT * FROM `shows` WHERE `show` = '{$_POST["Show"]}';";
$result = mysqli_query($mysqli, $query) or die("Query:{$query} <br>Error:".mysqli_error($mysqli));

are you sure that there is something in $_POST["Show"] even though that shouldn't matter

 

Scott.

[grahamb314]

Thanks again,

 

I put something and show and get

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'show FROM shows WHERE show= '123'' at line 1

 

Let me try your suggestion

[grahamb314]

Okay, it now adds post the table in the DB, but it will still accept duplicates like 123 will still be accepted even if 123 is already in the DB

 

Its gotta be something small,

 

 

[ratcateme]

i copied your code and added a little output what happens now

require_once 'mysql_connect.php';
$query = "SELECT * FROM `grahamb_test`.`shows` WHERE `show` = '{$_POST["Show"]}';";
$result = mysqli_query($mysqli, $query) or die("Query:{$query} <br>Error:".mysqli_error($mysqli));
echo "rows = ".mysqli_num_rows($result);
if (mysqli_num_rows($result) == 1) {
echo "Show name already exists";
} else {
$sql = mysqli_query($mysqli, "INSERT INTO `grahamb_test`.`shows` (`id`, `show`) VALUES (NULL,'" . $_POST["Show"] . "')");
}

[grahamb314]

Looking good...

 

rows = 1Show name already exists

 

What was the problem you think? - So I don't do it again

[ratcateme]

i am not sure what database does mysql_connect.php connect to. the only other thing i changed was i added `grahamb_test`. to the first query

 

Scott.

[grahamb314]

Interesting!  mysql_connect.php just connects to the database (With user and pass) - It's a seperate php page that I call in all Database related php pages

 

All I can say now is thank's for your help. It's people like you that keep people wanting to learn! - If it wasnt for forums like this, I'd have given up on PHP a while ago!

 

Thanks again!

 

Graham