Mysql data as variable value?

Kymaera

Okay, here's the deal. I've got a mysql table named "accepting" and one field named "accepting". I'm trying to use an if statement to check the data in the table.
Config.php contains the database connection info, etc.

include "config.php";
$acceptdata = "mysql_query(SELECT accepting FROM accepting;)";
if ($acceptdata = "1") {
print "<font color=#00FF33>Accepting";
} else {
print "<font color=#FF0033>Not Accepting";
}

That doesn't seem to work.

Kymaera

Well I knew it had to be redesigned. I was hoping somebody could show me how to do that. Thanks

Kymaera

o_O I want the broken code fixed.

Kymaera

The only one with the bad attitude is you. I came here looking for help, and you attacked me.

drag0n

http://www.phpbuilder.com/board/showthread.php?s=&threadid=10235043

Kymaera

drag0n, telling me to "go learn php" seems extremely mature on your part. I'm sure you are a very intelligent, upstanding person with limitless knowledge of PHP, and were never new to the programming language. But if you have no constructive advice, I suggest you don't post.

By the way, if anyone wants to help me, I would be very appreciative.

drag0n

You have to bring out the data and you have to query
You have to query the db for info that you need and ur db query is wrong.. like u query for db for that specfic ID and to see if it has one for accpeting... so ur code needs tyo be re designed
exactly u didnt understand anything I said so again: learn php: www.php.net 😉

Kymaera

So what would the correct query be? Assuming the id is 1, would it be like: SELECT accepting FROM accepting WHERE id=1?

drag0n

id is the id of the row like the fields would be :
ID ACCEPTING
Rows: 1 1
2 2
You select the row by id and check wheathere accpeting is 1 😉

Kymaera

Yes, I understand that. This table has one row, and I added the id field. So the id of that row is 1. And the accepting field is either 1 or 0. That query finds that row. Now how would I grab the data out of the accepting field after I select the row?

HalfaBee

I seem to have missed something here.

if ($acceptdata = "1") {

Dosen't do what you expect.
It sets $acceptdata to a value of "1" not a comparison.
It should be
if ($acceptdata == "1") {

You might also want to look at the www.php.net/mysql_fetch_array function.

HalfaBee

drag0n

Originally posted by drag0n
You have to bring out the data

Kymaera

Yes!

if ($acceptdata = "1") {

was the problem, halfabee. Thank you.

HalfaBee

I am thought they needed a little push in the right direction.
I am trying to teach my 11 year old php and it is quite frustrating for both of us 🙂

HalfaBee

Kymaera

lol. Good to get them started early. I started PHP about a month ago, and i'm having fun learning it.

HalfaBee

Kymaera, your code will only return a result resource, not the actual value in the DB.

HalfaBee

drag0n

$data = mysql_fetch_array($res);

$data[id];
$data[accepting];

those will echo out the values

Kymaera

Oh, sorry. I forgot to tell you I figured the mysql_fetch_array function out already. Thanks a lot though.