Kim,
Won't redefining the variable destroy the MySQL Resource, resulting in a resource invalid error, like he has?
$pic = mysql_query($sql,$connection); //Defines variable
$pic //Redefining variable, dumps old contents = //calling on an invalid MySQL Resource, because it doesn't exist mysql_fetch_array($pic);
My comments are in green.
Cole
Kim Steinhaug wrote:
Well, your query looked confusing to me, but break it up like this
First : select Then, name your coloumns, seperate with commas, or use wildcard for all * Define from what table you need to select FROM database_table Add some select statements with WHERE WHERE and the statements itself 1=1
What did I just write now? Looking at your example I guess : 1. you need the coloumns : Rune, username 2. your table : RuneRunner 3. your select criteria : UserID=3
I would then write your SQL query as this :
$sql = "SELECT Rune, username FROM RuneRunner WHERE UserID=3";
Then you could do your code, $pic = mysql_query($sql,$connection); $pic = mysql_fetch_array($pic);
You would also want to add a database abstraction layer, as for debugging purposes this will make your life much easier.
Your original code looks to correct if you rewrite it abit, like this :
SELECT RuneRunner_1.Rune, RuneRunner_1.username FROM RuneRunner RuneRunner_1 WHERE (RuneRunner_1.User_ID=3);
Not tested but it should work fine. On the other hand, the referencename after the table name should be a short one to make the statement much easier to read, somthing like :
SELECT rr.Rune, rr.username FROM RuneRunner rr WHERE (rr.User_ID=3);
Then again, since you only have one table in your statement there is no need for the reference name, and since your WHERE clause is a single statement there is no need for the paranthese either. So you are still back to :
SELECT Rune, username FROM RuneRunner WHERE User_ID=3;
-- -- Kim Steinhaug ---------------------------------------------------------------------- There are 10 types of people when it comes to binary numbers: those who understand them, and those who don't. ---------------------------------------------------------------------- www.steinhaug.com - www.easywebshop.no - www.webkitpro.com ----------------------------------------------------------------------
"Water_foul" <phpdbnews@xxxxxxxxxxxxxxxxxxx> wrote in message
news:20040623000943.72840.qmail@xxxxxxxxxxxxxxx
why doesn't this work:
$pic=mysql_query('SELECT Rune, username FROM RuneRunner
RuneRunner_1 WHERE (User_ID = 3)',$connection);
$pic=mysql_fetch_array($pic);
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