[SOLVED] Query problem

[SirChick]

I dont know why but i cannot get my query to work... it just say's "ID not found" but its using the session to find the ID so it must be a syntax issue but i cannot see where it is. Can you see the problem ? :

 

 

$FindHouseRents = mysql_query("SELECT * FROM houses.*, userregistration.* FROM houses, userregistration
				WHERE houses.HouseID ='{$row['HouseID']}' userregistration.UserID='{$_SESSION['Current_User']}'");

// Fetch the row from the database
if (!($secondrow = mysql_fetch_assoc($FindHouseRents))) {
echo "ID not found!";
exit;
}

[Jessica]

You have errors in your SQL, which you'd see if you had debugging or error checking.

Try this:

<?php
$sql = 'SELECT houses.*, userregistration.* FROM houses, userregistration WHERE houses.HouseID ='.$row['HouseID'].' AND userregistration.UserID='.$_SESSION['Current_User'];
//Do print $sql; here to preview your query. your first on has a TON of errors.
$FindHouseRents = mysql_query($sql) OR die(mysql_error().' with query: '.$sql);
?>

[SirChick]

This is what im getting :

 

 

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\homeloginvariables.php on line 145

ID not found!

 

Relating to this for some reason :S

 

// Fetch the row from the database
if (!($secondrow = mysql_fetch_assoc($FindHouseRents))) {
echo "ID not found!";
exit;

 

I don't see the mistake there though ^ ?

[Jessica]

Did you get any errors? Post the new code including the SQL and query code.

[SirChick]

$sql = 'SELECT houses.*, userregistration.* FROM houses, userregistration WHERE houses.HouseID ='.$row['HouseID'].' AND userregistration.UserID='.$_SESSION['Current_User'];
//Do print $sql; here to preview your query. your first on has a TON of errors.
$FindHouseRents = mysql_query($sql) OR die(mysql_error().' with query: '.$sql);
}
// Fetch the row from the database
if (!($secondrow = mysql_fetch_assoc($FindHouseRents))) {
echo "ID not found!";
exit;
}

 

 

All that returns was :

 

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\homeloginvariables.php on line 145

ID not found!

[Jessica]

What's with the closing bracket after your mysql_query line? If it's within a conditional statement it's probably not being run.

[SirChick]

oh sos thats cos its in an If its right end of it though

[Jessica]

Yeah, but the next part of the code isn't. So it looks like it's not entering the IF.

[SirChick]

Ok this is all of it (sorry for the delay) just had to do a quick emergency on a security thing!

 

 

Code:

 

// Fetch the row from the database
If (!($houserow = mysql_fetch_assoc($GetHouseInfo))) {
$sql = 'SELECT houses.*, userregistration.* FROM houses, userregistration WHERE houses.HouseID ='.$row['HouseID'].' AND userregistration.UserID='.$_SESSION['Current_User'];
//Do print $sql; here to preview your query. your first on has a TON of errors.
$FindHouseRents = mysql_query($sql) OR die(mysql_error().' with query: '.$sql);
}
// Fetch the row from the database
if (!($secondrow = mysql_fetch_assoc($FindHouseRents))) {
echo "ID not found!";
exit;
}

[Jessica]

Yeah, but the next part of the code isn't. So it looks like it's not entering the IF.

 

Why don't you move the rest of the related code INTO that if? Then you won't get this error.

[SirChick]

So you mean:

 

 

 

// Fetch the row from the database
If (!($houserow = mysql_fetch_assoc($GetHouseInfo))) {



$sql = 'SELECT houses.*, userregistration.* FROM houses, userregistration WHERE houses.HouseID ='.$row['HouseID'].' AND userregistration.UserID='.$_SESSION['Current_User'];
//Do print $sql; here to preview your query. your first on has a TON of errors.
$FindHouseRents = mysql_query($sql) OR die(mysql_error().' with query: '.$sql);

// Fetch the row from the database
if (!($secondrow = mysql_fetch_assoc($FindHouseRents))) {



echo "ID not found!";



exit;
}
}

[Jessica]

Yes...

[SirChick]

Ok although it makes sense how come

 

if (!($secondrow = mysql_fetch_assoc($FindHouseRents))) {

 

wont work outside the if cos surely $FindHouseRents still passes through?

[Jessica]

How could it? If you define a variable inside of a conditional statement, and the statement is never entered, the variable isn't defined.

[SirChick]

Well i thought that the variable was set here:

 

$FindHouseRents = mysql_query($sql) OR die(mysql_error().' with query: '.$sql);

 

and then at any time in the script i could of done :

 

if (!($secondrow = mysql_fetch_assoc($FindHouseRents))) {

 

etc

 

or do other stuff with $FindHouseRents variable etc. wldnt of thought it need to be in the main if.

[Jessica]

If you define a variable inside of a conditional statement, and the statement is never entered, the variable isn't defined.

 

Your code was in two different conditional statements.

 

Look.

Write this code:

<?php
$x = 1;
$y = 0;
if($x==2){
   $y = 5;
}
if($x==1){
   print $y;
}
?>

You can read it and see that $y will be == 0, NOT 5.

Now, if you never define $y, you'll either get an error or nothing, depending on your settings.

<?php
$x = 1;
if($x==2){
   $y = 5;
}
if($x==1){
// y doesn't exist!
   print $y;
}
?>

 

This is similar to what you were doing, as far as I can tell.

[SirChick]

oh right that does make more sense now! thanks jesi

 

Wasnt doubting you.. just interested in these things as im still learning

[Jessica]

No problem. Anytime you can't get something to work just simplify it till it does work and you'll find your problem