Query was empty?!?

[vajona]

I'm not sure what i have done wrong here...

all it says is "Query was empty"

 

<?php
SESSION_START();

?>

//some html

<?php

if($_SESSION['email']){

$mysql = "SELECT id,firstname FROM users WHERE 'id'='".$_SESSION['email']."'";
$res = mysql_query($sql) or die(mysql_error());

if(mysql_num_rows($res) == 0){
	session_destroy();
	echo"Please <a href=\"/index.html\">Login</a> or <a href=\"/register.php\">Join Us.</a>";
}else{
	$row = mysql_fetch_assoc($res);
	echo "welcome back, <a href=\"./index.php?act=profile&id=".$row['id']."\">".$row['firstname']."</a>\n";

}
}else {
echo"Please <a href=\"/index.html\">Login</a> or <a href=\"/register.php\">Join Us.</a>";

}
?>

[revraz]

Change the single quotes around 'id' to backticks `id`

[vajona]

Changed to `id` , but I'm still getting the error.

[revraz]

echo $mysql and post the exact error

[revraz]

You have the wrong variable name here

 

$mysql = "SELECT id,firstname FROM users WHERE 'id'='".$_SESSION['email']."'";

$res = mysql_query($sql) or die(mysql_error());

 

Use either $mysql or $sql, but not both.

[vajona]

well every time i use only sql or mysql it logs me out.

[revraz]

Then you have another problem, but the code above, you are referencing a variable that is empty.

[vajona]

true, why would it log me out though? it's the same session i have been using

[revraz]

If it logs you out, then

 

if(mysql_num_rows($res) == 0){

 

must be true.  So see why that is true.

[vajona]

Ya, i am really confused because the results for the id for that session isn't 0.

[vajona]

If i change to $sql = "SELECT * FROM users WHERE `id`='specific number'"; it works. why isn't the session working here?

[revraz]

Is your  $_SESSION['email'] a real email or a ID?

[vajona]

i believe it is real.

[revraz]

Not sure what you mean "you believe".. echo it and see what it is.

Echo your query.

[vajona]

Are you asking me if the session tests email=pass to have access to a page? if so than yes it does.

[revraz]

No, Echo This

 

$mysql = "SELECT id,firstname FROM users WHERE 'id'='".$_SESSION['email']."'";

echo $mysql;

exit;

[vajona]

SELECT id,firstname FROM users WHERE 'id'='name@blah.edu' i received the logged in email.

[revraz]

now copy and past that into PHPMYADMIN and run it as a query.

Also, you still have the single quotes in there you need to remove.

[vajona]

update:

 

i changed my script to

 

$mysql = "SELECT * FROM users WHERE `email`=".$_SESSION['email']."";

 

and when i echo it out i get

 

SELECT * FROM users WHERE `id`=name@blah.edu

 

when i go to PHPMYADMIN and run it as a query i get

 

#1064 - 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 '@blah.edu LIMIT 0, 30' at line 1

 

however; if i simply put " " around the email in the query window

 

SELECT * FROM users WHERE `email` = "name@blah.edu"

 

gives me the correct value which is greater than 0

 

so my question now is how do i get it to look like that when i echo it out. i really hope that makes sense

 

also i would like to thank you for helping. i know it's like pulling teeth. i am trying my hardest to learn everything i can about how to script.

[BlueSkyIS]

$mysql = "SELECT * FROM users WHERE `email` = '".$_SESSION['email']."'";

[vajona]

Thank you so much guys. It is working now. To bad you don't live around me id take you out for food to properly thank you.