Not getting expected results from SQL query - php

Question

this is my table
Field Type Null Key Default Extra
id_key int(11) NO PRI NULL auto_increment
tbl_users_username varchar(255) YES UNI NULL
tbl_users_password varchar(32) YES NULL
tbl_users_identifier varchar(32) YES NULL
tbl_users_token varchar(32) YES NULL
tbl_users_access_type int(1) YES NULL
tbl_users_timeout int(10) YES NULL
this is my code
$query = "SELECT * FROM tbl_users where tbl_users_username = '$_POST[email_address]'" ;
$result = mysql_query($query);
if($result)
{
echo "TRUE";
}
else
{
echo "FALSE";
}
There are no problems connecting to the database. The problem arises when I run the query. No matter if the email is in the table or not it returns a $result. What am I missing here? It is supposed to echo true if email address exists in table, false if it is not in table.

Answer 1

The result will be returned nevertheless(if you made a valid query) use mysql_num_rows to check if the mail exists instead.
Actually use PDO if you can

Answer 2

I suggest that you print out your query to ascertain that it is 100% the query you expect it to be.
In addition to that, You are checking:
$result = mysql_query($query);
if($result)
This checks if a resource was returned from your query. It your query is valid then as per the documentation, you should always have a resource.
Suggested code:
$result = mysql_query($query);
if($result)
{
$num = mysql_num_rows($result);
if ($num > 0)
{
//do logic processing here...
echo "TRUE";
}
}
else
{
die('Invalid query: ' . mysql_error());
}
Code Explanation:
Fetch the resource from the query
If valid resource was returned, continue and check the number of records that were fetched
from this query.
If the resource was not returned, the else part echo out a failure message.
I would strongly suggest that you look into using PDO library.
Briefly the Advantages of PDO:
Allows you to use prepared statement with ease (prepared statement are great for security!!!)
PDO can connect to various different Databases including MySQL, but others as well.
Its quite easy to use in my opinion and it quite easy to pick for new comers etc.
Stack overflow usage tips:
Always use the search in the top right hand corner. Numerous people have come across problems that may could help you.
Always have a look at the How to Ask Question FAQ
Feedback & always ask further questions if required!

Answer 3

$_POST[email_address] should use single or double quotes i guess:
$_POST["email_address"] or $_POST['email_address']
$query = "SELECT * FROM tbl_users where tbl_users_username = '" . $_POST['email_address'] . "'" ;
However, as #FabioCosta says, you have to fetch the result or count the number of returned rows to achieve what you are aiming at:
$query = "SELECT 1 FROM tbl_users where tbl_users_username = '" . $_POST['email_address'] . "'";
$result = mysql_query($query);
if(mysql_num_rows($result) === 1) {
echo "TRUE";
} else {
echo "FALSE";
}

Answer 4

If you use PHP's mysql_num_rows function, you should be able to return the number of rows, and your logic will work. So, try the following:
$num = mysql_num_rows($result);
if ($num > 0) {
echo "TRUE";
}else{
echo "FALSE";
}

Answer 5

$query = "SELECT *
FROM tbl_users
WHERE tbl_users_username = '" . mysql_real_escape_string($_POST['email_address']) . "'";
$result = mysql_query($query) or die(mysql_error());
if($result)
{
echo "TRUE";
}
else
{
echo "FALSE";
}
This should fix any problems you're running into, or at least give you a handy error message to work off of.

Answer 6

The following query will select the tbl_users_username value based on the value in the posted email_address' value. So, use this query:
$query =
"SELECT * FROM tbl_users where tbl_users_username = '".$_POST['email_address']."';";