Unable to select table.. php & mysql - php

Question

I created a table & when Submit button is hit after filling the form, those particulars should be inserted into the DB TABLE. But idk what's wrong with the code, it's echoing "Unable to select table".. My code is as follows:
<?php
if ( isset ( $_POST['submit'] ) )
{
mysql_connect("localhost","root","1234");
mysql_select_db("my_db")or die( "Unable to select database</span></p>");
$name1 = $_POST['name1'];
$email = $_POST['email'];
$password = $_POST['password'];
$confirmpassword = $_POST['confirmpassword'];
$gender = $_POST['gender'];
$place = $_POST['place'];
$college = $_POST['college'];
$result=MYSQL_QUERY("INSERT INTO USERS3 (id,name1,email,password,confirmpassword,gender,college,place)".
"VALUES ('NULL', '$name1', '$email', '$password', '$confirmpassword', '$gender', '$place', '$college')")or die( "<p><span style=\"color: red;\">Unable to select table</span></p>");
mysql_close();
echo "<p><span style=\"color: red;\">Thank You;</span></p>";
}
else
{
// close php so we can put in our code
?>
<form id="form1" action="" method="post">
Name:
<input type="text" name="name1" /><br/>
E-mail:
<input type="text" name="email" /><br/>
Password:
<input type="password" name="password" /><br/>
Confirm Password:
<input type="password" name="confirmpassword" /><br/>
Gender:
<input type="radio" name="gender" />
Male
<input type="radio" name="gender" />
Female
<br/>
Location:
<input type="text" name="place" /><br/>
College:
<input type="text" name="college" /><br/>
<input id="submit1" class="submit" type="submit" name="submit" value="Submit"/><br/>
<input type="reset" value="reset" />
</form>
<?php
} //close the else statement
?>

Answer 1

PHP doesn't recognise function with capital letters. Use small-case characters:
mysql_query( "INSERT INTO USERS3 ( id, name1, email,
`password`, confirmpassword, gender, college, place)
VALUES ('NULL', '$name1', '$email',
'$password', '$confirmpassword', '$gender', '$place',
'$college')") or die( "<p><span style=\"color: red;\">Unable to select table</span></p>");
mysql_close();

Answer 2

If 'id' is an autoincrement field, you shouldn't be passing any value for it - the database will deal with that.

Answer 3

Everything looks fine.Only problem(s) could be:
The table name might be incorrect.Especially, i suspect your correct table name is users3 instead of USERS3 .
The ideal way to check is:
Add following statement before $result=.....
echo "INSERT INTO USERS3 (id,name1,email,password,confirmpassword,gender,college,place)".
"VALUES ('NULL', '$name1', '$email', '$password', '$confirmpassword', '$gender', '$place', '$college')";
Run/Load the page.
Whatever is printed - copy it.
Go to phpmyadmin and select SQL Tab.Paste what you copied in previous step and run it.
phpmyadmin will display the exact error.