A null value becomes 0 in Mysql and PHP - php

Question

I have 2 drop-down lists with different names, and I'm trying to query in just one field.
I'm using a jQuery function wherein if item 1 is selected, the drop-down list 1 will be displayed, and if the item 2 is selected, the drop-down list 2 will be displayed.
This is how I populated my drop-down list from the mysql database and tables:
<div id="minquep">
<label>Branch</label>
<SELECT name="user_min">
<OPTION VALUE="0">Choose a branch
<?=$minq_options?>
</SELECT>
</div>
<div id="albury">
<label>Branch</label>
<SELECT name="user_branch">
<OPTION VALUE="0">Choose a branch
<?=$al_options?>
</SELECT>
And this is how I insert queries into mysql by filling out the form with drop-down lists in it:
if (isset($_REQUEST['Submit'])) {
$sql = "INSERT INTO $db_table(branch) values ('".mysql_real_escape_string(stripslashes($_REQUEST['user_branch'])).",".mysql_real_escape_string(stripslashes($_REQUEST['user_min']))."')";
if($_REQUEST['user_branch']= ""){
($_REQUEST['user_branch']) = NULL;
}
if($result = mysql_query($sql ,$db)) {
echo '<script type="text/javascript">alert("The user has been added successfully!\n");return true;</script>';
echo "<meta http-equiv=\"refresh\" content=\"0;URL=add_user.php\">";
}
else {
echo "ERROR: ".mysql_error();
}
}
The testing scenario is that, I choose the value under <select name="user_min">.
So I assume that sql will just bypass the result for user_branch because it is null. But it does prints '0' instead, after the insert query. For example, if the inserted ($_REQUEST['user_min']) value is "Brisbane" and the ($_REQUEST['user_branch']) value is null (because I didn't selected any value under the user_branch drop-down list), the branch field should just become "Brisbane", knowing that user_branch is NULL. But it does print "BRISBANE" with 0, like 0, Brisbane in my mysql table.
How can I fix this?
I already tried putting an if condition, it did not work.
if($_REQUEST['user_branch']= ""){
($_REQUEST['user_branch']) = NULL;
}
I've also tried changing the user_min into same name user_branch, but it does not get the selected value, instead of Brisbane it just prints '0'

Answer 1

A few things -
you are setting user_branch = NULL after the $sql.
You are using quotes around value. It's fine non-Null values -you need to check if there is NULL value then don't use quotes in your $sql.
You are using single "=" in your IF statement. This is failing your If statement. Change it to if($var == '')
Update
Some suggestions based on your existing code - however there are other best practices to achieve what you are trying to achieve....
$user_branch = $_REQUEST['user_branch'];
$user_min = $_REQUEST['user_min'];
//you should validate above values first
if$user_branch == "" || $user_min == "") {
$db_value = "NULL";
} else
{
$db_value = "'".mysql_real_escape_string(stripslashes($user_branch.','.$user_min))."'";
}
$sql = "INSERT INTO $db_table(branch) values (".$db_value.")";

Answer 2

If you don't want 0, then change:
<option value="0">Chose a Branch</option>
To:
<option value="">Chose a Branch</option>
Otherwise it will get passed through the form as 0
Also, your PHP code won't work as you want it to:
if (isset($_REQUEST['Submit'])) {
// do this before the query
if($_REQUEST['user_branch']== ""){ // note the extra = in there, so you aren't assigning the variable, you are compairing
$_REQUEST['user_branch'] = NULL; // no need for brackets round the variable
}
$sql = "INSERT INTO $db_table(branch) values (".mysql_real_escape_string(stripslashes($_REQUEST['user_branch'])).",".mysql_real_escape_string(stripslashes($_REQUEST['user_min'])).")"; // you had single quotes around everything inside the VALUES() function
if($result = mysql_query($sql ,$db)) {
echo '<script type="text/javascript">alert("The user has been added successfully!\n");return true;</script>';
echo "<meta http-equiv=\"refresh\" content=\"0;URL=add_user.php\">";
}
else {
echo "ERROR: ".mysql_error();
}
}

Answer 3

$sql = "INSERT INTO reg(Name ,Email) values ('".mysql_real_escape_string(stripslashes($_REQUEST['Name']))."','".mysql_real_escape_string(stripslashes($_REQUEST['Email']))."')";
This is my code. Here I found error of Undefined index.