php parameters passed inside function is not working

Question

In php/wordpress I have made a function. I want to pass some parameteres inside the function so that it will show result according to that. So far now my function code is like this

$user_id = get_current_user_id();
function check_user_access($role, $action = NULL ) {
    if( $role == 'subscriber') {
        if( $action = 'check_customer' ) {
            $check_customer = $wpdb->get_var("SELECT COUNT(id) FROM `table1` WHERE `user_id` = $user_id");
            return $check_customer;
        }

        if( $action = 'check_users' ) {
            $check_users = $wpdb->get_var("SELECT COUNT(id) FROM `table2` WHERE `user_id` = $user_id");
            return $check_users;
        }
    }
}

Now I am using this function like this

$role = 'subscriber';
$check_customers = check_user_access($role, $action = 'check_users' );
if( $check_users <=1 ) {
    //do something;
}
if( $check_users > 1 ) {
    //do something other;
}

But its showing the result of $action = 'check_customer'. Means its working for the first block condition. Can someone tell me how to solve this? Am I doing something wrong?

You're using = instead of ==, which means you're assigning the value instead of comparing it. The result of an assignment is always true, so your first nested if condition will always pass. — David
– David, Commented Jul 16, 2015 at 13:32
@NewUser: Yes, use == for comparison. Just like you do in your first if statement. — David
– David, Commented Jul 16, 2015 at 13:34
I don't really like reading Yoda Conditions but I can kinda see the point - they sidestep this issue completely... en.wikipedia.org/wiki/Yoda_conditions — CD001
– CD001, Commented Jul 16, 2015 at 13:35

Meenesh Jain · Accepted Answer · 2015-07-16 13:35:54Z

2

change your

 if( $action = 'check_customer' ) {}

to

if( $action == 'check_customer' ) {}

= means Assignment Operator

== means Comparison Operator

refer - from here

answered Jul 16, 2015 at 13:35

Meenesh Jain

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