I'm confused how to use isset and empty, I'm trying to write a simple api, there should be error saying which param is missing and return error too if the param is null.
What's wrong my statement below?
$email= isset($_POST['email']) ? mysqli_real_escape_string($mysqli, $_POST['email']) : '';
if(empty($email)) {
echo 'email cannot be empty!'
}
You don't actually need to use both isset and empty, because empty already does it.
No warning is generated if the variable does not exist. That means empty() is essentially the concise equivalent to !isset($var) || $var == false.
More details are here: http://php.net/manual/en/function.empty.php
So your code could look this way, for example:
if (empty(trim($_POST['email']))) {
echo "Email cannot be empty!\n";
// you should add return or raise exception here
// or even exit
exit;
}
$email = trim($_POST['email']);
if (!filter_var($email, FILTER_VALIDATE_EMAIL)) {
echo "Your email {$email} is invalid\n";
// you should add return or raise exception here
// or even exit
exit;
}
$email = mysqli_real_escape_string($email);
Change $email to :
$email= isset($_POST['email']) && !empty($_POST['email']) ? mysqli_real_escape_string($mysqli, $_POST['email']) : '';
So the $_POST['email'] is not containing empty value.
Try to use next code snippet
if( false === input_filter(INPUT_POST, FILTER_VALIDATE_EMAIL) ) {
echo "error message";
}
$email = input_filter(INPUT_POST, FILTER_SANITIZE_EMAIL );
// to do somethink with email...
I think you are trying to check $email value which is null. You should check $fname in if statement instead.
try this to check for all missing fields:
$required_fields = array('email', 'name', 'password');
$err_msgs=array();
foreach($required_fields as $field) {
if(empty( $_POST[$field]) ){
err_msgs[]= $field . ' cannot be empty!';
}
}
if (!empty($err_msgs)) {
echo json_decode($err_msgs);
}
Edit: removed isset() after reading Axalix's answer.
$emailvariable ?