I have the following code;
// check phone exist or not $query = "SELECT * FROM user WHERE phone_number=".$phone; $result = mysqli_query($conn,$query); $count = mysqli_num_rows($result); if($count!=0){ $error = true; $phoneError = "Provided Phone Number($phone) is already in use."; }
How can i send the result of the query as json feed to my colleague who we are working on the same project?
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Answer
You need to create an array, where you can add status like fail/success
and corresponding messages based on that.
A sample code is given below:-
$final_message = []; if($count > 0){ $final_message['status'] = 'fail'; $final_message['message'] = "Provided Phone Number($phone) is already in use."; }else{ $final_message['status'] = 'success'; } echo json_encode($final_message);