在php中捕获错误并将其作为json返回给html

Question

is it possible to catch error in a php file, lets say in the connection section, and return it to the HTML file as json for printing.

This is my try:

<?php
    $srevernme = "localhost";
    $username = "root";
    $password = "";
    $dbname = "db";

    //create connection
    $conn = new mysqli($srevernme,$username,$password,$dbname);

    $errors = array();  // array to hold connection errors

    //check connection                                              //<-----POSSIBLE ERROR
    if ($conn->connect_error) {
         $data['success'] = false;
         $errors['error_info'] = "connection failed:" . $conn->connect_error ."Please try later.";
         $data['errors'] = $errors;
         echo json_encode($data);
        //die("connection failed:" . $conn->connect_error);
    }



if ($_SERVER['REQUEST_METHOD'] == "POST") 
    {        
        if (isset($_POST["fnameInput"]) && isset($_POST["lnameInput"]) && isset($_POST["addressInput"]) && isset($_POST["cityInput"]) && isset($_POST["zipcodeInput"]))
        {
            // prepare and bind
            $stmt = $conn->prepare("INSERT INTO users (first_name, last_name) VALUES (?, ?)");

            if ($stmt == FALSE) {                                       //<-----POSSIBLE ERROR          
                $data['success'] = false;
                $errors['error_info'] = "Connection failed: Cannot create connection to sql DB. Please try later.";
                $data['errors'] = $errors;
                echo json_encode($data);
                //die("Connection failed: Cannot create connection to sql DB. Please try later.");
                }

           if (empty($errors)) 
           { 
                    //mark as success
                    $data = array(); //array for saving the data
                    $data['success'] = true;

                    //get wanted data....

                   $data['wanted_data'] = json_encode($some_data);
                   echo json_encode($data);

                   $stmt->close();
            }
    }
    $conn->close();

The relevant sections are marked with //<-----POSSIBLE ERROR. To be clear- data represents the data I want to echo to the html file and it had a key named success that his values are true if everything is ok or false if there is an error.

EDIT: In the current state the php file send request in this format when I shot down MySQL server (from chrome inspector):

<br />
<b>Warning</b>:  mysqli::mysqli():  in <b>C:\xampp\htdocs\projectegister.php</b> on line <b>7</b><br />
<br />
<b>Warning</b>:  mysqli::prepare(): Couldn't fetch mysqli in <b>C:\xampp\htdocs\ex3egister.php</b> on line <b>26</b><br />
{"success":false,"error_info":"Connection failed: Cannot create connection to sql DB. Please try later."}<br />
<b>Fatal error</b>:  Call to a member function bind_param() on null in <b>C:\xampp\htdocs\ex3egister.php</b> on line <b>39</b><br />

Thanks!

Answer 1

Ohh, is that what you mean?

It seems the JSON arrives just fine. What you mean is to disable php errors? In this case, you can turn off php's standard error reporting, so it only shows your custom errors. (Note that this means you will not get an error that you aren't manually taking care of)

see http://php.net/manual/en/function.error-reporting.php for more information.

// Turn off all error reporting
error_reporting(0);

This will prevent php from echo-ing its own errors, thus only returning your (valid) json_encoded errors.

edit: You'll want to return json_encode($errors) instead of json_encode($data) in case of an error.

edit#2: I would really advise you to not return anything until you've reached the end of your script, so it's less confusing. Simply put an if statement at the end.

if(!empty($errors)){
 echo json_encode($errors);
 return false; //don't go on
}
else{
 echo json_encode($data);
 return false; //don't go on
}

In case of an error, simply add to the $errors array, as it seems to me that's what you're wanting to do in the first place.