donglu5612
2015-05-07 08:47
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Json对象从PHP返回json_encode(),但在我的javascript中无效的json对象

I echo a simple associative array read from a mySQL table back to my jquery b.m.o the json_encode($array) method. However, my .ajax call fails back in the jquery and it seems due to the object not being a valid json object. I did some debugging in chrome and under the network tab verified the response preview- does indeed look to be in perfect json format:

{"UserName":"DehanL","UserPassword":"admin","UserEmail":"dehan@rocketmail.com"}

Here is my php:

<html>
<body>

<?php



$servername = "localhost";
$username = "root";
$password = "";
$dbName ="dbPodiumPro";

$inUsername = $_POST["name"];
$inPassword = $_POST["password"];


// Create connection
$conn = new mysqli($servername, $username, $password, $dbName);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

//Use the dbPodiumPro
mysqli_select_db($conn,"dbPodiumPro");


$sql = "SELECT * FROM users WHERE UserName='$inUsername' AND UserPassword ='$inPassword' ";

$result = mysqli_query($conn,$sql);

// Fetch one row
$row=mysqli_fetch_assoc($result);

//Check to see if the username and password combo exists
if(!empty($row['UserName']) AND !empty($row['UserPassword'])) 
    { 

    //$_SESSION['UserName'] = $row['UserPassword']; 
    //echo $_SESSION; 

    echo json_encode($row);

    } 

else { echo "Be gone imposter!"; }


$conn->close();


?>

</body>
</html>

Then the jquery:

$.ajax({
        type: 'POST', // define the type of HTTP verb we want to use (POST for our form)
        url: 'php/loginrequest.php', // the url where we want to POST
        data: formData, // our data object
        dataType: 'json' // what type of data do we expect back from the server

    })
    // using the done promise callback
    .done(function (data) {
        console.log("ajax done callback");
    })
    .fail(function (data) {
        console.log(data, "ajax failed callback");

        var IS_JSON = true;
        try {
            var json = $.parseJSON(data);
        } catch (err) {
            IS_JSON = false;
        }
        console.log(IS_JSON);
    });
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