PHP数据返回空无效

时间:2017-06-15 16:17:12

标签: javascript php jquery

所以我将密码传递给php页面并返回json数据 但是,如果它什么都不返回没什么好玩的。我的失败登录永远不会触发。有人能说出我做错了什么。

<script>
$(document).ready(function(){
var password = "";
//PRESSING AN INPUT KEY
$('.sixPinInput').keyup(function (k) {
    if (this.value.length == this.maxLength) {
        password = password+this.value;
        password = password.substring(0,6);
        $(this).next('.sixPinInput').focus();
        $(".pinIncorrectMessage").css("display", "block");
        $(".pinIncorrectMessage").html(password);
    }
    //PRESSING DELETE
    if (k.keyCode == 8) {
    $(this).prev().val("").focus();
    password = password.slice(0, -1);
    $(".pinIncorrectMessage").html(password);
    }
});
//PRESSING LAST INPUT KEY
$("#f").keyup(function() {
    if($("#f").val().length > 0) {
        $.post("http://heber/QC/includes/getUserInfo.php", {user: password}, function(data, status) {
            //IF SUCCESSFUL LOGIN
            password = "";
            $(".sixPinInput").val("");
            if(data) {
                $(data).each(function(index, value) {

                    alert(value.firstName + value.lastName)
                    $(".cover").fadeOut(200);
                    $(".sixPinInputContainer").fadeOut(200);
                    $("#pageBody").css("overflow", "scroll");
                })
            }
            //IF FAILED LOGIN
            if(!data) {
                $(".cover").fadeOut(200);
                $(".sixPinInputContainer").fadeOut(200);
                $("#pageBody").css("overflow", "scroll");
            }
        })
    }
})
});
</script>

这是我的PHP

<?php include "db.php"; ?>
<?php
if (isset($_POST['user'])) {
$password = $_POST['user'];
$query = $connect->prepare("
SELECT firstName, lastName, color1, color2, authorizationLevel FROM employee WHERE password = ? LIMIT 1");
if(!$query) {
    echo "Query Failed because " . mysqli_error($connect);
}
$query->bind_param("s", $password);
$query->execute();
$result = $query->get_result();
if ($result) {
    while($row = mysqli_fetch_assoc($result)) {
        $firstName = $row['firstName'];
        $lastName = $row['lastName'];
        $color1 = $row['color1'];
        $color2 = $row['color2'];
        $authorizationLevel = $row['authorizationLevel'];

        session_start();
        $_SESSION['firstName'] = $firstName;
        $_SESSION['lastName'] = $lastName;
        $_SESSION['color1'] = $color1;
        $_SESSION['color2'] = $color2;
        $_SESSION['authorizationLevel'] = $authorizationLevel;

        $array[] = array (
            'firstName'=>$firstName,
            'lastName'=>$lastName,
            'color1'=>$color1,
            'color2'=>$color2,
            'authorizationLevel'=>$authorizationLevel
        );  
    } 
header('Content-type: application/json');
echo json_encode($array);
} 
}
?>

0 个答案:

没有答案