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);
}
}
?>