JavaScript范围和Ajax调用无法正常工作
[英]Javascript scope and Ajax call doesn't work correctly
问题描述
I have some problems. 
我有一些问题。
First of all, my php file shoutbox.php, it's used for to log in an user. 首先，我的php文件shoutbox.php，用于登录用户。
session_start();
if(! isset($_SESSION["loggedIn"])) {
include "login.html";
}
else {
if(isset($_POST["submit"]) && $_POST["submit"] == "Log In") {
if(! isset($_POST["username"]) || $_POST["username"] == "" || ! isset($_POST["password"]) || $_POST["password"] == "") {
$GLOBALS["logErr"] = "Please, filln all fields!";
echo json_encode(false);
}
else if(isset($_POST["username"]) && $_POST["username"] != "" && isset($_POST["password"]) && $_POST["password"] != "") {
if(isset($_POST["select"]) && $_POST["select"] != "") {
session_start();
$_SESSION["loggedIn"] = TRUE;
$_SESSION["username"] = $_POST["username"];
$_SESSION["password"] = $_POST["password"];
$_SESSION["type"] = $_POST["select"];
$_SESSION["ip"] = get_ip_address(); /* Return current ip */
$user = array(
"username" => $_SESSION["username"],
"islogged" => $_SESSION["loggedIn"],
"password" => $_SESSION["password"],
"ip" => $_SESSION["ip"],
"type" => $_SESSION["type"]
);
echo json_encode($user);
}
}
}
else if(isset($_POST["logout"]) && $_POST["logout"] == "Log Out") {
session_start();
unset($_SESSION["loggedIn"]);
unset($_SESSION["username"]);
unset($_SESSION["password"]);
unset($_SESSION["type"]);
unset($_SESSION["ip"]);
echo json_encode(false);
}
if(isset($_SESSION["loggedIn"]) && $_SESSION["loggedIn"] == TRUE) {
echo json_encode(true); /* user is logged in */
}
else {
echo json_encode(false); /* user is not logged in */
}
}
It returns false or true and if is true it returns user array encoded. 它返回false或true，如果为true，则返回用户数组编码。
This is the Ajax call, I have to call 2 responses, false or true and if response is true, also user array.. 这是Ajax调用，我必须调用2个响应，即false或true，如果响应为true，则还要调用用户数组。
function sb_UserIsLoggedIn() {
$.ajax({
url: 'shoutbox.php',
type: 'POST',
dataType: 'json',
success: function(response, user) {
if(response) {
function user(user) {
CurrUserInfo = {
u_name: user["username"],
u_psw: user["password"],
ip: user["ip"],
typelog: user["type"],
logged: user["islogged"]
};
if(CurrUserInfo.typelog == "anonimous") {
OnlineListObj.onlineList.anonimous.push(user["username"]);
}
else if(CurrUserInfo.typelog == "visible") {
OnlineListObj.onlineList.visible.push(user["username"]);
}
OnlineListObj["onlineList"]["total"] = OnlineListObj.onlineList.anonimous.length + OnlineListObj.onlineList.visible.length;
OnlineListObj["onlineList"]["phrase"] = lang["there_are"] + OnlineListObj.onlineList.total + lang["online"] + OnlineListObj.onlineList.anonimous.length + lang["anonimous"] + OnlineListObj.onlineList.visible + lang["visibles"];
LoggedIn = true;
window.alert(LoggedIn);
window.alert(OnlineListObj["onlineList"]["total"]);
window.alert(OnlineListObj["onlineList"]["phrase"]);
window.alert(CurrUserInfo.ip);
window.alert(CurrUserInfo.typelog);
window.alert(CurrUserInfo.logged);
window.alert(CurrUserInfo.u_name);
window.alert(CurrUserInfo.u_psw);
}
}
else {
window.location.href = "shoutbox.php";
}
},
error: function(response) {
sb_Error("Unknown error, try again");
console.log(lang["ajax_error"]);
console.log(response.responseText);
LoggedIn = false;
},
});
}
Alerts say undefined for these: 警报说这些没有定义：
window.alert(CurrUserInfo.ip);
window.alert(CurrUserInfo.typelog);
window.alert(CurrUserInfo.logged);
window.alert(CurrUserInfo.u_name);
window.alert(CurrUserInfo.u_psw);
This because the call to user array fails...but a call works correctly, it returns true or false.. Why does it work and the other (to user array) not? 这是因为对用户数组的调用失败了……但是一个调用可以正常工作，返回true或false。为什么它起作用，而另一个（对用户数组）不起作用？ I'd like ajax call true or false and user array. 我想ajax调用true或false和用户数组。
Other problem: 其他问题：
In code about sb_isUserLoggedIn() I defined a var, LoggedIn... 在有关sb_isUserLoggedIn（）的代码中，我定义了一个变量LoggedIn ...
But if you call LoggedIn in another function, returns undefined: 但是，如果您在另一个函数中调用LoggedIn，则返回undefined：
/* Checks if the user is online */
function sb_isOn() {
sb_UserIsLoggedIn();
if(LoggedIn != false) {
return true;
}
else {
return false;
}
}
As I written, this code returns an error, LoggedIn is undefined. 如我所写，此代码返回错误，LoggedIn未定义。
How can I resolve all these problems? 我该如何解决所有这些问题？
Thanks in advance. 提前致谢。
1 个解决方案
解决方案1
0 已采纳 2015-04-05 00:50:49
It looks to me that the PHP makes two calls to echo json_encode() when it is successful. 在我看来，PHP成功执行了两次调用以echo json_encode() 。 One to echo the user array and one to echo true . 一种回应用户数组，另一种回应true 。 I think that results in invalid JSON. 我认为这会导致JSON无效。
I suggest you write the PHP so it makes one and ONLY one call to echo json_encode() . 我建议您编写PHP，以便它仅使一次调用echo json_encode() 。 You could have it return the following for success: 您可以让它返回以下成功信息：
echo json_encode(array("success" => true, "user" => $user));
And the following for failure: 以及以下失败信息：
echo json_encode(array("success" => false, "message" => 'Please, fill in all fields!'));
Then in the Javacript, you can have something like the following: 然后，在Javacript中，您可以看到以下内容：
success: function(result) {
if (result.success) {
var user = result.user;
// Do something with user here.
} else {
var message = result.message;
// Do something with message here.
}
}
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