Yas*_*sir 6 javascript variables scope function
我有一个变量mutedUser,我想坚持到另一个函数.我在click事件之外持久存在变量时遇到了一些麻烦.拥有它的最佳方法是什么,以便"返回mutedUser"将根据if语句的条件保持"静音"字符串添加?谢谢!
*console.log是我检查持久性停止的位置
this.isUserMuted = function isUserMuted(payload) {
var mutedUser = '';
// If mute button is clicked place them into muted users list
// check for duplicates in list
$("#messages-wrapper").off('click', '.message button.muteButton');
$("#messages-wrapper").on('click', '.message button.muteButton', function(e) {
$('#unMute').show();
//create userId reference variable
var chatUserID = parseInt($(this).parent().parent().attr("data-type"));
//store userId in muted user object
mutedUsers[chatUserID] = {};
mutedUsers[chatUserID].id = chatUserID;
mutedUsers[chatUserID].muted = true;
if (mutedUsers[chatUserID] !== null && mutedUsers[chatUserID].id === payload.a) {
console.log("user is now muted");
mutedUser += ' muted';
console.log(mutedUser + 1);
}
console.log(mutedUser + 2);
});
return mutedUser;
};
Run Code Online (Sandbox Code Playgroud)
如果我理解你要做的事情(通过查看代码),这将是最好的方法:
// If mute button is clicked place them into muted users list
// check for duplicates in list
$("#messages-wrapper").off('click', '.message button.muteButton');
$("#messages-wrapper").on('click', '.message button.muteButton', function(e) {
$('#unMute').show();
//create userId reference variable
var chatUserID = parseInt($(this).parent().parent().attr("data-type"));
//store userId in muted user object
mutedUsers[chatUserID] = {};
mutedUsers[chatUserID].id = chatUserID;
mutedUsers[chatUserID].muted = true;
});
this.isUserMuted = function isUserMuted(payload) {
var mutedUser = '';
if (mutedUsers[payload.a] !== null) {
mutedUser += ' muted';
}
return mutedUser;
};
Run Code Online (Sandbox Code Playgroud)
如果提供的用户在该数组中mutedUsers,代码将保留数组和isUserMuted函数检查.在您提供的代码中,每次isUserMuted调用函数时都会附加一个新的事件处理程序.
该isUserMuted功能甚至可以缩短为:
this.isUserMuted = function isUserMuted(payload) {
return mutedUsers[payload.a] !== null ? ' muted' : '';
};
Run Code Online (Sandbox Code Playgroud)