問題與JavaScript可變範圍

Question

我有一個變量mutedUser，我想堅持到另一個功能。我在點擊事件之外持續存在變量時遇到了一些問題。什麼是最好的方式，以便「返回mutedUser」將保持基於if語句條件的「靜音」字符串添加？謝謝！

*本的console.log的是我檢查，看看那裏的持久性停止

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; 
}; 

Answer 1

如果我理解你正在試圖做的（通過查看代碼）是什麼，這將是最好的方法：

// 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; 
}; 

的代碼保留數組mutedUsers和isUserMuted函數檢查提供的用戶是否在該數組中。在您所提供的代碼，你會附加一個新的事件處理程序，每次isUserMuted函數被調用..

的isUserMuted功能甚至可以縮短爲：

this.isUserMuted = function isUserMuted(payload) { 
    return mutedUsers[payload.a] !== null ? ' muted' : ''; 
}; 

Answer 2

編輯

對不起，我錯了。另一種方式是在變量傳遞，即

this.isUserMuted = function isUserMuted(payload, isMuted) { 
    isMuted = ''; 
    // 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"); 
      isMuted += ' muted'; 
      console.log(mutedUser + 1); 
     } 
     console.log(mutedUser + 2); 
    }); 
    return isMuted; 
}; 

Answer 3

你不能。如果你從函數返回一個字符串，它總是按值傳遞，即複製;它的價值不會再改變。您需要返回一個可以訪問局部變量的當前值的函數，或者一個屬性正在更改的對象。

正如你似乎已經有一個對象，選擇＃2將適合在這裏很好：

function User() { // or whatever you have 
    … 

    var user = this; 
    // If mute button is clicked place them into muted users list 
    // check for duplicates in list 
    $("#messages-wrapper").on('click', '.message button.muteButton', function(e) { 

     $('#unMute').show(); 

     //store userId in muted user object 
     mutedUsers[user.id] = user; 
     user.muted = true; 
    }); 
    this.muted = false; 
    this.isUserMuted = function() { 
     return this.muted ? ' muted' : ''; 
    } 
}