如果子字符串爲空並避免控制檯出錯，如何跳過？

Question

抓住子：

var hash = document.location.hash; 

// create an object to act like a dictionary to store each value indexed by its key 
var partDic = {}; 

// remove the leading "#" and split into parts 
var parts = hash.substring(1).split('&'); 

// If you just want the first value, whatever it is, use this. 
// But be aware it's a URL so can be set to anything in any order, so this makes little sense 
// var string = parts[0].split('=')[1]; 

// build the dictionary from each part 
$.each(parts, function(i, v) { 
    // do the "=" split now 
    var arr = v.split("="); 

    // decode to turn "%5B" back into "[" etc 
    var key = decodeURIComponent(arr[0]); 
    var value = decodeURIComponent(arr[1]); 

    // store in our "dictionary" object 
    partDic[key] = value; 
}); 

// Set a delay to wait for content to fully load 
setTimeout(function() { 
    var ag = partDic["comboFilters[Agencies]"].substring(1); 
    $('.Agency .dropdown-toggle').html(ag).append(' <span class="caret"></span>'); 
    var cl = partDic["comboFilters[Clients]"].substring(1); 
    $('.Client .dropdown-toggle').html(cl).append(' <span class="caret"></span>'); 
    var yr = partDic["comboFilters[Years]"].substring(1).slice(1); 
    $('.Year .dropdown-toggle').html(yr).append(' <span class="caret"></span>'); 
}, 1000); 

但是，如果沒有一個子，我越來越：

Uncaught TypeError: Cannot read property 'substring' of undefined 

Suggested answer in another question

var cl = (partDic["comboFilters[Clients]"] && partDic["comboFilters[Clients]"].length>0)?partDic["comboFilters[Clients]"].substring(1):''; 

但我仍然得到同樣的錯誤

Answer 1

你可以防守，並檢查一個關鍵前在使用它之前派：

if("comboFilters[Agencies]" in partDic) { 
     var ag = partDic["comboFilters[Agencies]"].substring(1); 
     $('.Agency .dropdown-toggle').html(ag).append(' <span class="caret"></span>'); 
    } 

或只是一個空字符串維護它：

var ag = (partDic["comboFilters[Agencies]"] || "").substring(1); 

Answer 2

你可以嘗試檢查它的類型：

var cl = (typeof partDic["comboFilters[Clients]"] === 'string')?partDic["comboFilters[Clients]"].substring(1):''; 

注意，你應該添加此檢查所有的變量：ag，cl，yr

Answer 3

也許事情就像嘗試：

var parts = (hash && hash.substring(1).split('&')) || []; 

Answer 4

可以使用子方法之前檢查一個條件..

if((!hash) || (!hash.substring(1)){ 
return false; 
}