JS對象數組刪除最后一個空元素

Question

我有那個javascript對象：

MyObject = [
   {
     "name" : "aaa",
     "firstname" : "aaaaa"
   },
   {
     "name" : "bbb",
     "firstname" : "bbbb"
   },
   {
     "name" : "cccc",
     "firstname" : "" <--------- firstname is empty, but the element is not in the last
   },
   {
     "name" : "ddd",
     "firstname" : "dddd"
   },
   {
     "name" : "eeee",
     "firstname" : ""  <--------- firstname is empty
   },
   {
     "name" : "fffff",
     "firstname" : ""  <--------- firstname is empty
   },
]

我想刪除具有“ firstname”為空的最新行 ...（一種修整）...我不想刪除具有“ firstname”為空的所有行...，而是那些最新行中的人。 （在底部）

因此，結果將是：

MyObject = [
   {
     "name" : "aaa",
     "firstname" : "aaaaa"
   },
   {
     "name" : "bbb",
     "firstname" : "bbbb"
   },
   {
     "name" : "cccc",
     "firstname" : ""
   },
   {
     "name" : "ddd",
     "firstname" : "dddd"
   }
]

謝謝

Answer 1

您可以在數組末尾彈出，只要名字為空

 var MyObject = [{ "name": "aaa", "firstname": "aaaaa" }, { "name": "bbb", "firstname": "bbbb" }, { "name": "cccc", "firstname": "" }, { "name": "ddd", "firstname": "dddd" }, { "name": "eeee", "firstname": "" }, { "name": "fffff", "firstname": "" }]; for (var i=MyObject.length;i--;) if (MyObject[i].firstname==="") MyObject.pop(); else break; console.log(MyObject) 

 .as-console-wrapper {max-height: 100%!important; top: 0!important;} 

Answer 2

嘗試這個：

 var MyObject = [ { "name" : "aaa", "firstname" : "aaaaa" }, { "name" : "bbb", "firstname" : "bbbb" }, { "name" : "cccc", "firstname" : "" }, { "name" : "ddd", "firstname" : "dddd" }, { "name" : "eeee", "firstname" : "" }, { "name" : "fffff", "firstname" : "" } ]; var i = MyObject.length; while(true) { if (!MyObject[--i].firstname) { MyObject.pop(); } else { break; } } console.log(MyObject);