在前面

Question

添加里面的Json新對象，我有一個JSON對象，像這種

[{"SubLoc":"a","Description":"A","Equipment":""},{"SubLoc":"b","Description":"B","Equipment":""},{"SubLoc":"c","Description":"C","Equipment":""},{"SubLoc":"d","Description":"D","Equipment":""}] 

我要在前面加一個多個屬性，使得JSON看起來像

[{"SubLoc":"Select","Description":"Select","Equipment":""},{"SubLoc":"a","Description":"A","Equipment":""},{"SubLoc":"b","Description":"B","Equipment":""},{"SubLoc":"c","Description":"C","Equipment":""},{"SubLoc":"d","Description":"D","Equipment":""}] 

我試着不印字像這個 -

$scope.JsonVar.unshift({SubLoc:'Select', Description:'Select'}); 

但它給我造成這樣的...

[{"SubLoc":"Select","Description":"Select","Equipment":""}[{"SubLoc":"a","Description":"A","Equipment":""},{"SubLoc":"b","Description":"B","Equipment":""},{"SubLoc":"c","Description":"C","Equipment":""},{"SubLoc":"d","Description":"D","Equipment":""}]] 

Answer 1

unshift()方法返回數組的新長度，而不是數組本身。所以你不能將結果重新分配給$scope.JsonVar。就像這樣使用它：

$scope.JsonVar.unshift({SubLoc:'Select', Description:'Select'});