假设我有一个这样的数组:
var someData = [ // count: 5
[ something, something, something ], // count: 3
[ something ], // count: 1
[ something, something, something, something ], // count: 4
[ ], // count: 0
[ something, something ] // count: 2
];
我需要像这样制作数组:
var myArray = [];
var firstCount = someData.lenght; // count: 5
for( a = 0; a < firstCount; a++ ) {
allObjects[a] = firstObject;
var secondCount = someData[a].lenght; // each has different
for( b = 0; b < secondCount; b++ ) {
allObjects[a] = secondObject;
}
}
我期待它如何:
allObjects = [
[ firstObject, secondObject, secondObject ], // count: 3
[ firstObject ], // count: 1
[ firstObject, secondObject, secondObject, secondObject ], // count: 4
[ ], // count: 0
[ firstObject, secondObject ], // count: 2
];
我可能在这个例子中犯了很少的(格式)错误,但我希望我的目标很明确 - 我需要将对象推送到已经在数组中的对象数组但它似乎取代了值而不是在其中添加新值。
我在这里缺少什么?
答案 0 :(得分:1)
我可能错了,但是根据您的预期结果,您似乎想要将其推送到数组的第二个索引。在这种情况下,您可以将内部for循环更改为
<div class="intpol3"><%= link_to 'Confirm my account',
confirmation_url(@resource, confirmation_token: @token) %></div>
注意b = 1和allObjects [a] [b]
的变化