我使用了与其他方法不同的方法,因为我做了两个假设:
1/您不知道密钥的名称,但每个项目只有一个密钥 2/每个项目的密钥都可以不同 我会给你第二个选择,第二个假设是:2/所有项目都有只有一把钥匙,但对所有人来说都是一样的
第一个选项:
var items = [
{name:"Foo"},
{name:"Bar"},
{name:"foo"},
{name:"bar"},
{name:"foobar"},
{name:"barfoo"}
];
// object keys very simple shim
Object.keys = Object.keys || function(o) {
var result = [];
for(var name in o) {
if (o.hasOwnProperty(name))
result.push(name);
}
return result;
};
// function to get the value of every first keys in an object
// just remember that saying "first key" does not make real sense
// but we begin with the assumption that there IS ONLY ONE KEY FOR EVERY ITEM
// and this key is unknown
function getFirstKeysValues(items) {
var i = 0, len = items.length, item = null, key = null, res = [];
for(i = 0; i < len; i++) {
item = items[i];
key = Object.keys(item).shift();
res.push(item[key]);
}
return res;
}
console.log(getFirstKeysValues(items)); //["Foo", "Bar", "foo", "bar", "foobar", "barfoo"]
第二个选项将使用映射,因为我们相信每个孩子都拥有相同的密钥(我不会使用这个,因为我不太喜欢 .map - 兼容性):
var items = [
{name:"Foo"},
{name:"Bar"},
{name:"foo"},
{name:"bar"},
{name:"foobar"},
{name:"barfoo"}
];
// object keys very simple shim
Object.keys = Object.keys || function(o) {
var result = [];
for(var name in o) {
if (o.hasOwnProperty(name))
result.push(name);
}
return result;
};
// function to get the value of every first keys in an object
// just remember that saying "first key" does not make real sense
// but we begin with the asumption that there IS ONLY ONE KEY FOR EVERY ITEM
// and this key is unknown but the same for every child
function getFirstKeysValues(items) {
var key = items.length > 0 ? Object.keys(items[0]).shift() : null;
items = items.map(function (item) {
return item[key];
});
return items;
}
console.log(getFirstKeysValues(items));