Question

我有以下JSON结构，我需要迭代data.list的嵌套值。当我使用以下declare type ab_rec is record ( a2 table2.a%type , b2 table2.b%type , a1 table1.a%type , b1 table1.b%type ); type ab_nt is table of ab_rec; l_recs ab_nt; begin WITH with_b AS ( Select A, B from Table1 ) SELECT * bulk collect into l_recs FROM (Select A, B from Table2) a, with_b b WHERE a.A = b.A(+) order by a.A; ..... end;进行硬编码时，我能够获得嵌套值，但是当我尝试迭代整个data.list对象时，我无法获得嵌套值。

console.log(data["list"][0]["My website is https://www.test.com"][0][0].command);

Answer 1

您可以使用对象中的第一个键。

var data = { list: [{ "The first website is https://www.w3.org/": [[{ command: "This is dummy content", new: false, message: "This was fun to make" }]] }, { "The second website is https://www.mozilla.org": [[{ command: "This is the second command", new: true, message: "Lorem ipsum" }]] }], verified: false },
    i;

for (i = 0; i < data.list.length; i++) {
    console.log( data.list[i][Object.keys(data.list[i])[0]][0][0].command);
}

Answer 2

因为列表中的项是对象，所以您在单独的循环中迭代它们。此外，您必须考虑到它是一个对象，因此您应该使用object属性作为索引而不是整数。

以下应该有效：

for (var i = 0; i < data.list.length; i++) {
        // this doesn't work
        for (var property in data.list[i]) {
            console.log(data.list[i][property][0][0].command);
        }
}

在JSON中访问嵌套数组对象和值

2 个答案: