Question

我有这段代码，但出于某种原因，我一直在收到colours.undefined

var varray = new Array("WH", "BL");
var fname = "colours",
    lname = "size_code_id";
var arr = new Array();
var mquery = {
    "$or": []
};
for (var i = 0; i < varray.length; i++); {
    var query = {
        "$and": []
    };
    var s1 = fname.concat(".").concat(varray[i]);
    var s2 = fname.concat(".").concat(varray[i]).concat(".").concat(lname);
    var sub1 = {}, sub2 = {};

    sub1[s1] = {
        $exists: 1
    };
    sub2[s2] = "S";

    query["$and"].push(sub1);
    query["$and"].push(sub2);

    arr.push(query);
};

我想获得colours.WH.size_code_id，但不断获得colours.undefined.size_code_id

我缺少什么，任何建议非常感谢

Answer 1

在

之后删除;

for (var i = 0; i < varray.length; i++) ; {

实施例

var varray = new Array("WH", "BL");

var fname = "colours",
    lname = "size_code_id";

var arr = new Array();

var mquery = {
    "$or": []
};

for (var i = 0; i < varray.length; i++) {
    var query = {
        "$and": []
    };
    var s1 = fname.concat(".").concat(varray[i]);
    var s2 = fname.concat(".").concat(varray[i]).concat(".").concat(lname);
    var sub1 = {}, sub2 = {};

    sub1[s1] = {
        $exists: 1
    };
    sub2[s2] = "S";

    query["$and"].push(sub1);
    query["$and"].push(sub2);

    arr.push(query);
};

console.log(arr)

javascript，继续在循环中获取未定义

1 个答案: