Question

我有一个json变量

 var a = { Account: [ 'Name', 'Site', 'BillingState', 'Phone', 'Type', 'Owner.Alias' ],

联系人：['Name1'，'Site1']}

我正在像这样迭代循环

app.post('/', function (req, res) {
a = req.body.table_structure
b = createSelectQuery(a)  res.send(b) // this return only "FROM CONTACT" })

我已经创建了另一个函数，希望我可以获得所需的字符串，但循环变为异步

function createSelectQuery(a){
        var selectQuery = "Select " 
        var obj = JSON.parse(a);
        for (var c in obj){
            for (var d in obj[c]){
                selectQuery += obj[c][d] + ', '
            }
            selectQuery = "FROM "+c
        }
        return selectQuery
    }

我期待结果“来自帐户的SELECT Name，..等但是得到了仅从帐户作为输出

Answer 1

你需要使用回调来使你的循环同步

Answer 2

我相信因为在JSON文件＆＃39;帐户＆＃39;是数组，你尝试循环到它，就好像它是一个对象。试试：

function createSelectQuery(a){
        var selectQuery = "Select " 
        var obj = JSON.parse(a);
        for (var c in obj){
            for (var d=0; d < obj[c].length; d++){
                selectQuery += obj[c][d] + ', '
            }
            selectQuery = "FROM "+c
        }
        return selectQuery
    }

Answer 3

试试这个

function createSelectQuery(a){
    var selectQuery = "Select "; 
    var obj = JSON.parse(a);
    for (var c in obj) {
        selectQuery += obj[c].join(", ") + " FROM " + c;
    }
    return selectQuery;
}

Answer 4

您可以尝试使用arr.join（）连接字符串，然后使用Object.keys追加字符串，而不是循环。这样就可以了解

var selectQuery = "Select ";
    var a = { Account: [ 'Name', 'Site', 'BillingState', 'Phone', 'Type', 'Owner.Alias' ]};
    var obj = a;
    var arr = obj['Account'];
    var joinStr = arr.join(',').toString();
    selectQuery = selectQuery +" "+joinStr +" FROM "+Object.keys(a);
    console.log(selectQuery);

如何在nodejs中迭代循环

4 个答案: