使用jquery选择sqlite多表

我正在尝试创建一个android应用程序，我必须执行以下操作：

我有两个表格，分别是带有条形码和金额列的bill和带有条形码和payamount列的table billpayments

如何从表格billpayments amount中选择barcore=表格bill barcode

这是我的密码

function getItem1(flatname) {
var total = 0;
var payment = 0;
var previous = 0;
$.mobile.notesdb.transaction(function(t) {
    t.executeSql('SELECT barcode, buildingcode, buildingaddress, flatname, flatdescription, entryseason, period, amount FROM bill WHERE flatname = ?',
    [flatname],
    function(t, result) {
        var i,
            len = result.rows.length,
            row,
            rowb;


        for (i = 0; i < len; i += 1) {
            row = result.rows.item(i);
            var html = '<input type="checkbox" name="code_'+ i +'" id="code_'+ i +'" value="' + row.amount + '"/><label for="code_'+ i +'">' + row.period +'..........'+ row.amount+'</label></br>';
            $('#Code').append($(html));

            total = total + row.amount;


        function previousPayments() {
            $.mobile.notesdb.transaction(function(t) {
                t.executeSql('SELECT barcode, amount FROM billpayments WHERE barcode=?',[row.barcode],
                    function(t, resultpayment) {
                        var ii,
                            lenb = resultpayment.rows.length,
                            rowb;
                        for (ii = 0; ii < lenb; ii += 1) {
                            var rowb = resultpayment.rows.item(ii);
                            $('#displayflat article').append($('<p>' + rowb.barcode + '</p>'));
                            previous = previous + rowb.amount;
                        }
                    });
            });
        }

        previousPayments();
        $('#displayflat article').append($('<p>' + row.barcode + '</p>'));
        }

---

连接条形码列上的表。这应该可以解释连接。我想你想要一个内在的连接或者只是连接，同样的事情

SELECT column_name(s)
FROM table_name1         /* billpayments */
INNER JOIN table_name2   /* bills */
ON table_name1.column_name=table_name2.column_name

查看网站，确保你得到了你想要的东西