sqlite的多表選擇與jQuery

我想創建一個Android應用程序，我必須做follwoing：sqlite的多表選擇與jQuery

我有條形碼和金額列和表billpayments條碼和payamount列兩個表法案。

如何從表格中選擇金額= barcore = table bill barcode？

這是我的代碼

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>')); 
     }

來源

2012-01-29 kosbou

加入上的條形碼列的表。 http://www.w3schools.com/sql/sql_join.asp這應該解釋加入。我認爲你想要一個內部聯接（或者只是加入，同樣的事情）。

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

退房的網站，以確保你得到你想要的。

來源

2012-01-29 19:47:09 AJcodez

sqlite的多表選擇與jQuery

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