2015-05-25 70 views
0

我有一個內聯編輯的數據表腳本,它的工作完美。麻煩的是,Ajax沒有發送數據到POST到PHP。我不知道該怎麼做,因爲我認爲我的代碼是正確的。內聯工作,輸入顯示當我按下編輯,當我按下保存它叫我test.php但它不發佈數據。這是我的js。Datatable內聯編輯不工作,因爲它應該

JS

function editRow(oTable, nRow) { 
      var aData = oTable.fnGetData(nRow); 
      var jqTds = $('>td', nRow); 
      jqTds[0].innerHTML = '<input type="text" class="form-control input-small" id="nume" value="' + aData[0] + '">'; 
      jqTds[1].innerHTML = '<input type="text" class="form-control input-small" id="prenume" value="' + aData[1] + '">'; 
      jqTds[2].innerHTML = '<input type="text" class="form-control input-small" value="' + aData[2] + '">'; 
      jqTds[3].innerHTML = '<input type="text" class="form-control input-small" value="' + aData[3] + '">'; 
      jqTds[4].innerHTML = '<input type="text" class="form-control input-small" value="' + aData[4] + '">'; 
      jqTds[5].innerHTML = '<input type="text" class="form-control input-small" value="' + aData[5] + '">'; 
      jqTds[6].innerHTML = '<input type="text" class="form-control input-small" value="' + aData[6] + '">'; 
      jqTds[7].innerHTML = '<input type="text" class="form-control input-small" value="' + aData[7] + '">'; 
      jqTds[8].innerHTML = '<a class="edit" href="">Save</a> <a class="cancel" href="">Cancel</a>'; 

     } 

     function saveRow(oTable, nRow) { 
      var jqInputs = $('input', nRow); 
      oTable.fnUpdate(jqInputs[0].value, nRow, 0, false); 
      oTable.fnUpdate(jqInputs[1].value, nRow, 1, false); 
      oTable.fnUpdate(jqInputs[2].value, nRow, 2, false); 
      oTable.fnUpdate(jqInputs[3].value, nRow, 3, false); 
      oTable.fnUpdate(jqInputs[4].value, nRow, 4, false); 
      oTable.fnUpdate(jqInputs[5].value, nRow, 5, false); 
      oTable.fnUpdate(jqInputs[6].value, nRow, 6, false); 
      oTable.fnUpdate(jqInputs[7].value, nRow, 7, false); 
      oTable.fnUpdate('<a class="edit" href="">Edit</a>', nRow, 8, false); 
      oTable.fnDraw(); 
     } 


     table.on('click', '.edit', function (e) { 
      e.preventDefault(); 

      /* Get the row as a parent of the link that was clicked on */ 
      var nRow = $(this).parents('tr')[0]; 

      if (nEditing !== null && nEditing != nRow) { 
       /* Currently editing - but not this row - restore the old before continuing to edit mode */ 
       restoreRow(oTable, nEditing); 
       editRow(oTable, nRow); 
       nEditing = nRow; 
      } else if (nEditing == nRow && this.innerHTML == "Save") { 
       /* Editing this row and want to save it */ 
       saveRow(oTable, nEditing); 
       nEditing = null; 
       $.ajax({ 
       type: 'POST', 
       url: 'test.php', 
       data: {num: $('#nume').val()}, 
       success: function(data) 
       { 
        $("#content").html(data); 
        console.log(data); 
       } 
      }); 
      } else { 
       /* No edit in progress - let's start one */ 
       editRow(oTable, nRow); 
       nEditing = nRow; 
      } 
     }); 
+0

檢查元素以查看使用數據表進行渲染後實際製作了哪些結構......此外,這不是一個常見問題嘗試使feedle得到更好的幫助。 – mansoor

+0

結構很好,這裏是我按下編輯後生成的html ' – chris227

回答

0

Ajax調用之前試試這個,它應該工作。

var data = {}; 
       $.each(jqInputs, function(index, val) { 
        // data.push(val.value); 
        data[val.id] = val.value; 
       }); 
       // console.log(data); 
       $.ajax({ 
        url: 'controller/ajax/bars/update_bar_timing', 
        type: "POST", 
        data: {'jqInputs':data}, 
        success: function(result){ 
        } 
       }); 
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