Question

我正在尝试以JSON格式获取存储在表中的数据，但是当它使用.each来获取数据时，最新值一直在重复。

我已经显示了实际输出，该输出应该与表行中的数据不同

function getPackTableDetails() {
                var packDetailsRow = [];
                var tableRowCount = $('#addPackTable tr').length;
                var returnJSONFormat = "";
                    $('#addPackTable tr').each(function(index) {
                        if(index != 0) {
                            var packDetailsObj = {};
                            ($(this).find("td")).each(function(idx) {
                                switch(idx) {
                                    case 0:
                                        packDetailsObj.table_pack_name = $('#table_pack_name').val();
                                    break;
                                    case 1:
                                        packDetailsObj.table_pack_type = $('#table_pack_type').val();
                                    break;
                                    case 2:
                                        packDetailsObj.base_amount = $('#base_amount').val();
                                    break;
                                    case 3:
                                        packDetailsObj.tax_amount = $('#tax_amount').val();
                                    break;
                                    case 4:
                                        packDetailsObj.total_amount = $('#total_amount').val();
                                    break;
                                }
                            });
                            packDetailsRow.push(packDetailsObj);
                        }
                    });
                    // console.log(JSON.stringify(packDetailsRow));
                    return packDetailsRow;
            }

当我使用输入类型文本时，我使用值来存储数据，然后使其检索到以下jquery代码

Output:
{
  "pack_details": [
    {
      "table_pack_name": "Demo Pack Name5",
      "table_pack_type": "Demo Pack Name5",
      "base_amount": "200.00",
      "tax_amount": "0.00",
      "total_amount": "200.00"
    },
    {
      "table_pack_name": "Demo Pack Name5",
      "table_pack_type": "Demo Pack Name5",
      "base_amount": "200.00",
      "tax_amount": "0.00",
      "total_amount": "200.00"
    },
    {
      "table_pack_name": "Demo Pack Name5",
      "table_pack_type": "Demo Pack Name5",
      "base_amount": "200.00",
      "tax_amount": "0.00",
      "total_amount": "200.00"
    }
  ]
}

实际输出：

{
  "pack_details": [
    {
      "table_pack_name": "Demo Pack Name5",
      "table_pack_type": "Demo Pack Name5",
      "base_amount": "20.00",
      "tax_amount": "10.00",
      "total_amount": "30.00"
    },
    {
      "table_pack_name": "Demo Pack Name6",
      "table_pack_type": "Demo Pack Name6",
      "base_amount": "200.00",
      "tax_amount": "0.00",
      "total_amount": "200.00"
    },
    {
      "table_pack_name": "Demo Pack Name7",
      "table_pack_type": "Demo Pack Name7",
      "base_amount": "400.00",
      "tax_amount": "40.00",
      "total_amount": "440.00"
    }
  ]
}

Answer 1

因为您每次迭代都分配相同的值

switch(idx) {
  case 0:
    packDetailsObj.table_pack_name = $('#table_pack_name').val();
    break;
  case 1:
    packDetailsObj.table_pack_type = $('#table_pack_type').val();
    break;
  case 2:
    packDetailsObj.base_amount = $('#base_amount').val();
    break;
  case 3:
    packDetailsObj.tax_amount = $('#tax_amount').val();
    break;
  case 4:
    packDetailsObj.total_amount = $('#total_amount').val();
    break;
}

Answer 2

应遵循的几件事是id应该是元素唯一的，并且两个元素都不应具有相同的id。检查this以进一步阅读。您可以改为使用name属性读取值，如下所示。

$('#addPackTable tr').each(function (index) {
  if (index != 0) {
    var packDetailsObj = {};
    ($(this).find("td")).each(function (idx) {
      switch (idx) {
        case 0:
          packDetailsObj.table_pack_name = $(this).find('[name="table_pack_name"]').val();
          break;
        case 1:
          packDetailsObj.table_pack_type = $(this).find('[name="table_pack_type"]').val();
          break;
        case 2:
          packDetailsObj.base_amount = $(this).find('[name="base_amount"]').val();
          break;
        case 3:
          packDetailsObj.tax_amount = $(this).find('[name="tax_amount"]').val();
          break;
        case 4:
          packDetailsObj.total_amount = $(this).find('[name="total_amount"]').val();
          break;
      }
    });
    packDetailsRow.push(packDetailsObj);
  }

想要从输入中获取数据并以JSON构造

2 个答案: