Question

我有两个表，每个表都有唯一的ID例如：

＆＃13;

<table id="cvo">
  <tr id="1">
    <td>C</td>
  </tr>
  <tr id="2">
    <td>C</td>
  </tr>
  <tr id="3">
    <td>O</td>
  </tr>
  <tr id="4">
    <td>V</td>
  </tr>
  <tr id="5">
    <td>C</td>
  </tr>
</table>
<table id="prices">
  <tr id="1">
    <td>
      <input type="text" class="editbox " value="0.30" />
    </td>
  </tr>
  <tr id="2">
    <td>
      <input type="text" class="editbox " value="1.90" />
    </td>
  </tr>
  <tr id="3">
    <td>
      <input type="text" class="editbox " value="4.40" />
    </td>
  </tr>
  <tr id="4">
    <td>
      <input type="text" class="editbox " value="9.40" />
    </td>
  </tr>
  <tr id="5">
    <td>
      <input type="text" class="editbox " value="0.95" />
    </td>
  </tr>
</table>

＆＃13;

我希望JS根据值（C V或O）总结价格表中的值

所以在上面的例子中，JS会返回： C：3.15 V：9.40 O：4.40。

我用.find（）

尝试了以下内容

var sum = 0.0;
$('#prices > tbody  > tr').each(function() {

    var price = $(this).closest('tr').find('.editbox').val();

    var amount = Number(price)

    sum+=amount;

});

console.log(sum)

Answer 1

试试这个

function GetPriceValues()
{

    var tableCvo = $('#cvo');
    var tablePrices = $('#prices');

    var trs = tableCvo.find('tr');

    var prices = {};
    $.each(trs, function(i, tr){
        var oTr = $(tr);
        var trid = oTr.attr('id');
        var text = oTr.text();
        var price = tablePrices.find('#'+trid+' .editbox').val();

        var priceObj = prices[text];
        if(!priceObj){
            prices[text] = Number(price);
        }else{
            prices[text] = prices[text]+Number(price);
        }
    });

    var keys = Object.keys(prices);
    $.each(keys, function(i, key){
        console.log(key+":"+prices[key]);
    });

}

Answer 2

var sum = {};
var prices = $("#prices input")
$("#cvo td").each(function(index){ // for each letter
  if(!sum.hasOwnProperty($(this).text())){ // create sum of the letter if not created
    sum[$(this).text()]=0;
  }
  sum[$(this).text()]+=Number($(prices.get(index)).val()); // add the price related to the current index
})

console.log(sum)

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<table id="cvo">
  <tr id="1">
    <td>C</td>
  </tr>
  <tr id="2">
    <td>C</td>
  </tr>
  <tr id="3">
    <td>O</td>
  </tr>
  <tr id="4">
    <td>V</td>
  </tr>
  <tr id="5">
    <td>C</td>
  </tr>
</table>
<table id="prices">
  <tr id="1">
    <td>
      <input type="text" class="editbox " value="0.30" />
    </td>
  </tr>
  <tr id="2">
    <td>
      <input type="text" class="editbox " value="1.90" />
    </td>
  </tr>
  <tr id="3">
    <td>
      <input type="text" class="editbox " value="4.40" />
    </td>
  </tr>
  <tr id="4">
    <td>
      <input type="text" class="editbox " value="9.40" />
    </td>
  </tr>
  <tr id="5">
    <td>
      <input type="text" class="editbox " value="0.95" />
    </td>
  </tr>
</table>

Answer 3

var sum = {};

$('.editbox').each(function() {
    var id = $(this).closest('tr').attr('data-id');
    code = $('#' + id).text(); //ignore TD and just return text inside
    if (!sum[code]) { sum[code] = 0; }
    sum[code] += parseFloat($(this).val());
});

var result = '', i;
for (i in sum) {
    result += i + ':' + sum[i] + ' ';
}

您在第二个表中使用重复的ID，它应该与第一个表匹配：

<table id="prices">
    <tr data-id="1">

仅在理论上有效，没有测试，对不起。

使用jQuery在嵌套表中总计值

3 个答案: