jquery每个函数的结果怎么办?

wvt8vs2t  于 2021-06-18  发布在  Mysql
关注(0)|答案(1)|浏览(248)

我必须构建一个表单,我一直在处理jquery,但是如果你能给我一些建议的话,我会被困在一个步骤中,
所以让我解释一下,我有20种产品,有offer1,offer2,而月费这一切到现在都是完美的工作,我只是需要在你选择了具体产品的金额,当你按下下一步按钮再次显示结果,但只有在那里的金额已填写,所以基本上只有产品的客户感兴趣的是他们,这是我的主要问题,我知道我必须在jquery中处理每个函数,但我不知道如何处理它?
我已经附上了一张图片的形式,以便您可以看到一个例子形式的图片
编辑:这是我目前的代码:

function convert(value) {
        return "" + ((Number(value) || 0).toFixed(2).replace(/(\d)(?=(\d{3})+ 
        (?!\d))/g, "$1,")) + "";
    }

    function testing(id, normaloffer, talesalesoffer, montlyoffer) {
        var normaoffertotal = normaloffer * $('#amount' + id).val();
        var talesalestotal = talesalesoffer * $('#amount' + id).val();
        var montlytotal = montlyoffer * $('#amount' + id).val();
        $(".normaloffertotal" + id).html(normaoffertotal);
        $(".telesalesoffertotal" + id).html(talesalestotal);
        $(".montlytotal" + id).html(montlytotal);
        $('#hnoffer' + id).val(normaoffertotal);
    }

    console.log(convert(10496.470000000001));

    function normaloffertotalcalc() {
        var sumnormaltotal = 0;
        $('span[id^="normaloffertotalspan"]').each(function () {
            var text = $(this).text();
            sumnormaltotal += parseFloat(text, 10);
        });
        $('#sumnormaltotal').html(sumnormaltotal);

        var sum_telesale = 0;
        $('span[id^="telesalesoffertotalspan"]').each(function () {
            var text = $(this).text();
            sum_telesale += parseFloat(text, 10);
        });
        $('#sumtelesaletotal').html(sum_telesale);

        var sum_montly = 0;
        $('span[id^="montlyofferspan"]').each(function () {
            var text = $(this).text();
            sum_montly += parseFloat(text, 10);
        });
        $('#summontlytotal').html(sum_montly);

        console.log('Normal Total Offer ' + convert(sum_normal));
        console.log('TeleSales Total Offer ' + convert(sum_telesale));
        console.log('Montly Total Offer ' + convert(sum_montly));
    }
</script>

这是html

<?php
    $count = 0;
    $query = mysqli_query($con, "SELECT * FROM products");
    while ($row = mysqli_fetch_array($query)) {
    $count++;
?>
<tr>
    <td align="center"><?= $row[1] ?></td>
    <td align="center"><input type="text" placeholder="0" id="amount<?= $count ?>"
                                              name="amount<?= $count ?>" onblur="normaloffertotalcalc()"
                                              onkeyup="testing(<?= $count ?>, <?= $row[2] ?>, <?= $row[3] ?>, <?= $row[4] ?>)"
                                              style="-webkit-appearance: none; background: none; text-align: center; border: none; width: 40px;">
     </td>
     <td align="center">£<?= $row[2] ?></td>
     <td align="center">£<span class="normaloffertotal<?= $count ?>"
                                              id="normaloffertotalspan">0</span> <input type="hidden"
                                                                                        class="normaloffertotalspan"
                                                                                        id="hnoffer<?= $count ?>">
      </td>
      <td align="center">£<?= $row[3] ?></td>
      <td align="center">£<span class="telesalesoffertotal<?= $count ?>" id="telesalesoffertotalspan">0</span>
      </td>
      <td align="center">£<?= $row[4] ?></td>
      <td align="center">£<span class="montlytotal<?= $count ?>" id="montlyofferspan">0</span></td>
                </tr>
      <?php } ?>
dxxyhpgq

dxxyhpgq1#

我找到了解决这个问题的办法

$('input[type=text]').each(function(){
            var val = parseInt($(this).val());

            if(val > 0){
                var id = $(this).attr('id');
            if(id){
                console.log(id);
            }
                console.log(val);

            }
        });

这就是我所做的代码,并且工作得非常完美:)
谢谢所有给我建议的人
还有一件事,你知道我怎样才能从mysql只导出id为2,7,1的产品吗?例如:)

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