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我有一个包含天数差异的文本字段。我需要用它来计算我页面上的一些东西。但是，由于某种原因，undefined当我执行console.log(). 有任何想法吗？

<script>
        $(document).on('change', '#unitselector', function()
        {
            var unit = $('select option:selected').text();
            var rate = $(this).val();
            var tax = $('select option:selected').attr('label');
            var depart=$('#departdate').val();
            var arrival = $('#arrivaldate').val();
            var dataString = {unit:unit, depart:depart, arrival:arrival};
            console.log(dataString);
            //console.log($('.extracharges').serialize());
            $.ajax({
                type: "POST",
                url: "classes/unit_info.php",
                data: dataString,
                cache: false,
                success: function(html)
                {
                    $('.unitinfolist').html(html);
                }
            });

            var days = $('#days').attr('name');
            var days2 = $('#days').attr('value');
            console.log('days: '+days);
            console.log('days: '+ days2 );
            $('.rentalcharges').find('#rent').val(rate*days);
        });
    </script>

和 PHP

$arrive = $_POST['arrival'];
$leave = $_POST['depart'];

$a = date_create($arrive);
$d = date_create($leave);
$diff = date_diff($d,$a);

echo '<input type="text" id="days" value="'.$diff->d.'" name="'.$diff->d.'" />';
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1 回答 1

1

ajax 调用是异步的，因此即使您在 ajax 调用之后编写了日志记录代码，您也会在值从服务器返回之前对其进行记录。

将您的代码移动到 ajax 成功处理程序中，如下所示：

<script>
        $(document).on('change', '#unitselector', function()
        {
            var unit = $('select option:selected').text();
            var rate = $(this).val();
            var tax = $('select option:selected').attr('label');
            var depart=$('#departdate').val();
            var arrival = $('#arrivaldate').val();
            var dataString = {unit:unit, depart:depart, arrival:arrival};
            console.log(dataString);
            //console.log($('.extracharges').serialize());
            $.ajax({
                type: "POST",
                url: "classes/unit_info.php",
                data: dataString,
                cache: false,
                success: function(html)
                {
                    $('.unitinfolist').html(html);
                    var days = $('#days').attr('name');
                    var days2 = $('#days').attr('value');
                    console.log('days: '+days);
                    console.log('days: '+ days2 );
                    $('.rentalcharges').find('#rent').val(rate*days);
                }
            });

        });
    </script>
于 2013-10-30T23:59:42.403 回答