php - 如何获取ajax值并存储在PHP变量中？

Question

custom.js 文件:

$(document).ready(function() {
    $("#company_name").keyup(function() {
        $.ajax({
            type: "POST",
            url: "http://localhost/capms_v2/ca_autocomplete/getcompanyName",
            data: {
                keyword: $("#company_name").val()
            },
            dataType: "json",
            success: function(data) {
                //alert(data);
                if (data.length > 0) {
                    $('#DropdownCompany').empty();
                    $('#company_name').attr("data-toggle", "dropdown");
                    $('#DropdownCompany').dropdown('toggle');

                } else if (data.length == 0) {
                    $('#company_name').attr("data-toggle", "");
                }
                $.each(data, function(key, value) {
                    if (data.length >= 0)
                        $('#DropdownCompany').append('<li role="displayCountries" ><a role="menuitem DropdownCompany" id=' + value['company_id'] + ' Address1=' + value['company_address1'] + ' Address2=' + value['company_address2'] + ' city=' + value['company_city'] + ' state=' + value['company_state'] + ' pincode=' + value['company_zip'] + '  class="dropdownlivalue">' +
                            value['company_name'] + '</a></li>');
                });
            }
        });
    });
    $('ul.txtcountry').on('click', 'li a', function() {
        $('#company_name').val($(this).text());
        $('#company_id').val($(this).attr("id"));
        // $('#company_address1').val($(this).text());

        $('#tableCityID').html($(this).attr("id"));
        $('#tableCityName').html($(this).text());
        $('#Address1').html($(this).attr("Address1"));
        $('#Address2').html($(this).attr("Address2"));
        $('#city').html($(this).attr("city"));
        $('#state').html($(this).attr("state"));
        $('#pincode').html($(this).attr("pincode"));
    });
});

我在span id="tableCityID"中获取id，但是如果我存储该值并将该值传递给mysql，它就不会获取该值

$com = '<span id="tableCityID">'; 

如果我回显选择查询

echo $sql="select * from ca_job WHERE job_status!='Closed' AND job_customer_name = '".$com."'";

我得到的结果是未完成的单个代码

select * from ca_job WHERE job_status!='Closed' AND job_customer_name = '15

如果有人遇到这个问题，请帮助我。提前致谢。

Answer 1

只需使用 </span>

像这样

$com = '<span id="tableCityID"></span>';