[英]How to get ajax value and store in PHP variable?
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,则不会获取该值
$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."'";
我得到未完成的单个代码的结果
select * from ca_job WHERE job_status!='Closed' AND job_customer_name = '15
如果有人遇到此问题,请帮助我。 提前致谢。
只需使用</span>
像这样
$com = '<span id="tableCityID"></span>';
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.