Javascript 将数据ajax发布到php
我想发布一些点击删除时的值。请检查错误Javascript 将数据ajax发布到php,javascript,jquery,ajax,Javascript,Jquery,Ajax,我想发布一些点击删除时的值。请检查错误 $.ajax({ type: "POST", var checkid = $('#delete').click(function(){ $(this).val();});, url: "survey-command.php", data: { checkid: checkid, } }).done(function( msg ) {
$.ajax({
type: "POST",
var checkid = $('#delete').click(function(){ $(this).val();});,
url: "survey-command.php",
data: { checkid: checkid, }
}).done(function( msg ) {
alert( "Data Saved: " + msg );
});
我不知道如何将值传递给这个checkid变量我认为您的代码应该是这样的:
$('#delete').click(function(){
var checkid= $(this).val(); //assuming $('#delete') is an input.
// otherwise use $('#delete').html();
$.ajax({
type: "POST",
url: "survey-command.php",
data: { checkid: checkid, }
}).done(function( msg ) {
alert( "Data Saved: " + msg );
});
});
用这个
$('#delete').click(function(){
var checkid = $(this).val();
$.ajax({
type: "POST",
url: "survey-command.php",
data: { checkid: checkid, }
}).done(function( msg ) {
alert( "Data Saved: " + msg );
});
});
当您需要在按钮上触发post事件时,请从$.ajax函数中删除单击。单击此按钮是您需要使用的正确代码:
$('#delete').click(function(){
$.ajax({
type: "POST",
url: "survey-command.php",
data: { checkid: $(this).val();, }
}).success(function( msg ) {
alert( "Data Saved: " + msg );
});
});
那有什么问题?为什么你认为它不起作用?哈哈哈,Levuva PatelGr8,我们能把这个对话转到whats应用程序吗?;-)