ajax成功函数从php接收信息
我无法使用ajax从php脚本接收数据。我有一个表单将信息发送到php脚本。这只是一个测试。如果它起作用,我想从数据库接收信息ajax成功函数从php接收信息,php,ajax,callback,Php,Ajax,Callback,我无法使用ajax从php脚本接收数据。我有一个表单将信息发送到php脚本。这只是一个测试。如果它起作用,我想从数据库接收信息 $('.search-member-submit-btn').click(function () { var firstname = $("#search-member-firstname").val(); var lastname = $("#search-member-lasttname").val(); $.ajax({//AJA
$('.search-member-submit-btn').click(function () {
var firstname = $("#search-member-firstname").val();
var lastname = $("#search-member-lasttname").val();
$.ajax({//AJAX request
type: "POST",
url: "/website/include/process/send_membersearchrequest_process.php",
async: true,
data: {firstname: firstname, lastname: lastname},
success: function (data) {
alert(data);
$(".search-member-result-address").html(data);
},
});
});
HTML:
苏晨
PHP
我宁愿这样做:
Javascript
$.ajax({//AJAX request
type: "POST",
url: "/website/include/process/send_membersearchrequest_process.php",
async: true,
dataType:'json'
data: {"firstname": firstname, "lastname": lastname},
success: function (data) {
alert(data.result);
$(".search-member-result-address").html(data.result);
},
});
PHP
像这样,您向服务器发送一个json,并得到一个json作为回报。添加响应类型,如:
$.ajax({//AJAX request
type: "post",
url: "/website/include/process/send_membersearchrequest_process.php",
async: true,
data: {firstname: firstname, lastname: lastname},
dataType: 'json', // You have to add this line
success: function (data) {
alert(data.response);
$(".search-member-result-address").html(data.response);
},
});
以及您的PHP文件:
<?php
$lastname = $_POST['lastname'];
echo json_encode(array('response' => $lastname));
exit();
?>
虽然这是一种很好的风格,但它不能解释为什么他的原始代码不起作用。AJAX调用中缺少数据类型:“json”
。如果他的原始、更简单的代码不起作用,为什么您认为这会起作用?数据应该是数据。响应.thx以获得支持。现在它会提醒[对象]。为什么不显示lastname字符串?alert(data.response);代替警报(数据);检查开发人员工具的网络选项卡,查看AJAX调用的原始响应。alert()
是否触发?它显示了什么?
<?php
$lastname = $_POST['lastname'];
$res = array("result" => $lastname);
echo json_encode($res);
?>
$.ajax({//AJAX request
type: "post",
url: "/website/include/process/send_membersearchrequest_process.php",
async: true,
data: {firstname: firstname, lastname: lastname},
dataType: 'json', // You have to add this line
success: function (data) {
alert(data.response);
$(".search-member-result-address").html(data.response);
},
});
<?php
$lastname = $_POST['lastname'];
echo json_encode(array('response' => $lastname));
exit();
?>