我试图用get请求调用一个php脚本,并使用数据地址,但结果显示了我调用的页面的来源。
我呼叫的页面在这里这是我的ajax函数,它将获得这个页面
$( document ).ready(function() {
var address = document.getElementById("address");
$.ajax({
url: '/r10database/checkSystem/ManorWPG.php',
type: 'GET',
data: 'theaddress='+address.value,
cache: false,
success: function(output)
{
alert('success, server says '+output);
}, error: function()
{
alert('Something went wrong, saving system failed');
}
});
});
$( document ).ready(function() {
var address = document.getElementById("address");
$.ajax({
url: '/r10database/checkSystem/ManorWPG.php',
type: 'GET',
data: 'theaddress='+address.value,
cache: false,
success: function(output)
{
alert('success, server says '+output);
}, error: function(error)
{
alert (error); // this is the change from the question
}
});
});
将dataType设置为带大括号的json
data: {theaddress:address.value},
dataType:'json',
success: function(output)
{
alert('success, server says '+output);
}, error: function(xhr)
{
alert (xhr.status);
}
并将ManorWPG.php中的数据获取为$_get['headdress']
**如果失败,则共享xhr.status。