当进行 ajax 调用时,请参见下面的示例 success
确实会返回 201 状态。你如何更好地处理这些,即 success 函数中的 200、201?
$.ajax({
type: "POST",
dataType: "json",
url: "http://api.domain.com/sms",
data: {
// Send value in mobile input field.
mobile: $("#mobile").val(),
},
// On successful AJAX call do the following.
success: function(data) {
$('#messageText').text('SMS successfully sent');
},
error: function(jqXhr) {
data = JSON.parse(jqXhr.responseText);
}
});
最佳答案
使用statusCode
对象:
var handle200 = function(data, textStatus, jqXHR) {
alert('200'); // success codes have the success signature
};
var handle201 = function(data, textStatus, jqXHR) {
alert('201'); // success codes have the success signature
// test it if you are in doubt:
console.log(data);
console.log(textStatus);
console.log(jqXHR);
};
var handle404 = function(jqXHR, textStatus, errorThrown) {
alert('404'); // failing codes have the error signature
});
var request = $.ajax({
type: 'POST',
url: '/myresource/posttarget',
data: { name: 'john' },
statusCode: {
200: handle200,
201: handle201,
404: handle404
}
});
关于javascript - 如何处理ajax 201,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12410051/