在出现成功消息后尝试重新加载页面,但未重新加载。
我有这个代码
$.ajax({
type: "post",
url: '/action.cfm?method=quote',
data: datastring,
success: function(data) {
var valid = $.trim(data);
if (valid.toLowerCase().indexOf("error") == '-1') {
localStorage.setItem("swal", swal({
title: "Good job!",
text: 'Thanks',
type: "success",
showConfirmButton: true
}).then(function() {
location.reload();
}));
localStorage.getItem("swal");
} else {
swal("Oops", data, "error");
}
}
});但是我在这上面得到了一个错误
(index):389 Uncaught TypeError: Cannot read property 'then' of undefined
at Object.success ((index):389)
at fire (jquery-1.12.4.js:3232)
at Object.fireWith [as resolveWith] (jquery-1.12.4.js:3362)
at done (jquery-1.12.4.js:9840)
at XMLHttpRequest.callback (jquery-1.12.4.js:10311)发布于 2019-01-21 02:19:39
确保你使用的是新版本的type (下面,在jsfiddle中,你可以找到2.1.2) -我收到了关于使用不推荐使用的showConfirmButton和jsfiddle属性的警告。以下是在jsfiddle中工作的略微修改的代码:
$(document).ready(function() {
$.ajax({
type: "post",
url: '/echo/json/',
data: {},
success: function(data) {
var valid = $.trim(data);
if (valid.toLowerCase().indexOf("error") == '-1') {
localStorage.setItem("swal", swal({
title: "Good job!",
text: 'Thanks',
icon: "success",
button: true
}).then(function() {
console.log('sth');
location.reload();
}));
localStorage.getItem("swal");
} else {
swal("Oops", data, "error");
}
}
});
});https://jsfiddle.net/wlecki/rwc58h17/
但总的来说,如果你必须将它存储在浏览器的本地存储中,我建议你只在那里存储一个甜蜜警报选项,如果你想要显示一个甜蜜警报,就使用它们:
$(document).ready(function() {
$.ajax({
type: "post",
url: '/echo/json/',
data: {},
success: function(data) {
var valid = $.trim(data);
if (valid.toLowerCase().indexOf("error") == '-1') {
localStorage.setItem("swal", JSON.stringify({
title: "Good job!",
text: 'Thanks',
icon: "success",
button: true
}));
swal(JSON.parse(localStorage.getItem("swal"))).then(() => location.reload());
} else {
swal("Oops", data, "error");
}
}
});
});https://stackoverflow.com/questions/54277772
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