Javascript AJAX如何隐藏处于警报状态的div
我是ajax新手,我做得还不错,但问题是我让ajax向服务器发送一些数据,然后获取一些数据库。我希望它隐藏error div,然后执行警报,但当我运行代码时,警报首先显示,然后区域被设置为隐藏Javascript AJAX如何隐藏处于警报状态的div,javascript,ajax,Javascript,Ajax,我是ajax新手,我做得还不错,但问题是我让ajax向服务器发送一些数据,然后获取一些数据库。我希望它隐藏error div,然后执行警报,但当我运行代码时,警报首先显示,然后区域被设置为隐藏 if(response.status == "SUCCESS"){ $('#error').hide('fast'); alert("Thanks for submitting. We will get back to you as asoon as possible.\n\n" +
if(response.status == "SUCCESS"){
$('#error').hide('fast');
alert("Thanks for submitting. We will get back to you as asoon as possible.\n\n" + response.result);
}
增加一个延迟
window.setTimeout(function() { alert('...') },100)
增加一个延迟
window.setTimeout(function() { alert('...') },100)
听起来您希望在显示警报之前等待动画完成 隐藏回调中的调用警报:
$('#error').hide('fast', function() { alert(...); });
听起来您希望在显示警报之前等待动画完成 隐藏回调中的调用警报:
$('#error').hide('fast', function() { alert(...); });
您需要一个回调函数,如下所示:
if(response.status == "SUCCESS"){
$('#error').hide('fast',function()
{
alert("Thanks for submitting. We will get back to you as asoon as possible.\n\n" + response.result);
});
}
您需要一个回调函数,如下所示:
if(response.status == "SUCCESS"){
$('#error').hide('fast',function()
{
alert("Thanks for submitting. We will get back to you as asoon as possible.\n\n" + response.result);
});
}
可以提供要隐藏的回调,该回调在动画完成后发生:
$('#error').hide('fast', function() {
alert("Your alert string");
});
阅读官方文件中的更多内容:
您可以提供一个要隐藏的回调,该回调在动画完成后发生:
$('#error').hide('fast', function() {
alert("Your alert string");
});
阅读官方文件中的更多内容: