show() is invalid after ajax success
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PHP中文网 2017-05-19 10:40:54
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The console can print login success and 222, but the element always flashes on the page and disappears. What's going on?

PHP中文网
PHP中文网

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给我你的怀抱

When there is a button tag in the form, and the form is submitted asynchronously using ajax, the page will be refreshed
Reason: When the button exists, the form will be submitted again, so the page is refreshed
Solution: Just replace the button tag with an a tag

伊谢尔伦

Either your selector did not select the right one, or other Js is interfering at the same time

大家讲道理

Probably covered. You can find out by turning on debugging and looking at the style of the right element.

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