Solution to invalid visited style in css pseudo-class

Incorrect writing method is to write visited after hover and active, for example:

.ui-page-theme-a .digilinx-ui-btn{background:#00a325;border-color:#00891f;color:#FFF;text-shadow: 0 1px 0 #00a325;}
.ui-page-theme-a .digilinx-ui-btn:hover{background:#00891f;border-color:#00721a;color:#FFF;text-shadow: 0 1px 0 #00891f;}
.ui-page-theme-a .digilinx-ui-btn:active{background:#00721a;border-color:#005c15;color:#FFF;text-shadow: 0 1px 0 #00721a;}
.ui-page-theme-a .digilinx-ui-btn:visited{background:#00a325;border-color:#00891f;color:#FFF;text-shadow: 0 1px 0 #00a325;}

The result: the style of the clicked button or link will always be fixed on visited, while the styles of hover and active will not Overwritten

The correct way to write it is to write visited in front, as follows:

.ui-page-theme-a .digilinx-ui-btn{background:#00a325;border-color:#00891f;color:#FFF;text-shadow: 0 1px 0 #00a325;}
.ui-page-theme-a .digilinx-ui-btn:visited{background:#00a325;border-color:#00891f;color:#FFF;text-shadow: 0 1px 0 #00a325;}
.ui-page-theme-a .digilinx-ui-btn:hover{background:#00891f;border-color:#00721a;color:#FFF;text-shadow: 0 1px 0 #00891f;}
.ui-page-theme-a .digilinx-ui-btn:active{background:#00721a;border-color:#005c15;color:#FFF;text-shadow: 0 1px 0 #00721a;}

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