我有一个下面的javascript代码，单击下面的超链接后可以正常工作。
<a href="delete_event.php?event_id=110" onClick="return ConfirmDelete()" class="list-group-item">Delete Event</a>
<script>
function ConfirmDelete() {
var ans=confirm("Are you sure to delete this event?");
if(!ans) {
return false;
}
}
</script>
现在，我有一个按钮的HTML代码，当用户单击下面的按钮时，我想执行与上面相同的功能。
<button type="button" class="btn btn-info" data-toggle="popover">Delete Event</button>
那么，告诉我我需要对为超链接编写的代码进行哪些更改？
谢谢。
只需添加一个与锚点相同的onclick属性，而无需 return
<button type="button" class="btn btn-info" data-toggle="popover" onClick="ConfirmDelete('delete_event.php?event_id=110')">Delete Event</button>
function ConfirmDelete(url) {
var ans=confirm("Are you sure to delete this event?");
if(!ans) {
return false;
}
window.location.href = url;
}
并且由于您还添加了jQuery标签
<button type="button" class="btn btn-info" data-toggle="popover" data-url="delete_event.php?event_id=110">Delete Event</button>
<script>
$( "button.btn-info" ).bind( "click", function(){
ConfirmDelete($( this ).attr( "data-url" )) ;
} )
function ConfirmDelete(url) {
var ans=confirm("Are you sure to delete this event?");
if(!ans) {
return false;
}
window.location.href = url;
}
</script>
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