//html
<div class="modal-footer">
<button type="button" class="btn btn-primary" id="adduserteam" data-toggle="popover" data-placement="top" data-content="select atleast one employee!" data-trigger="focus">Add</button>
<button type="button" class="btn btn-primary" id="removeuserteam" data-toggle="popover" data-placement="left" data-content="select atleast one employee!" data-trigger="focus">Remove</button>
</div>
// on click function//
$('#removeuserteam').on('click',function() {
if(teamuser.rows('.selected').data().length > 0){
$.ajax({
url : "../../rest/teamRest/removeFromTeam",
type : "POST",
dataType : "json",
contentType : "application/json",
data: JSON.stringify(test.emp),
success : function(data) {
teamuser.ajax.reload();
alert("success deleteRule: ");
$('#addteammodal').modal('hide');
},
error : function(request,status,error) {
teamuser.ajax.reload();
alert("error deleteRule: ");
}
});
$('.edit').removeClass('edit');
}
else
$('[data-toggle="popover"]').popover('show');
});
在这段代码中,我有 onclick 函数,按钮的 html 部分,如果我单击“removeuser”按钮,即使弹出窗口正在打开,我也会有一个具有相同配置的按钮
最佳答案
原因是两个按钮
都有
data-toggle="popover"
因此两个弹出窗口都会打开。
而是通过 id
或 $(this)
进行操作:
$('#adduserteam').click(function(){
$(this).popover('show');
})
$('#removeuserteam').click(function(){
$(this).popover('show');
})
更新问题后:
$('#removeuserteam').on('click',function() {
var $this = $(this);
if(teamuser.rows('.selected').data().length > 0){
$.ajax({
url : "../../rest/teamRest/removeFromTeam",
type : "POST",
dataType : "json",
contentType : "application/json",
data: JSON.stringify(test.emp),
success : function(data) {
teamuser.ajax.reload();
alert("success deleteRule: ");
$('#addteammodal').modal('hide');
},
error : function(request,status,error) {
teamuser.ajax.reload();
alert("error deleteRule: ");
}
});
$('.edit').removeClass('edit');
}
else
$this.popover('show');
});
关于javascript - 单击单个按钮即可打开 2 个弹出窗口,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30451877/