在我的HTML中
$(document).click(function(){
alert('Document Clicked');
})<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<body>
<button type='button'>CLICK [NO ALERT]</button>
<button type='button'>ME[NO ALERT]</button>
</body>在我的代码中,如果单击按钮,如何防止警报显示,但除了按钮之外的任何内容都可以被警告.
div.stars {
display: inline-block;
}
input.star { display: none; }
label.star {
float: right;
padding: 10px;
font-size: 36px;
color: #444;
transition: all .2s;
}
input.star:checked ~ label.star:before {
content: '\f005';
color: #FD4;
transition: all .25s;
}
input.star-5:checked ~ label.star:before {
color: #FE7;
text-shadow: 0 0 20px #952;
}
input.star-1:checked ~ label.star:before { color: #F62; }
label.star:hover { transform: rotate(-15deg) scale(1.3); }
label.star:before {
content: '\f006';
font-family: FontAwesome;
}<div class="stars">
<form action="">
<input class="star star-5" id="star-5" type="radio" name="star"/>
<label class="star …表单示例让我说我有两个表
users posts
[ id - username ] [ id - username - content ]
[ 1 - AXBAXBA ] [ 1 - AXBAXBA - contentA ]
[ 2 - AXAAXAA ] [ 2 - AXAAXAA - contentB ]
[ 3 - AXDAXDA ] [ 3 - AXDAXDA - contentC ]
这种设计更好吗？因为我将快速使用一个查询
SELECT username, content FROM posts
或者更适合加入这种桌子设计
users posts
[ id - username ] [ id - userid - content ]
[ 1 - AXBAXBA ] [ 1 - …$('[data-toggle=popover]').popover();
$('#button').click(function(){
alert('Clicked');
});<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/4.0.0-alpha.6/css/bootstrap.min.css" integrity="sha384-rwoIResjU2yc3z8GV/NPeZWAv56rSmLldC3R/AZzGRnGxQQKnKkoFVhFQhNUwEyJ" crossorigin="anonymous">
<script src="https://code.jquery.com/jquery-3.1.1.slim.min.js" integrity="sha384-A7FZj7v+d/sdmMqp/nOQwliLvUsJfDHW+k9Omg/a/EheAdgtzNs3hpfag6Ed950n" crossorigin="anonymous"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/tether/1.4.0/js/tether.min.js" integrity="sha384-DztdAPBWPRXSA/3eYEEUWrWCy7G5KFbe8fFjk5JAIxUYHKkDx6Qin1DkWx51bBrb" crossorigin="anonymous"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/4.0.0-alpha.6/js/bootstrap.min.js" integrity="sha384-vBWWzlZJ8ea9aCX4pEW3rVHjgjt7zpkNpZk+02D9phzyeVkE+jo0ieGizqPLForn" crossorigin="anonymous"></script>
<button type="button" class="btn btn-lg btn-danger" data-toggle="popover" data-html='true' title="Popover title" data-content="<button id='button' type='button'>buton</button>">Click to toggle popover</button>在这个弹出窗口中，当我单击#button它的内部时，它会执行alert();任何操作，但是如果我#button从弹出窗口中取出并单击它，它会正常工作。
$array = ['a', 'b', 'c', 'd'];
$vars = ['a', 'f', 'g'];
foreach ($vars as $var) {
if (in_array($var, $array)) {
return true;
} else {
return false;
}
}
我怎样才能检查其中是否$vars存在$array？只有其中一个需要,true如果所有其他false都不是问题,但如果有超过1个true值,
例如,['a', 'b', 'c', 'g']我希望函数停在第一个true值并结束该过程.