我的场景:我正在显示传单地图.地图有几个图块,每个图块可能包含一个或多个图标.以下是瓷砖的外观:
<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" pointer-events="none" width="256" height="256" viewBox="0 0 256 256" class="leaflet-tile leaflet-tile-loaded" style="width: 256px; height: 256px; transform: translate3d(455px, -4px, 0px); opacity: 1;">
<g></g>
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</svg>
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导致Chrome,Firefox(通过devtools检查):
但是,在Safari中,不会渲染图标.元素在那里,但缺少图片.截图来自browserstack for safari,ios6.突出显示的蓝色框是元素(再次通过devtools检查),显示元素处于适当位置,具有正确的尺寸,但没有显示图像:
我已经试过:
我没有尝试过:
除了更换svg渲染器之外,我还能尝试其他任何建议吗?
最低可重复的示例:https: //jsfiddle.net/tocxvxy3/3/
我有一个分割字符串的功能(为了清晰起见粘贴在最后).单独使用时,此功能可按预期工作.例:
SELECT value
FROM dbo.mg_fn_Split('2#1','#')
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返回
-- value --
-- 2 --
-- 1 --
-----------
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但是当在"WHERE IN"子句中使用时,就像在这个例子中一样(稍后在tableA上更多):
SELECT * FROM TableA WHERE TableA.id IN
(
SELECT value
FROM dbo.mg_fn_Split('2#1','#')
)
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我收到错误:"传递给LEFT或SUBSTRING函数的长度参数无效."
这里使用TableA作为示例.使用不同的表(假设它们具有id列)有时会返回正确的结果,而在其他表上我得到错误.
我假设它与执行顺序有关,但我仍然没有看到什么可能"破坏"该功能.
我正在寻找一个"正在发生什么"的解释,而不是"使用它".我知道我可以使用连接来获得结果.
功能定义:
-- Description: Returns a table containing the results of a string-split operation.
-- Params:
-- DelimitedList: The string to split
-- Delimiter: The delimiter char, defaults to ','
-- Columns:
-- Position - The char index of the item
-- Value - The actual …Run Code Online (Sandbox Code Playgroud) 小提琴重现:调整大小直到出现垂直滚动条:http : //jsfiddle.net/yxf1v60x/2/embedded/result/
预期行为:鼠标滚轮在黄色区域上使用时滚动。
观察到的行为:在 Chrome 中不起作用,在 Firefox 中工作正常。
HTML代码:
<div class="parent"><input /><div class="child"><div class="subChild"><input /></div></div></div>
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CSS代码:
html,body{
overflow:hidden;
height: 100%;
width: 100%;
}
.parent {
pointer-events:none;
position: absolute;
z-index:10;
top:0px;
left:0;
right:0;
bottom:0px;
background-color:#ff0000;
height: auto;
overflow-y:auto;
}
.child {
pointer-events:auto;
position: absolute;
z-index:11;
top:50px;
left:0;
right:0;
bottom: 0;
max-height: 100%;
background-color:#0000FF;
}
.subChild{
height: 500px;
background-color:yellow;
width: 100%;
}
}
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所以我不确定这里发生了什么:指针事件显然适用于黄色区域,因为相关的输入元素是可点击的。通过键盘滚动也可以(尝试向上/向下翻页,重点放在黄色区域)。
a) 这是 Chrome 的错误吗?
b) 我可以在保持父子绝对定位的同时避免它吗?我不希望在父元素上有指针事件。
我在他们访问时弹出的网站上有FancyBox,里面有一些信息.我想添加一些用户可以点击的按钮,并设置一个cookie,大约一个月左右不会显示该消息.
当涉及到这样的事情时,我很无用,所以如果有人能够告诉我该怎么做,那将是非常棒的.
这是我到目前为止所拥有的.在底部,我添加了我认为可能是建议的cookie("noShow")的锚,但我不确定它是否会像它一样工作.我已经加载的所有的jQuery的脚本此之前的fancybox,经过这些加载jquery.cookie.js了.如果重要的话,我正在使用FancyBox 2的最新下载.
<script type="text/javascript">
function openFancy() {
setTimeout( function() {$('#autoStart').trigger('click'); },1000);
}
$(document).ready(function() {
openFancy();
$('#autoStart').fancybox();
});
</script>
<!-- This is the popup itself -->
<a id="autoStart" style="display:none" href="#autoFancybox"></a>
<div style="display: none;">
<div id="autoFancybox" style="width: 800px">
<div>
<!-- My content for the Fancybox is here -->
<br />
<p style="font-size:10px" align="right">
<a id="noShow" href="#noShow">Don't me show this message again</a>
</p>
</div>
</div>
</div>
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谢谢,利亚姆.
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