小编aan*_*s77的帖子

我有以下PHP代码:

echo "<div style='float:left;'>";
echo "<table>";
echo "<tr>";
echo "<th></th>";
echo "<th colspan='4'>Laks beholdt</th>";
echo "</tr>";
echo "<tr>";
echo "<th>Uke</th>";
echo "<th>&lt;3 kg</th>";
echo "<th>3-7 kg</th>";
echo "<th>&gt;7 kg</th>";
echo "<th>Totalt</th>";
echo "</tr>";

while ($row = mysql_fetch_array($result, MYSQL_ASSOC) or die(mysql_error()))
{
   echo "<tr>";
   echo "<td>" . $row['Uke'] . "</td>";
   echo "<td style='text-align:right; padding-right:10px;'>" . number_format($row['SumSmall'], 1,
      ",", " ") . " kg</td>";
   echo "<td style='text-align:right; padding-right:10px;'>" . number_format($row['SumMedium'], 1,
      ",", " ") . " kg</td>";
   echo "<td style='text-align:right; padding-right:10px;'>" . …

我在CSS中将以下嵌入式SVG定义为背景图像。

div {
  border: 1px solid black;
  background-image: url("data:image/svg+xml;charset=utf8,<svg xmlns='http://www.w3.org/2000/svg' version='1.1' preserveAspectRatio='none' viewBox='0 0 10 10'> <path d='M2 10 L8 0 L10 0 L10 10' fill='%238A92DF'></path></svg>");
  background-repeat: no-repeat;
  background-position: center center;
  background-size: 100%;
}

它在Chrome，Firefox和Edge中可以正常运行，但在Internet Explorer 11中却无法运行。为什么？

JSfiddle在这里。

考虑这个包装为JSON的JavaScript对象:

vars = $.parseJSON('{"timestamp":1402720347,"AktiveSprak":{"en":{"ID":"en","Norsk_navn":"Engelsk","Lokalt_navn":"English","Bilde":"flagg_en.png","Aktivt":"1"},"no":{"ID":"no","Norsk_navn":"Norsk","Lokalt_navn":"Norsk","Bilde":"flagg_no.png","Aktivt":"1"}}}')

我想迭代对象AktiveSprak,但我无法弄清楚如何.

现在我可以做vars.AktiveSprak但是为了使用jQuery的.each函数,对象需要像这样包装在jQuery中$(vars).现在,为什么我不能做这些:

$(vars).AktiveSprak.each(function(){})
$(vars)[0].AktiveSprak.each(function(){})
$(vars).find("AktiveSprak").each(function(){})
$(vars)[0].find("AktiveSprak").each(function(){})