背景:我有一个包含两个嵌入式脚本的网页。但是,一次只能工作一个。我有一个用于导航栏和筛选表的工具。
我尝试过的事情:
将脚本放入一个脚本标签。
重命名功能。
将脚本分别嵌入head和body标签。
预期的最终结果:如何使它们一次都起作用?
我在下面提供了一段代码。它们位于HTML文件中的结束正文标记处。
<!DOCTYPE html>
<html>
<head>
<meta name="viewport" content="width=device-width, initial-scale=1">
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/font-awesome/4.7.0/css/font-awesome.min.css">
<style>
* {
box-sizing: border-box;
}
#myInput {
background-image: url('/css/searchicon.png');
background-position: 10px 10px;
background-repeat: no-repeat;
width: 100%;
font-size: 16px;
padding: 12px 20px 12px 40px;
border: 1px solid #ddd;
margin-bottom: 12px;
}
#myTable {
border-collapse: collapse;
width: 100%;
border: 1px solid #ddd;
font-size: 18px;
}
#myTable th, #myTable td {
text-align: left;
padding: 12px;
}
#myTable tr {
border-bottom: 1px solid #ddd;
}
#myTable tr.header, #myTable tr:hover {
background-color: #f1f1f1;
}
body {
margin: 0;
font-family: Arial, Helvetica, sans-serif;
}
.topnav {
overflow: hidden;
background-color: #333;
}
.topnav a {
float: left;
display: block;
color: #f2f2f2;
text-align: center;
padding: 14px 16px;
text-decoration: none;
font-size: 17px;
}
.topnav a:hover {
background-color: #ddd;
color: black;
}
.active {
background-color: #4CAF50;
color: white;
}
.topnav .icon {
display: none;
}
@media screen and (max-width: 600px) {
.topnav a:not(:first-child) {display: none;}
.topnav a.icon {
float: right;
display: block;
}
}
@media screen and (max-width: 600px) {
.topnav.responsive {position: relative;}
.topnav.responsive .icon {
position: absolute;
right: 0;
top: 0;
}
.topnav.responsive a {
float: none;
display: block;
text-align: left;
}
}
</style>
</head>
<body>
<div class="topnav" id="myTopnav">
<a href="#home" class="active">Home</a>
<a href="#news">News</a>
<a href="#contact">Contact</a>
<a href="#about">About</a>
<a href="javascript:void(0);" class="icon" onclick="myFunction()">
<i class="fa fa-bars"></i>
</a>
</div>
<div style="padding-left:16px">
<h2>Responsive Topnav Example</h2>
<p>Resize the browser window to see how it works.</p>
</div>
<input type="text" id="myInput" onkeyup="myFunction()" placeholder="Search for names.." title="Type in a name">
<table id="myTable">
<tr class="header">
<th style="width:60%;">Name</th>
<th style="width:40%;">Country</th>
</tr>
<tr>
<td>Alfreds Futterkiste</td>
<td>Germany</td>
</tr>
<tr>
<td>Berglunds snabbkop</td>
<td>Sweden</td>
</tr>
<tr>
<td>Island Trading</td>
<td>UK</td>
</tr>
<tr>
<td>Koniglich Essen</td>
<td>Germany</td>
</tr>
<tr>
<td>Laughing Bacchus Winecellars</td>
<td>Canada</td>
</tr>
<tr>
<td>Magazzini Alimentari Riuniti</td>
<td>Italy</td>
</tr>
<tr>
<td>North/South</td>
<td>UK</td>
</tr>
<tr>
<td>Paris specialites</td>
<td>France</td>
</tr>
</table>
<script>
function myFunction() {
var input, filter, table, tr, td, i;
input = document.getElementById("myInput");
filter = input.value.toUpperCase();
table = document.getElementById("myTable");
tr = table.getElementsByTagName("tr");
for (i = 0; i < tr.length; i++) {
td = tr[i].getElementsByTagName("td")[0];
if (td) {
if (td.innerHTML.toUpperCase().indexOf(filter) > -1) {
tr[i].style.display = "";
} else {
tr[i].style.display = "none";
}
}
}
}
</script>
<script>
function myFunction() {
var x = document.getElementById("myTopnav");
if (x.className === "topnav") {
x.className += " responsive";
} else {
x.className = "topnav";
}
}
</script>
</body>
</html>
答案 0 :(得分:0)
考虑了SomePerformance所说的内容后,我设法缩小了问题的范围。
进行这些调整后,两个功能均按预期工作。