当我单击下一页的后退按钮时,复选框值不应重置。
它应该和我检查或未检查的一样。下面是第一页和下一页的代码。
第一页
<!DOCTYPE html>
<html>
<body>
<form>
<input type="checkbox" name="code" value="ECE">ECE<br>
<input type="checkbox" name="code" value="CSE">CSE<br>
<input type="checkbox" name="code" value="ISE">ISE<br>
<br>
<input type="button" onclick="dropFunction()" value="save">
<br><br>
<script>
function dropFunction() {
var branch = document.getElementsByName("code");
var out = "";
for (var i = 0; i < branch.length; i++) {
if (branch[i].checked == true) {
out = out + branch[i].value + " ";
window.location.href="next.html";
}
}
}
</script>
</form>
</body>
</html>
下一页
<html>
<head>
<title>Welcome to </title>
</head>
<body color="yellow" text="blue">
<h1>welcome to page</h1>
<h2>here we go </h2>
<p> hello everybody<br></p>
</body>
<image src="D:images.jpg" width="300" height="200"><br>
<button onclick="goBack()">Go Back</button>
<script>
function goBack() {
window.location.href="first.html";
}
</script>
</body>
</html>
完整解决方案:示例。首先将id添加到您的复选框中:
<input type="checkbox" name="code" value="ECE" id='1'>ECE<br>
<input type="checkbox" name="code" value="CSE" id='2'>CSE<br>
<input type="checkbox" name="code" value="ISE" id='3'>ISE<br>
<input id="spy" style="visibility:hidden"/>
然后更改您的dropFunction:
function dropFunction() {
var branch = document.getElementsByName("code");
var out = "";
localStorage.clear();
for (var i = 0; i < branch.length; i++)
if (branch[i].checked == true)
localStorage.setItem(branch[i].id, true);
for (var i = 0; i < branch.length; i++) {
if (branch[i].checked == true) {
out = out + branch[i].value + " ";
window.location.href="next.html";
}
}
}
并在first.html中添加一些新的javascript代码:
window.onload = function() {
var spy = document.getElementById("spy");
if(spy.value=='visited')
for(var i=1;i<=3;i++)
if(localStorage.getItem(i))
document.getElementById(i).checked=true;
spy.value = 'visited';
}