php如何给select下拉框赋值
a.php页面顶部有部分html代码,有个表单,表单有select、input、button,
表单提交给自己:form action="" method="post"
现在有个问题,每次页面提交后,select的值都回到默认值,想保持所选择的值不变。设计的代码如下:
button 的onclick事件使用js代码
<script> <br />function check() {<br /> var t = document.getElementById("bbsty"); //bbsty是select的id<br /> var seleValue=t.options[t.selectedIndex].value; //获取select的值 <br /> document.form1.submit(); <br /> for(i=0;i<t.length;i++){ //给select赋值 <br /> if(t.options[i].value== selvalue){ <br /> t.options[i].selected=true; <br /> }<br /> }<br />}<br /></script>
我在页面提交前取select值了,提交后再给select赋回去,但是没能成功,于是我想到使用隐藏域做中间变量。
<script><br /> var sele = document.getElementById("bbsty"); <br /> var seleValue=sele.options[sele.selectedIndex].value;//获取select的值<br /> document.getElementById("sel").value=seleValue; //将本页面的bbsty(select)值赋值给上面的隐藏域<br /></script>
function check() {
document.form1.submit();
var selvalue = document.getElementById("sel");
var t = document.getElementById("bbsty");
for(i=0;i
t.options[i].selected=true;
}
}
}
但为什么还是不能成功啊!请高人给看看,我的思路对不对,代码哪块有问题,该咋实现这个功能啊?
------解决方案--------------------
思路问题,
<br />document.form1.submit(); <br />
这东西执行之后,不在于你用hidden的input或者是用js变量保存的值,后边的东西都是白瞎的,因为你页面已经跳转了,等于你重新加载了当前的页面。
所以,如果你想让
<br />document.form1.submit(); <br />
后边的代码执行,必须保证当前页面没有被重新加载,这才应该是你的思路,而不是通过那种方式保存值的问题。
提示:你可以改一下form的target,让form提交到一个宽高为0的iframe里,这样,你本身页面不会被重新加载。这样才能解决你的问题
其实这个问题还有很多其他办法解决,建议有时间去想一想。
------解决方案--------------------
方法很多,这是一种
<?php<br />$bbsty = '';<br />if($_POST) {<br /> $bbsty = $_POST['bbsty'];<br />}<br />?><br /><form method=post><br /><select name=bbsty><br /><option value='a'>a</option><br /><option value='b'>b</option><br /><option value='c'>c</option><br /><option value='d'>d</option><br /></sclect><br /><input type=submit value=ok><br /></form><br /><script><br />document.forms[0].bbsty.value = '<?php echo $bbsty; ?>';<br /></script><br />

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