<html>??? <body>?? ?<select name="" id="dj">?? ??? ??? ?<?php for($i=1;$i<=7;$i++){ ?>?? ??? ???? <option value="<?php? echo $i ; ?>"><?php echo $i; ?></option>?? ??? ???? <?php } ?>?? ??? ?</select>?? ?</body>?? ? <script>?? ??? document.getElementById('dj').onchange = function (){?? ? ?? ?window.location.href = '?page='+this.value;?? ??? }?? ? </script>?? ?</html>
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灬紫羽
TA貢獻107條經驗 獲得超71個贊
注意看我代碼后面給的注釋：
<html>
<script?type="text/javascript"?src="http://localhost/test/js/jquery.js"></script><!--注意此處的引用jquery.js文件,改成你自己的文件路徑即可-->
<body>
<select?name=""?id="dj">
????<?php?for($i=1;$i<=7;$i++){??>
????????<option?value="<?php??echo?$i?;??>"?onclick="javascript:test()"><?php?echo?$i;??></option>
????<?php?}??>
</select>
</body>
<script>
//????document.getElementById('dj').onchange?=?function?(){
//????????window.location.href?=?'?page='+this.value;//你的window.location.href本身就是跳轉的意思;要想不跳轉就用異步發送的方式;
//????}
????//get方式發送
????var?page?=?document.getElementById('dj').value;
????function?test(){
????????$.ajax({
????????????type:'GET',
????????????url:'?page='+page,
????????????success:function(data){
????????????????alert(page);
????????????},
????????????error:function(){
????????????????alert(errno);
????????????}
????????})
????}
</script>
</html>
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PHP菜鳥222我想達到分頁的效果，改為#則不跳轉了
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