三个值是从数据库读出来,不让他为空就Ok
代码:
$arr_app_type =explode(";",$rs["app_types"]);
$sql = "select * from app_type where id in (10,4,6) order by id";
$mysql->query($sql);
while($row = $mysql->fetch_array()){
?>
"
/>
}
?>
回复讨论(解决方案)
不知道你想问什么
当不选中其一时,提交不成功?
在form 的onsubmit 事件中调用js函数验证就是了
参考一下
<script>function foo(){ var chs=document.getElementsByName('app_types[]'); var counter=0; for(var i=0;i<chs.length;i++){ if(chs[i].checked) counter++; } if(counter==0) return false; return true;}</script><form action="" method="post" onsubmit="return foo()"><input id="app" name="app_types[]" type="checkbox" value="1"><input id="app" name="app_types[]" type="checkbox" value="2"><input id="app" name="app_types[]" type="checkbox" value="3"><input type='submit'></form>
谢谢了,稍做了一下逻辑就搞定了
谢谢了,高手们

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