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Home Backend Development PHP Tutorial php中select获取val的值如何传给sql语句

php中select获取val的值如何传给sql语句

Jun 23, 2016 pm 02:18 PM

点击左边出现右边的数据
代码如下:帮忙看哈
<script> <br /> <br /> document.getElementById("sel").value="3"; <br /> document.getElementById("sel").fireEvent("onchange"); <br /> function demo() <br /> { <br /> var d=document.getElementById("sel").value; <br /> document.getElementById("tex").value=d; <br /> alert(d); <br /> <br /> } <br /> </script>



require_once'admin/config.php';
if($conn){
$stmt=odbc_exec($conn,"SELECT  DISTINCT [类别] FROM [网站商品视图] WHERE [类别] like 'A%'");
if($stmt){
$arr=array();
while($row=odbc_fetch_array($stmt)){
$arr[]=$row;
}
for($i=0;$i {
if($i%4==0){
echo "

";
}
?>



}
}
}
?> 

回复讨论(解决方案)

php接收到的数据是什么?
print_r($_GET);
print_r($_POST);

print_r($_POST); 

直接在原页面执行无刷新  怎么把alert(d)的值传给sql语句。

无刷新的话就得用ajax来做了,把select框的值传给php,php返回json数据给js,js完成页面映射。

有没有实例啊!

接受数据是A-Z


html实现:


PHP获取:
$sel = $_POST['sel'];

那样还是得有提交按钮。意思是点击左边右边就能出现

那样还是得有提交按钮。意思是点击左边右边就能出现
通过表单提交,那么你select的值就可以传递到服务端。

跪求解答。。。

那样还是得有提交按钮。意思是点击左边右边就能出现

ajax啊  你左边点击时触发js函数  用ajax将选的值传到服务器  获取对应的数据拿回来放到右边

重写onchange事件处理函数,函数里发个ajax请求将sel的值给服务端,得到结果后展示给右边.
ajax网上例子很多,找找吧.

onchange事件怎样函数里发个ajax请求将sel的值给服务端啊。ajax我还不太懂

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