you can add a data into this table
<script> <br> function loadxmldoc () <br> { <br> var xmlhttp; <br> xmlhttp=new xmlhttprequest (); <br> xmlhttp.open("post","insert.php",true); <br> xmlhttp.setrequestheader("content-type","application/x-www-form-urlencoded"); <br> xmlhttp.send("name&sex&birth"); <br> } <br> </script>
name:
sex:
birth:
以上是网页代码，求助如何获取表单中的数据并用xmlhttp.send发送给insert.php
新人初学，还请多多关照
回复讨论(解决方案)
insert.php的代码是这样的
error_reporting(E_ALL ^E_DEPRECATED);
$con = mysql_connect("localhost:/tmp/mysql.sock","chris","6661573");
if (!$con)
echo "shibai";
else echo "connect mysql success";
mysql_select_db("birth", $con);
$sql="insert shengri (name,sex,birth) values ('$_POST[name]','$_POST[sex] ','$_POST[birth]')";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
echo "1 record added";
mysql_close($con)
?>
求大神们给看看
把你前台的页面换成下面的
you can add a data into this table
<script> <br> function loadXMLDoc () <br> { <br> var xmlhttp; <br> var name = document.getElementById("name").value; <br> var sex = document.getElementById("sex").value; <br> var birth = document.getElementById("birth").value; <br> <br> xmlhttp=new XMLHttpRequest (); <br> xmlhttp.onreadystatechange = function() <br> { <br> if (xmlhttp.readyState == 4 && xmlhttp.responseText) <br> { <br> alert(xmlhttp.responseText); <br> } <br> } <br> xmlhttp.open("POST","./insert.php",true); <br> xmlhttp.setRequestHeader('content-type','application/x-www-form-urlencoded'); <br> xmlhttp.send("name="+name+"&sex="+sex+"&birth="+birth); <br> } <br> </script>
name:
sex:
birth:
非常感谢楼上的回答，很管用。
