如何将getElementsByTagName内容插入sql表?

时间:2015-06-12 09:29:10

标签: mysql xmlhttprequest getelementbyid

我想解析XML页面,并将内容插入到SQL数据库中,但我不知道如何在插入语法中调用变量!

这是我的代码:



<?php
$server = "localhost";
$user = "root";
$password = "";
$database = "database";

$conn = mysqli_connect($server, $user, $password, $database);

if (!$conn) { die ("Connection Failed:" .mysql_connect_error()); }
echo "Connected Successfully "; 

$sql = "INSERT INTO tableest (stepid, stepname, stepstatus) VALUES ('???', '???', '???')";

?>

<!DOCTYPE html>
<html>
<head>
<script>
if (window.XMLHttpRequest)
{ // code for IE7+, Firefox, Chrome, Opera, Safari
    xmlhttp = new XMLHttpRequest();
}
else
{ // code for IE6, IE5
    xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.open("GET","example.xml",false);
xmlhttp.send();
xmlDoc = xmlhttp.responseXML; 


function displayTicket()
{
    var x = xmlDoc.getElementsByTagName("step");
    for (i = 0; i < x.length; i++) {
        id = (x[i].getElementsByTagName("id")[0].childNodes[0].nodeValue);
        name = (x[i].getElementsByTagName("name")[0].childNodes[0].nodeValue);
        skip = (x[i].getElementsByTagName("status")[0].childNodes[0].nodeValue);

        txt = "id:" + id + "<br>name: " + name + "<br>skip? : " + skip;
        document.getElementById("showid").innerHTML=txt;
    }
}
</script>
</head>

<body onload="displayTicket()">
    <div id='showid'></div>

</body>
</html>
&#13;
&#13;
&#13;

但我一直收到未定义的变量通知!

第二期:

在displayticket函数中使用for循环(如上所述)只给我一步的信息(并没有完成所有步骤)

但如果按以下方式使用(没有函数),则给出正确的输出:

&#13;
&#13;
var x = xmlDoc.getElementsByTagName("step");
for (i = 0; i < x.length; i++) {
    document.write(x[i].getElementsByTagName("id")[0].childNodes[0].nodeValue);
    document.write(x[i].getElementsByTagName("name")[0].childNodes[0].nodeValue);
    document.write(x[i].getElementsByTagName("status")[0].childNodes[0].nodeValue);
}
&#13;
&#13;
&#13;

任何想法如何解决这个问题?

所有帮助和输入都表示赞赏! 提前致谢

0 个答案:

没有答案