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我正在尝试创建一个具有自动填充功能的表单，其中用户输入员工 ID，并在他们离开员工 ID 字段时填充姓名、职位、状态等字段，类似于http中的表格功能://www.crackajax.net/popform.php#。数据稍后将被提交到另一个数据库以跟踪其他用户输入的信息。但是，当我组装文件时，我似乎无法让 onchange 功能工作。以下是表格：

表格本身：

<!DOCTYPE html>

<html lang="en">

<head>
    <title></title>
    <script type="text/javascript" src="scripts.js"></script>
    <script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
    <meta charset="utf-8" />

</head>

<body>

    <form id="entry" name="entry" method="post">


            <label for="employeeid" class="stdlabel">Employee ID:</label>
                <input type="text" id="employeeID" name="employeeID" onchange="geteedata()">
        <br>
            <label for="employeename" class="stdlabel">Employee Name:</label>
                <input id ="employeename" type="text" name="employeename">
        <br>
            <label for="employeestatus "class="stdlabel">Status:</label>
                <input type="text" name="employeestatus" >
        <br>
            <label for="employeetitle"class="stdlabel">Job Title:</label>
                <input type="text" name="employeetitle" >
        <br>
            <label for="loccode" class="stdlabel">Employee Loc Code:</label>
                <input type="text" name="loccode" >

    <input type="submit" name ="submit" value="Submit" style="margin-right:10px">
  </form>

 <?php
      if(isset($_POST['submit'])){

        $con = mysql_connect("localhost", "root", "xxx");
        if(!$con) {
            die("Cannot connect to database:" . mysql_error()); 
        }

        mysql_select_db("mydb", $con);

        $sql = "INSERT INTO casetracker (emplid,empname,empjobtitle,status, loccode) VALUES ('$_POST[employeeid]','$_POST[employeename]','$_POST[employeetitle]','$_POST[employeestatus]','$_POST[loccode]')";

        mysql_query($sql,$con) or die(mysql_error());
        mysql_close($con);

    ?>                
</body>

</html>

以下是 javascript 文件：

function geteedata() {
var url = "search.inc.php?param=";
var idValue = document.getElementById("employeeID").value;
var myRandom = parseInt(Math.random() * 99999999);  // cache buster
http.open("GET", url + escape(idValue) + "&rand=" + myRandom, true);
http.onreadystatechange = handleHttpResponse;
http.send(null);
}

function handleHttpResponse() {
if (http.readyState == 4) {
results = http.responseText.split(",");
document.getElementById('employeename').value = results[0];
document.getElementById('employeestatus').value = results[1];
document.getElementById('employeetitle').value = results[2];
document.getElementById('loccode').value = results[3];
}
   }

这里是 search.inc.php：

if(strlen($param)>0){
$result = mysql_query("SELECT * FROM employee_data WHERE Pernr ='$param'");
if(mysql_num_rows($result)==1) {
while($myrow = mysql_fetch_array($result)){
   $employeename = $myrow["Last_First"];
   $employeestatus = $myrow["Emp_Status"];
   $employeetitle = $myrow["Position_Text"];
   $loccode = $myrow["loccode"];
   $textout .= $employeename.",".$employeestatus.",".$employeetitle.",".$loccode;
}
} else {
$textout=" , , ,".$param;
}
}
echo $textout;

有人对我忘记、遗漏或做错了什么有想法吗？

提前致谢！

4

1 回答 1

0

看起来您在发送请求之前没有创建 XMLHttpRequest 对象的实例。

尝试以下操作：

function geteedata() {
var url = "search.inc.php?param=";
var idValue = document.getElementById("employeeID").value;
var myRandom = parseInt(Math.random() * 99999999);  // cache buster

var http = new XMLHttpRequest();
http.open("GET", url + escape(idValue) + "&rand=" + myRandom, true);
http.onreadystatechange = handleHttpResponse;
http.send(null);
}

查看MDN 使用 XMLHttpRequest了解更多信息。

于 2014-04-25T20:07:12.897 回答