Question

我正在尝试实施Ajax技术但是我没有得到什么问题请帮我找出问题，找出问题。
我认为问题出在此页面上。

 <html>
 <body>
  <script language = "javascript" type = "text/javascript">
     //Browser Support Code
     function ajaxFunction() {
        var ajaxRequest;

        try {        
           // Opera 8.0+, Firefox, Safari
           ajaxRequest = new XMLHttpRequest();
        } catch (e) {

           // Internet Explorer Browsers
           try {
              ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
           } catch (e) {

              try {
                 ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
              } catch (e) {
                 // Something went wrong
                 alert("Your browser broke!");
                 return false;
              }
           }
        }

        // Create a function that will receive data
        // sent from the server and will update
        // div section in the same page.
        ajaxRequest.onreadystatechange = function() {

           if(ajaxRequest.readyState == 4) {
              var ajaxDisplay = document.getElementById('ajaxDiv');
              ajaxDisplay.innerHTML = ajaxRequest.responseText;
           }
        }

        // Now get the value from user and pass it to
        // server script.
        var age = document.getElementById('age').value;
        var wpm = document.getElementById('wpm').value;
        var sex = document.getElementById('sex').value;
        var queryString = "&age = " + age ;

        queryString +=  "&wpm = " + wpm + "&sex = " + sex;
        ajaxRequest.open("GET", "http://localhost/ajax-example.php" + 
  queryString, true);
        ajaxRequest.send();
     }
  </script>

  <form name = 'myForm'>
     Max Age: <input type = 'text' id = 'age' /> <br />
     Max WPM: <input type = 'text' id = 'wpm' /> <br />
     Sex: 

     <select id = 'sex'>
        <option value = "m">m</option>
        <option value = "f">f</option>
     </select>

     <input type = 'button' onclick = 'ajaxFunction()' value = 'Query 
  MySQL'/>
  </form>

  <div id = 'ajaxDiv'>Your result will display here</div>
 </body>
 </html>

这是完成数据库代码的php页面数据库名称为ajaxtutorials，表名称为ajax_example

<?php
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "";
$dbname = "ajaxtutorials";

$conn= mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);

mysqli_select_db($conn, $dbname) or die(mysqli_error($conn));

$age = $_GET['age'];
$sex = $_GET['sex'];
$wpm = $_GET['wpm'];

$age = mysqli_real_escape_string($conn, $age);
$sex = mysqli_real_escape_string($conn, $sex);
$wpm = mysqli_real_escape_string($conn, $wpm);


$query = "SELECT * FROM ajax_example WHERE sex like '$sex'";

if(is_numeric($age))
$query .= " AND age <= $age";

if(is_numeric($wpm))
$query .= " AND wpm <= $wpm";

$qry_result = mysqli_query($conn, $query) or die(mysqli_error($conn));

$display_string = "<table>";
$display_string .= "<tr>";
$display_string .= "<th>Name</th>";
$display_string .= "<th>Age</th>";
$display_string .= "<th>Sex</th>";
$display_string .= "<th>WPM</th>";
$display_string .= "</tr>";

// Insert a new row in the table for each person returned
while($row = mysqli_fetch_array($qry_result)) {
   $display_string .= "<tr>";
   $display_string .= "<td>$row[name]</td>";
   $display_string .= "<td>$row[age]</td>";
   $display_string .= "<td>$row[sex]</td>";
   $display_string .= "<td>$row[wpm]</td>";
   $display_string .= "</tr>";
}

echo "Query: " . $query . "<br />";
$display_string .= "</table>";

echo $display_string;
?>

请帮助我识别问题，找出问题。
我认为问题出在此页面上。

单击按钮时出现错误

找不到对象在此服务器上找不到请求的URL。

Answer 1

请更改以下部分

  var queryString = "&age = " + age ;

收件人

var queryString = "?age = " + age ;

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我正在尝试实施Ajax技术，但找不到问题所在

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