Question

我现在遇到了更大的问题。每当我想选择摄入量和受试者时，它将分别显示未定义的索引：摄入量和未定义的索引：程序。即使我的数据库包含数据，当我选择其他选择时，它也没有显示主题的选项。我的代码无法检索它还是其他内容？Result Image: Errors需要帮助，谢谢

 <?php
    include "..\subjects\connect3.php";
    //echo "Connection successs";

    $query = "SELECT * FROM programmes_list";
    $result = mysqli_query($link, $query);
    ?>

    <form name = "form1" action="" method="post">
    <table>
    <tr>
    <td>Select Pragramme</td>
    <td><select id="programmedd" onChange="change_programme()">
    <option>select</option>
    <?php
    while($row=mysqli_fetch_array($result)){
        ?>
    <option value="<?php echo $row["ID"]; ?>"><?php echo $row["programme_name"]; ?></option>
    <?php
        }
        ?>
        </select></td>
        </tr>

        <tr>
            <td>Select intake</td>
            <td>
            <div id="intake">
            <select>
            <option>Select</option>
            </select>
            </div>
            </td>
        </tr>

        <tr>
            <td>Select Subjects</td>
            <td>
            <div id="subject">
            <select>
            <option>Select</option>
            </select>
            </div>
            </td>
        </tr>


    </table>
    </form>




    <script type="text/javascript">
    function change_programme()
    {
        var xmlhttp=new XMLHttpRequest();
        xmlhttp.open("GET","ajax.php?programme="+document.getElementById("programmedd").value,false);
        xmlhttp.send(null);

        document.getElementById("intake").innerHTML=xmlhttp.responseText;

    }


        function change_intake()
    {
        var xmlhttp=new XMLHttpRequest();
        xmlhttp.open("GET","ajax.php?intake="+document.getElementById("intakedd").value,false);
        xmlhttp.send(null);

        document.getElementById("subject").innerHTML=xmlhttp.responseText;
    }
    </script>



    //ajax.php
    <?php
        $dbhost = 'localhost' ;
        $username = 'root' ;
        $password = '' ;
        $db = 'programmes' ;

        $link = mysqli_connect("$dbhost", "$username", "$password");

        mysqli_select_db($link, $db);

        $programme=$_GET["programme"];
        $intake=$_GET["intake"];

    if ($programme!="")
    {
        $res=mysqli_query($link, "select * from intakes where intake_no = $programme");
        echo "<select id='intakedd' onChange='change_intake()'>";

        while($value = mysqli_fetch_assoc($res))
        {

        echo "<option value=".$value['ID'].">";
        echo $value["intake_list"];
        echo "</option>";
        }   
        echo "</select>";
    }

    if ($intake!="")
    {
        $res=mysqli_query($link, "select * from subject_list where subject_no = $intake");
        echo "<select>";

        while($value = mysqli_fetch_assoc($res))
        {

        echo "<option value=".$value['ID'].">";
        echo $value["subjects"];
        echo "</option>";
        }   
        echo "</select>";
    }

        ?>

Answer 1

您获取未定义索引通知的原因是change_programme() 仅在调用ajax.php时发送?programme=，{{1} } 仅在通话中发送change_intake()。但是，在ajax.php中，您正试图从?intake=获取两者。

因此，来自$_GET的调用会为change_programme()提供未定义的索引通知，因为它实际上并未在URL中提供。由于同样的原因，$_GET['intake']的调用会为change_intake()提供未定义的索引通知。

您可以通过检查是否已设置来解决此问题：

$_GET['programme']

这是：

的简写

$programme = isset($_GET["programme"]) ? $_GET["programme"] : "";
$intake = isset($_GET["intake"]) ? $_GET["intake"] : "";

作为旁注，从不信任您从网址获得的任何输入，尤其是，如果您要在查询中直接使用它。请首先通过if (isset($_GET["programme"])) { $programme = $_GET["programme"]; } else { $programme = ""; } if (isset($_GET["intake"])) { $intake = $_GET["intake"]; } else { $intake = ""; }传递$programme和$intake变量请使用绑定参数的prepared statement来减少SQL注入攻击的可能性

未定义的索引和可能缺少的检索php

1 个答案: