Question

我在这些变量中得到一个未定义的索引错误：id，subj，mid，fin。我已正确定义它们，我不知道我的代码有什么问题。我认为问题在于我定义了这4个变量的代码的位置。请帮忙？感谢。

echo "
    <form action=editgrades.php method=post>
        ID: <input type='text' name='id' maxlength='5' size='3'>
        Subject: <input type='text' name='subj' maxlength='3' size='3'>
        Midterm: <input type='text' name='mid' maxlength='3' size='3'>
        Finals: <input type='text' name='fin' maxlength='3' size='3'>
        <input type='submit' name='submit' value='Update'>
    </form>
";
$mysqli = mysqli_connect("localhost", "root", "", "school");
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
} else {
    $sql = "select * from studentgrades";
}
$id = $_POST['id'];
$subject = $_POST['subj'];
$midterm = $_POST['mid'];
$finals = $_POST['fin'];
$average = ($midterm + $finals) / 2;
if ($average >= 70) {
    $remarks = 'Passed';
} else {
    $remarks = 'Failed';
}
$res = mysqli_query($mysqli, $sql);
if ($res) {
    $sql1 = "
        update studentgrades set Subject = " . $subject . ",
        Midterm = " . $midterm . ",
        Finals = " . $finals . ",
        Average = " . $average . ",
        Remarks = " . $remarks . "
        where ID = " . $id . "
    ";
    $res1 = mysqli_query($mysqli, $sql1);
    if ($res1) {
        echo "
            Grades updated successfully.
            <br><br>
        ";
    }
}

Answer 1

这是因为您在提交表单之前尝试使用POST变量。

它没有抱怨$id。 “未定义的索引”表示您尝试使用不存在的数组的键，在本例中为$_POST数组。

在尝试使用索引之前检查索引是否已设置：

$id = isset($_POST['id']) ? $_POST['id'] : null;
$subject = isset($_POST['subj']) ? $_POST['subj'] : null;
etc...

或者只是关闭错误报告（不推荐）

error_reporting(E_ALL & ~E_NOTICE & ~E_DEPRECATED);

Answer 2

这是您的问题 - 您不检查是否设置了帖子。因此，如果您在提交表单之前调用此页面，那么您将尝试调用任何$ _POST var的未定义索引。

做这样的事情：

if(isset($_POST['id']))
{
    $mysqli = mysqli_connect("localhost", "root", "", "school");
    if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
    }
    else{
    $sql = "select * from studentgrades";
    }
    $id = $_POST['id'];
    $subject = $_POST['subj'];
    $midterm = $_POST['mid'];
    $finals = $_POST['fin'];
    $average = ($midterm+$finals)/2;
    if($average >= 70){
    $remarks = 'Passed';
    }
    else {
    $remarks = 'Failed';
    }
    $res = mysqli_query($mysqli, $sql);
    if ($res) {
    $sql1 = "
    update studentgrades set Subject = ".$subject.",
    Midterm = ".$midterm.",
    Finals = ".$finals.",
    Average = ".$average.",
    Remarks = ".$remarks."
    where ID = ".$id."
    ";
    $res1 = mysqli_query($mysqli, $sql1);
    if ($res1){
    echo "
    Grades updated successfully.
    <br><br>
    ";
    }
}

尽管isset（$ _ POST ['id']））可以用任意数量的方式替换，以验证表单是否已提交。

附注 - 您应该在输出任何html之前运行所有后期处理，以防您需要根据发布结果修改标题。如果此代码需要执行以下操作：

if($id == '')
{
    header('location: /');
    exit;
}

它会告诉你标题已经发送并给你另一个错误。

PHP未定义的索引/变量

2 个答案: