Question

我需要从MySQL数据库中插入和提取数据。我能够提取信息，但是当我尝试插入它时，它会给我很多错误信息。其中大多数我都能解决，但这个我似乎无法弄清楚：

Notice: Undefined variable: emailInput in C:\xampp\htdocs\imp02\week5\firstPHPDatabase.php on line 25
Notice: Undefined variable: aanmelding in C:\xampp\htdocs\imp02\week5\firstPHPDatabase.php on line 25
Notice: Undefined variable: ipadres in C:\xampp\htdocs\imp02\week5\firstPHPDatabase.php on line 25
Notice: Undefined variable: opmerkingen in C:\xampp\htdocs\imp02\week5\firstPHPDatabase.php on line 25
Notice: Undefined variable: aantalPersonen in C:\xampp\htdocs\imp02\week5\firstPHPDatabase.php on line 25

我的代码：

<?php
$databaseLink = mysqli_connect('localhost', 'root', '', 'newyearseveparty');
if (mysqli_connect_error())
    echo mysqli_connect_error();

$selectorQuery = "SELECT * FROM attendants";
echo "The query We use is $selectorQuery!";
$attendants = array();
if ($result = mysqli_query($databaseLink, $selectorQuery)) {
    while ($tableRow = mysqli_fetch_assoc($result)) {
        $attendants[] = $tableRow;
    }
} else {
    echo mysqli_error($databaseLink) . 'QUERY: ' . $$selectorQuery;
}
if (isset($_POST['value'])){
    $nameInput = $_POST['nameInput'];
    $emailInput = $_POST['emailInput'];
    $aanmelding = $_POST['aanmelding'];
    $ipadres = $_SERVER['REMOTE_ADDR'];
    $opmerkingen = $_POST['opmerkingen'];
    $aantalPersonen = $_POST['aantalpersonen'];
}
$sql = "INSERT INTO attendants (naam, email, komt, ipadres, opmerkingen, aantalpersonen)
VALUES('Justin', '$emailInput', '$aanmelding', '$ipadres','$opmerkingen', '$aantalPersonen')";

if (!mysqli_query($databaseLink,$sql))
{
    die('Error: ' . mysqli_error($databaseLink));
}
echo "1 record added";
mysqli_close($databaseLink);
?>
<!doctype html>
<html>
<head>
    <title></title>
    <meta name="description" content=""/>
    <meta charset="utf-8"/>
    <link rel="stylesheet" href=""/>
</head>
<body>
<?php

if (!empty($attendants)) {
    foreach ($attendants as $people) {
        echo '<ol>';
        echo 'Attendent';
        echo "<li>Naam: {$people['naam']}</li>";
        echo "<li>E-mail: {$people['email']}</li>";
        echo "<li>Komt: {$people['komt']}</li>";
        echo "<li>Datum van aanmelding: {$people['datum']}</li>";
        echo "<li>Ipadres: {$people['ipadres']}</li>";
        echo "<li>Eventuele opmerkingen: {$people['opmerkingen']}</li>";
        echo "<li>Aantal personen: {$people['aantalpersonen']}</li>";
        echo '</ol>';
    }
} else {
    echo "af er is iets fout gegaan, of er heeft nog niemand zich ingeschreven";
}
?>
<form action="<?php $_SERVER['PHP_SELF'] ?>" method="post">
    <fieldset>
        <label for="nameInput" class="labelstyle">Naam: </label>
        <input id="nameInput" name="inputFields" type="text" autofocus="auto"><br>
        <label for="emailInput" class="labelstyle">E-Mail</label>
        <input id="emailInput" name="inputFields" type="text"><br>
        <label for="aanmelding" class="labelstyle">Komt u, voor ja 1, voor nee 0</label>
        <input id="aanmelding" name="inputFields" type="text"><br>
        <label for="opmerkingen" class="labelstyle">opmerkingen</label>
        <input id="opmerkingen" name="inputFields" type="text"><br>
        <label for="aantalpersonen" class="labelstyle">aantal personen</label>
        <input id="aantalpersonen" name="inputFields" type="number"><br>
        <input type="submit" name="submit" value="submit">
    </fieldset>
</form>
</body>
</html>

我希望你们中的一些人能够很好地帮助我

Answer 1

输入元素的名称将作为$_POST数组的键发布。因此，请更改此设置或更改$_POST键。

<form action="<?php $_SERVER['PHP_SELF'] ?>" method="post">
    <fieldset>
        <label for="nameInput" class="labelstyle">Naam: </label>
        <input id="nameInput" name="nameInput" type="text" autofocus="auto"><br>
        <label for="emailInput" class="labelstyle">E-Mail</label>
        <input id="emailInput" name="emailInput" type="text"><br>
        <label for="aanmelding" class="labelstyle">Komt u, voor ja 1, voor nee 0</label>
        <input id="aanmelding" name="aanmelding" type="text"><br>
        <label for="opmerkingen" class="labelstyle">opmerkingen</label>
        <input id="opmerkingen" name="opmerkingen" type="text"><br>
        <label for="aantalpersonen" class="labelstyle">aantal personen</label>
        <input id="aantalpersonen" name="aantalpersonen" type="number"><br>
        <input type="submit" name="submit" value="submit">
    </fieldset>
</form>

Answer 2

您无法使用元素的 id 访问POST元素。要将元素作为$_POST['nameInput']访问，输入字段的 NAME 应为 nameInput 例如： -

<input id="nameInput" name="inputFields" type="text" autofocus="auto"><br>

应改为

<input id="nameInput" name="nameInput" type="text" autofocus="auto"><br>

和

如果您希望将表单提交到同一页面。您应该保留空白形式的操作属性，而不是action=''

和
提交按钮的名称为'submit'。所以如果表单已经提交，你应该像这样检查isset($_POST['submit'])

Answer 3

使用isset（$ _ POST ['value']）来处理这种情况

Answer 4

使用如下：＆＃34; $ _ POST [variableName]＆＃34;; 不要在帖子内给变量单引号。

Answer 5

您无法使用元素的POST访问id元素。要将元素作为$_POST['nameInput']访问，输入字段的name应为nameInput }

例如：

<input id="nameInput" name="inputFields" type="text" autofocus="auto"><br>

应改为

<input id="nameInput" name="nameInput" type="text" autofocus="auto"><br>

如果您希望将表单提交到同一页面。您应该将表单的action属性留空。（如action=''）

提交按钮的name为'submit'。所以如果表单已经提交，你应该像这样检查isset($_POST['submit'])

注意：未定义的变量：在第25行的C：\ xampp \ htdocs \ imp02 \ week5 \ firstPHPDatabase.php中

5 个答案: