Question

我喜欢在输入字段中填写标准值我有这个代码：

$stma = $conn->prepare("SELECT * FROM `users` WHERE ID = '".$_GET['gebruiker']."' ");
$stma->execute();
$row_count = $stma->rowCount(); // returns 1
foreach ($conn->query($stma) as $rows) {
    $Username = $rows['Username'];
}

/// my form
echo '<form method="POST" >
    <table>
    <th colspan="3"><h1>Gebruiker bewerken</h1></th>
        <tr>
            <th>
                <h3>Gebruikersnaam: </h3>
            </th>
            <td>
                <input style="width: 70%;" type="text" READONLY value="'.$Username.'" > 
                // the value must be filled in this input field
            </td>
        </tr>
        <tr>
            <th>
                <h3>Wachtwoord: </h3>
            </th>
            <td>
                <input style="width: 70%;" type="password"  name="wachtwoord" REQUIRED>
            </td>
        </tr>
        <tr>
            <th>
            </th>
            <td colspan="2">
                <input type="submit" name="bewerken" class="button" style="vertical-align:middle" value="Opslaan">
            </td>
        </tr>
        '.$error.'
    </table>
</form>';

代码没有填写我从数据库中获得的值我仍然得到一个空的表格字段我的查询返回1个结果行（我已选中）
有人看到我的错误吗？我没有看到我犯的错误（这是我的错误，它也适用于我的其他形式）

Answer 1

要确保它输出所有错误和警告（用于调试），这可能会有所帮助：

ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);

将上述代码放在文件顶部。您可能还希望阻止任何SQL注入：

$stma = $conn->prepare("SELECT * FROM `users` WHERE ID = ? ");
$stma->bindParam(1, $_GET['gebruiker'], PDO::PARAM_INT);
$stma->execute();
$stma->debugDumpParams(); // you could use this to check whether or not all parameters are set correctly
$row_count = $stma->rowCount(); // returns 1
foreach ($conn->query($stma) as $rows) {
    $Username = $rows['Username'];
}

Answer 2

以下是一个工作示例。

<强> PHP

try {
        $conn = new PDO('mysql:host=localhost;dbname=YourDBname', 'root', '');
    } catch (PDOException $e) {
        echo $e->getMessage();
    }

    $id = $_GET['gebruiker'];

    $sql = "SELECT * FROM `users` WHERE id = :id";
    $stm = $conn->prepare($sql);

    $stm->execute(['id'=>$id]);

    $user = $stm->fetchObject();

    $username = $user->username;

<强> HTML

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <title>Test</title>
</head>
<body>
    <form action="POST">
        <input type="text" value="<?php echo (isset($username)) ? $username : 'No value' ; ?>">
    </form>
</body>
</html>

如果您的网址设置了gebruiker，那么您只需执行以下操作：script.php？gebruiker = 1您可以将1替换为表格中存在的任何ID值。

Answer 3

please try this code
$stma = $conn->prepare("SELECT * FROM `users` WHERE ID = '".$_GET['gebruiker']."' ");

$stma->execute();

$row_count = $stma->rowCount(); // returns 1

foreach ($conn->query($stma) as $rows) {
  $Username = $rows['Username'];
}

**please replace this code**
$res = $conn->query("SELECT * FROM users WHERE ID = '".$_GET['gebruiker']."' ");

$allRows = $res->fetch_assoc(); 

$Username = $allRows['UserName'];

PHP表单输入值不起作用

3 个答案: