Question

使用一种形式。尝试让用户对数据库进行INSERT或UPDAT。

尝试var_dump（）;尝试过error_reporting（E_ALL）;

$stmt = $connection->prepare("SELECT * FROM profiles WHERE user=?");
$stmt->bind_param("s", $user);
/* execute prepared statement */
$stmt->execute();

$result = $stmt->get_result();

if (isset($_POST['text']))
{
    if ($result->num_rows)
        $text = ($_POST['text']);
    $birthday = ($_POST['birthday']);
    $gender= ($_GET['gender']);

    $stmt = $connection->prepare("UPDATE profiles SET
    user=?, text=?, birthday=?, gender=?
    WHERE user=?");

    $stmt->bind_param("ssss", $user, $text, $birthday, $gender);

/* execute prepared statement */
    $stmt->execute();
}
//using bound paramaters for profile
else
{
    $stmt = $connection->prepare("INSERT INTO profiles
    (user, text, birthday, gender)
    VALUES (?,?,?,?)");

    $stmt->bind_param("ssss", $user, $text, $birthday, $gender);

/* execute prepared statement */
    $stmt->execute();

/* close statement and connection */
    $stmt->close();
}

没有收到有关性别的未定义变量（单选按钮），但是该变量消失了。数据不插入数据库。有时我在列中看到null。

Answer 1

UPDATE查询有5个占位符，但是您仅在bind_param中绑定了4个参数。最后需要另外一个$user。

    $stmt = $connection->prepare("UPDATE profiles SET
    user=?, text=?, birthday=?, gender=?
    WHERE user=?");

    $stmt->bind_param("sssss", $user, $text, $birthday, $gender, $user);

但是无需设置user，因为您只是将其设置为相同的值，因此将其更改为：

    $stmt = $connection->prepare("UPDATE profiles SET
    text=?, birthday=?, gender=?
    WHERE user=?");

    $stmt->bind_param("ssss", $text, $birthday, $gender, $user);

如果user列是唯一键，则可以使用INSERT ... ON DUPLICATE KEY UPDATE将所有代码组合成一个查询。

$stmt = $connection->prepare("
    INSERT INTO profiles (user, text, birthday, gender)
    VALUES (?,?,?,?)
    ON DUPLICATE KEY UPDATE text = VALUES(text), birthday = VALUES(birthday), gender = VALUES(gender)");
$stmt->bind_param("ssss", $user, $texxt, $birthday, $gender);

Answer 2

您的第二个if语句缺少括号，以及@Barmar提到的其他错误。建议您在弄清楚代码为何无法正常工作之后，再查看INSERT ... ON DUPLICATE KEY UPDATE语句。

为什么要向表中插入null值是一个更明显的原因，是因为首先要设置变量$text，$birthday，$gender if语句的一半，它将不会执行，并且将是未初始化的变量，并且在null之后都等于else。

如果代码未输出任何错误，请使用echo或print函数让您自己了解代码的到达位置，这将使您知道代码在哪里停止工作。您还可以在选择的IDE中使用php调试插件，这将帮助您更快地解决问题。

这是您的代码版本，其中进行了一些修复和调整。花点时间看清楚@Barmar提到的内容，并确保在所有if语句中添加括号，甚至是一行语句，这样您就不会在学习时感到困惑。

/** Did the user give us a name to use?  If not, resubmit the form with a user name */
if (isset($_POST['user']))
{
    $user     = $_POST['user'];     /** Remember to Sanitize & Validate Structure of User Inputs */
    $text     = $_POST['text'];     /** Remember to Sanitize & Validate Structure of User Inputs */
    $birthday = $_POST['birthday']; /** Remember to Sanitize & Validate Structure of User Inputs */
    $gender   = $_POST['gender'];   /** Remember to Sanitize & Validate Structure of User Inputs  */

    $stmt = $connection->prepare("SELECT * FROM profiles WHERE user=?");
    $stmt->bind_param("s", $user);
    /* execute prepared statement */
    $stmt->execute();

    $result = $stmt->get_result();

    if ($result->num_rows)
    {
        $stmt = $connection->prepare("UPDATE profiles SET
            text=?, birthday=?, gender=?
            WHERE user=?");

        $stmt->bind_param("ssss", $text, $birthday, $gender, $user);

        /* execute prepared statement */
        if ($stmt->execute()) {
            echo "Record updated successfully";
        } else {
            echo "Error updating record: " . $stmt->error;
        }
    }
    else
    {
        //using bound paramaters for profile
        $stmt = $connection->prepare("INSERT INTO profiles
            (user, text, birthday, gender)
            VALUES (?,?,?,?)");

        $stmt->bind_param("ssss", $user, $text, $birthday, $gender);

        /* execute prepared statement */
        $stmt->execute();
     }

    /* close statement and connection */
    $stmt->close();
}
else
{
    echo "No User Inputs - Go ahead and submit the form";
}

使用if else语句更新或插入数据

2 个答案: