Question

我有一个像这样设置的mysqli数据库表

ID --- ---电子邮件密码。

我想要做的是在这里插入一行新信息。这就是我尝试过的：

编辑：这就是我连接数据库的方式，将连接保存在全局变量中：

global $db;
$db = new mysqli( 'localhost', 'username', 'password', 'database' );
if ( $db->connect_errno > 0 ) {
die( 'Unable to connect to database [' . $db->connect_error . ']' );

}

function juice_sign_up( $email, $password, $password_confirm )
{
    global $db;
    $emailCheck = 'SELECT email FROM user WHERE email = $email';
    if($emailCheck == 'NULL'){
        $hashPass = password_hash($password);
        INSERT INTO user VALUES ($email, $hashPassword);
}

这是该函数获取变量的地方：

    if( isset($_POST('password_confirm'))){
      juice_sign_up($_POST['email'], $_POST['password'], $_POST['password_confirm']);
    }

我是使用MYSQL的新手，还没有完全理解语法。任何帮助表示赞赏。

Answer 1

这里有一些问题。您不执行选择查询（您只是构建字符串）。并且插入语句就好像它是PHP代码一样，但这不是你执行语句的方式。此外，您至少需要将输入转义为insert语句，或者甚至使用参数绑定。

要执行查询，您可以使用mysqli->query。它返回一个查询结果对象（用于选择查询）或true（用于DML语句）。如果发生错误，它会返回false。

我在以下代码段的注释中解决了这些问题：

function juice_sign_up( $email, $password, $password_confirm )
{
    global $db;
    // The line below just assigns a string to $emailCheck. You still need to execute the query.
    $emailCheck = 'SELECT email FROM user WHERE email == $email';

    // Try to execute it.
    $queryResult = $db->query($emailCheck);

    // Check if the query succeeded and if a row is found.
    // For select queries an object is returned from which you can fetch the results.
    if ($queryResult !== false && $queryResult->fetch_object() === false)
    {
        $hashPass = password_hash($password);
        // Inserting should be done in a similar way. Build a query, and execute it.
        $email = $db->real_escape_string($email);
        $hashPassword = $db->real_escape_string($hashPassword);
        // Mind the escaping of illegal characters (above) and the quotes (below).
        $statement = "INSERT INTO user VALUES ('$email', '$hashPassword')";

        // Note: you won't get a result object for insert statements.
        $result = $db->query($statement);
        // Check the value of result to see if it worked.

    }
}

Answer 2

您无法执行类似的SQL查询。

试试这个：

$sql = 'INSERT INTO user VALUES ($email, $hashPassword);';
$db_con = new mysqli($host, $user, $password, $database);
$result = $db_con->query($sql);

您必须“获取”结果才能获得所需数据，请阅读文档如何使用它。：）

Answer 3

试试这个：

function juice_sign_up( $email, $password, $password_confirm )
{
   global $db;
   $emailCheck = mysqli_query($db, 'SELECT email FROM user WHERE email = $email');
   if($emailCheck == 'NULL'){
       $hashPass = password_hash($password);
       mysqli_query($db, INSERT INTO user VALUES ($email, $hashPassword));

}

MYSQLi将新行插入数据库

3 个答案: