Question

我正在尝试这段代码几个小时。我无法弄清楚我的错误是什么。

$conn = mysqli_connect('localhost', 'user', 'pass', 'mydb')
            or die( "Unable to connect");

$check = mysqli_query($conn, "select * from Users where Fuid='$fbid'",
            MYSQLI_STORE_RESULT);

$rows = mysqli_num_rows($check);


if (empty($rows)) { // if new user . Insert a new record        
    $query = "INSERT INTO Users (Fuid,Funame,Ffname,Femail)
            VALUES ('$fbid','$funame','$ffname','$femail')";

    mysqli_query($conn, $query);    
}

可能会有一个非常小的错误，但我只是被炸了。

Answer 1

这似乎很难做很多事情。试试这个......

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);    
$conn = new mysqli('localhost', 'user', 'pass', 'mydb');
$conn->set_charset('utf8'); // or whatever is appropriate

$check = $conn->prepare('SELECT 1 FROM `Users` WHERE Fuid = ?');
$check->bind_param('i', $fbid); // assuming $fbid is an integer
$check->execute();
if (!$check->fetch()) {
    $sql  = 'INSERT INTO `Users` (Fuid, Funame, Ffname, Femail) VALUES (?, ?, ?, ?)';
    $stmt = $conn->prepare($sql);
    $stmt->bind_param('isss', $fbid, $funame, $ffname, $femail);
    $stmt->execute();
}

Answer 2

使用

mysqli_query($conn, $query);

而不是

 mysqli_query($query, $conn);

Answer 3

你颠倒了参数！

mysqli_query($query, $conn);

应该是

mysqli_query($conn, $query);

试试这个：使用mysqli_connect_errno代替die

$conn = mysqli_connect('localhost', 'user', 'pass', 'mydb');
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}    

$check = mysqli_query($conn, "select * from Users where Fuid='$fbid'",
            MYSQLI_STORE_RESULT);

$rows = mysqli_num_rows($check);


if (empty($rows)) { // if new user . Insert a new record        
    $query = "INSERT INTO Users (Fuid,Funame,Ffname,Femail)
            VALUES ('$fbid','$funame','$ffname','$femail')";

    mysqli_query($conn,$query);    
}

mysqli_query（）期望参数1为mysqli，给定null

3 个答案: