获取非对象的属性num_rows

问题描述 投票:0回答:2

我有一个PHP脚本,它发送电子邮件,从表中选择数据,然后将一些数据插入表中。我在同一个文件中有3个查询。但在最后一个查询中,我收到错误消息

试图在第76行的C:\ xampp \ htdocs \ Revolutionen \ includes \ accept.php中获取非对象的属性'num_rows'

我找了任何拼写错误但是找不到任何错误。谁比我聪明,可以帮助我?呵呵。

完整代码:

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "emildeveloping5";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

mysqli_set_charset($conn,"utf8");

$sql = "UPDATE ansokningar SET besvarad=2 WHERE id=".$_GET['row_id'];

if ($conn->query($sql) === TRUE) {
     echo "Ansökningen godkänndes.";
} else {
    echo "Error updating record: " . $conn->error;
}

$sql2 = "SELECT email FROM ansokningar WHERE id=".$_GET['row_id'];
$result = $conn->query($sql2);

if ($result->num_rows > 0) {
    echo "";
    while($row = $result->fetch_assoc()) {
     $email = $row['email'];
    }

     $to = "$email";
     $subject = "Ansökan | Stockholm Emergency Roleplay";
     $txt = "
     <html>
     <head>
     </head>
     <body>
     <p>Din ansökan är godkänd!</p>
     <p>Intervju tider står i våran Discord.</p>
     <p>Mvh Stockholm Emergency Roleplay.</p>
     </body>
     </html>

     ";
     $headers = "From: [email protected]" . "\r\n" .
     "CC: [email protected]";

     mail($to,$subject,$txt,$headers);

     echo ('Email är skickat.');

} else {
    echo "Error" . $conn->error;
}

$sql3 = "SELECT steam FROM ansokningar WHERE id=".$_GET['row_id'];
$result2 = $conn->query($sql3);

if ($result2->num_rows > 0) {
    echo "";
    while($row = $result2->fetch_assoc()) {
        $steam = $row['steam'];
     echo "";
    }
     echo ('Steam HEX fångat.');

} else {
    echo "Error" . $conn->error;
}

$sql4 = "INSERT INTO whitelist (identifier, whitelisted) VALUES ('$steam', '1')";
$result3 = $conn->query($sql4);

if ($result3->num_rows > 0) {
    echo "";
    while($row = $result3->fetch_assoc()) {
     echo "";
    }
     echo ('Personen blev automatiskt whitelistad.');

} else {
    echo "Error" . $conn->error;
}

$conn->close();
?>```
php mysqli
2个回答
-1
投票

INSERT查询不返回结果。所以,当你这样做

$result3 = $conn->query($sql4);

$result3将是TRUEFALSE,而不是mysqli_result对象。您不应该尝试检查行数或调用$result3->fetch_assoc(),因为没有什么可以获取。只需检查它是否成功:

if ($result3) {
    echo 'Personen blev automatiskt whitelistad.';
} else {
    echo "Error" . $conn->error;
}

-1
投票

实际上问题是你试图访问while循环的变量。

if ($result2->num_rows > 0) 
{ 
 echo ""; 
 while($row =$result2->fetch_assoc())     { 
$steam = $row['steam']; 
echo ""; 
$sql4 = "INSERT INTO whitelist (identifier, whitelisted) VALUES.   ('$steam', '1')"; 
} 
echo 'Steam HEX fångat.'; } 
else { echo "Error" . $conn->error; }

然后$ sql4 =选择白名单查询

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