获取非对象的属性num_rows [英] Get property num_rows of non-object
问题描述
我有一个PHP脚本，可以发送电子邮件，从表中选择数据，然后将一些数据插入表中.我在同一文件中有3个查询.但是在最后一个查询中，我收到了错误消息
I have a PHP script that sends an email, selects data from a table and then inserts some data into a table. I have 3 querys in the same file. But on the last query I get the error message
尝试在第76行的C:\ xampp \ htdocs \ Revolutionen \ includes \ accept.php中获取非对象的属性"num_rows"
Trying to get property 'num_rows' of non-object in C:\xampp\htdocs\Revolutionen\includes\accept.php on line 76
我一直在寻找任何拼写错误，但找不到任何错误.有谁比我聪明，可以帮助我?呵呵.
Ive looked for any typo-errors but coulden't find any. Anyone that's smarter than me that could help me out? Hehe.
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>
";
$headers = "From: noreply@stockholmemergencyroleplay.se" . "\r\n" .
"CC: noreply@stockholmemergencyroleplay.se";
mail($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();
?>```
推荐答案
INSERT查询不会返回结果.因此，当您这样做
INSERT queries don't return results. So when you do
$result3 = $conn->query($sql4);
$result3将是TRUE或FALSE，而不是mysqli_result对象.您不应该尝试检查行数或调用$result3->fetch_assoc()，因为没有要获取的内容.只需检查它是否成功:
$result3 will be either TRUE or FALSE, not a mysqli_result object. You shouldn't try to check the number of rows or call $result3->fetch_assoc(), since there's nothing to fetch. Just check whether it was successful:
if ($result3) {
echo 'Personen blev automatiskt whitelistad.';
} else {
echo "Error" . $conn->error;
}
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