Question

我想用查询更新mysql表，但它显示

错误 - mysqli_query（）期望参数3为长

if($money >= 400) {

$query = "UPDATE users SET spam = spam + 1, money = money - 400 WHERE user_id=".$_SESSION['user'];
$query = "UPDATE users SET score = score + 2500 WHERE user_id=".$_SESSION['user'];

$update = mysqli_query($conn,$query,$query);
echo 'Spam.exe Upgraded for $400!!';

} else {

// Prints error
echo 'Not enough Money!!';

}

Answer 1

只需像这样单独调用2个查询：

if($money >= 400) {

$query1 = "UPDATE users SET spam = spam + 1, money = money - 400 WHERE user_id=".$_SESSION['user'];
$query2 = "UPDATE users SET score = score + 2500 WHERE user_id=".$_SESSION['user'];

$update1 = mysqli_query($conn,$query1); // call 1st query
$update2 = mysqli_query($conn,$query2); // call 2nd query
echo 'Spam.exe Upgraded for $400!!';

} else {

// Prints error
echo 'Not enough Money!!';

}

Answer 2

只需使用 mysqli_multi_query() ，即可使用字符串连接将两个更新查询作为单个查询运行：

$query = "UPDATE users SET spam = spam + 1, money = money - 400 WHERE user_id=".$_SESSION['user'].";";
$query .= "UPDATE users SET score = score + 2500 WHERE user_id=".$_SESSION['user'];

mysqli_multi_query($con,$query);

Answer 3

检查以下代码：

$query = "UPDATE users SET spam = spam + 1, money = money - 400 WHERE user_id=".$_SESSION['user'];
$query1 = "UPDATE users SET score = score + 2500 WHERE user_id=".$_SESSION['user'];

$update = mysqli_query($conn,$query);
$update1 = mysqli_query($conn,$query1);
echo 'Spam.exe Upgraded for $400!!';

Answer 4

mysqli_query的第三个参数是$resultmode。请参阅http://php.net/manual/en/mysqli.query.php。

只需使用mysqli_query($conn,$query).

即可

mysqli_query（）期望参数3很长

4 个答案: