php 致命错误：在非对象上调用成员函数 query()
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Fatal error: Call to a member function query() on a non object
提问by Weinz
I've got this error:
我有这个错误：
Fatal error: Call to a member function query() on a non-object in /Applications/XAMPP/xamppfiles/htdocs/login.php on line 8
致命错误：在第 8 行的 /Applications/XAMPP/xamppfiles/htdocs/login.php 中的非对象上调用成员函数 query()
The line is this:
这条线是这样的：
$res = $mysqli->query("SELECT * FROM user WHERE user='$user' and password='$pw'");
This is login.php:
这是登录.php：
$user = $_POST['user'];
$pass = $_POST['pass'];
$pw = md5($pass);
include_once('connect.php');
function check_login($user,$pw,&$result){
$res = $mysqli->query("SELECT * FROM user WHERE user='$user' and password='$pw'");
$cont = 0;
while($row = $res->fetch_object()){
$cont++;
$result = $row;
}
if($cont == 1){
return 1;
}
else{
return 0;
}
}
if(!isset($_SESSION['userid'])){
if(isset($_POST['login'])){
if(check_login($user,$pw,$result) == 1){
session_start();
$_SESSION['userid'] = $result->id_user;
header("location:index.php?var=ok");
}
else{
header('location:index.php?var=log');
}
}
}
And the code of connect.php :
和 connect.php 的代码：
$mysqli = new mysqli('localhost', 'root', 'pass', 'cms' );
if ($mysqli->connect_error) {
die('Error de Conexión (' . $mysqli->connect_errno . ') '
. $mysqli->connect_error);
}
What could be the problem? Problems connecting the database?
可能是什么问题呢？连接数据库有问题？
回答by helmbert
This is most likely a scopingissue. This means that the variable $mysqlithat you define in your included file is outside of the check_loginfunction's scope (i.e. is not known inside this function).
这很可能是一个范围界定问题。这意味着$mysqli您在包含文件中定义的变量在check_login函数的作用域之外（即在该函数内部是未知的）。
You could try to get the $mysqlivariable from global scope with
您可以尝试$mysqli从全局范围获取变量
function check_login($user,$pw,&$result){
global $mysqli;
$res = $mysqli->query("SELECT * FROM user WHERE user='$user' and password='$pw'");
// ...
Edit: Oh, and you should also mind the SQL injection vulnerabiliy in your code. Use prepared statementsto prevent this issue (or at least escape your input variables with a function like mysqli::real_escape_string).
编辑：哦，您还应该注意代码中的 SQL 注入漏洞。使用准备好的语句来防止这个问题（或者至少用一个函数来转义你的输入变量mysqli::real_escape_string）。
回答by Green Black
function check_login($user,$pw,&$result){
global $mysqli;
$res = $mysqli->query("SELECT * FROM user WHERE user='$user' and password='$pw'");
$cont = 0;
while($row = $re[...]
Mind the "global". This puts the var in the scope of your method. This is a quick and dirty solution, but will work in this case.
注意“全球”。这会将 var 置于您的方法范围内。这是一个快速而肮脏的解决方案，但在这种情况下会起作用。
回答by Tomasz Fija?kowski
variable $mysqli in function check_login is out of scope (it is declare outside the function) so it is null.
函数 check_login 中的变量 $mysqli 超出范围（它在函数外声明），因此它为空。
回答by Tucker
It looks like your $mysqli variable is not being set properly within the scope of your code thats executing the query. Can you confirm that $mysqli is indeed a mysqli object and not, say, set to null?
看起来您的 $mysqli 变量没有在执行查询的代码范围内正确设置。你能确认 $mysqli 确实是一个 mysqli 对象而不是，比如说，设置为 null？
You can check this by doing: echo print_r($mysqli);right before you call the ->query method in your code.
您可以通过执行以下操作来检查：echo print_r($mysqli);就在您在代码中调用 ->query 方法之前。
If it is not set properly, track the variable backward in your code until you see why
如果未正确设置，请在代码中向后跟踪变量，直到您明白原因
回答by Robot Boy
public function connectDB($DBServer, $DBUser, $DBPass, $DBName) {
global $con;
$con = new mysqli($DBServer, $DBUser, $DBPass, $DBName) or die ("Error occured");
return $con;
and then pass $conin all the functions that you define or run wherever.
然后传入$con您在任何地方定义或运行的所有函数。
