致命错误:在线调用非对象上的成员函数 query():$result = $conn->query($sql) 或 die(mysqli_error());
Fatal error: Call to a member function query() on a non-object on line: $result = $conn->query($sql) or die(mysqli_error());
谁知道出了什么问题以及如何解决?
Who knows whats wrong and how to fix it?
<?php
function dbConnect($usertype, $connectionType = 'mysqli') {
$host = 'localhost';
$db = 'phpsols';
if ($usertype == 'read') {
$user = 'psread';
$pwd = '123';
} elseif ($usertype == 'write') {
$user = 'pswrite';
$pwd = '123';
} else {
exit('Unrecognized connection type');
}
if ($connectionType == 'mysqli') {
return new mysqli($host, $user, $pwd, $db) or die ('Cannot open database');
} else {
try {
return new PDO("mysql:host=$host;dbname=$db", $user, $pwd);
} catch (PDOException $e) {
echo 'Cannot connect to database';
exit;
}
}
}
// connect to MySQL
$conn = dbConnect('read');
// prepare the SQL query
$sql = 'SELECT * FROM images';
// submit the query and capture the result
**$result = $conn->query($sql) or die(mysqli_error());**
// find out how many records were retrieved
$numRows = $result->num_rows;
?>
<!DOCTYPE HTML>
<html>
<head>
<meta charset="utf-8">
<title>Connecting with MySQLi</title>
</head>
<body>
<p>A total of <?php echo $numRows; ?> records were found.</p>
</body>
</html>
罪魁祸首很可能是这一行:
The culprit is most likely this line:
return new mysqli($host, $user, $pwd, $db) or die ('Cannot open database');
do xyz or die() 构造与 return 语句结合会导致有趣的行为(即整个事情被解释为 OR 表达式，因为 new mysqli 永远不会为假，永远不会处理死".).在此处查看类似案例.
The do xyz or die() construct leads to funny behaviour in conjuction with the return statement (i.e. the whole thing is interpreted as an OR expression and because new mysqli will never be false, the "die" is never processed.). See a similar case here.
改为这样做:
$result = new mysqli($host, $user, $pwd, $db) ;
if (!$result) die (....);
return $result;
另外，有点相关，我认为你永远不会发现 PDO 连接错误，因为:
Also, slightly related, I think you will never catch a PDO connection error because this:
return new PDO("mysql:host=$host;dbname=$db", $user, $pwd);
将总是退出函数，永远不会到达catch块.与您的实际问题一样，解决方案是首先将对象传递给 $result 变量.
will always exit the function, and never reach the catch block. As with your actual problem, the solution is to pass the object to a $result variable first.
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