php 在布尔值上调用成员函数 fetch()
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Call to a member function fetch() on boolean
提问by marjanos
I receive this error:
我收到此错误：
Fatal error: Call to a member function fetch() on boolean in C:\xampp\htdocs\repo\generator\model\database.php on line 34
致命错误：在第 34 行调用 C:\xampp\htdocs\repo\generator\model\database.php 中布尔值的成员函数 fetch()
When I run this code:
当我运行此代码时：
class database
{
private $user = 'root';
private $pass = '';
public $pdo;
public function connect() {
try {
$this->pdo = new PDO('mysql:host=localhost; dbname=generatordatabase', $this->user, $this->pass);
echo 'Po??czenie nawi?zane!';
}
catch(PDOException $e) {
echo 'Po??czenie nie mog?o zosta? utworzone: ' . $e->getMessage();
}
}
public function createTable() {
$q = $this->pdo -> query('SELECT * FROM article');
while($row = $q->fetch()) {
echo $row['id'].' ';
}
$q->closeCursor();
}
}
?>
回答by ajmedway
As per the PHP manual for PDO::query
根据PDO::query的 PHP 手册
PDO::query() returns a PDOStatement object, or FALSE on failure.
PDO::query() 返回一个 PDOStatement 对象，或者失败时返回 FALSE。
It looks like your query is failing (on line 33) and thus returning a BOOLEAN (false), likely because at that point in execution, PDO has not connected to a database that contains a table called article. In the connect() method I see that it tries to connect to a db called 'generatordatabase'; ensure this connection is being made prior to calling createTable(), otherwise ensure that it contains a table called 'article'.
看起来您的查询失败（在第 33 行）并因此返回 BOOLEAN (false)，可能是因为在执行时，PDO 尚未连接到包含名为article的表的数据库。在 connect() 方法中，我看到它尝试连接到名为“generatordatabase”的数据库；确保在调用 createTable() 之前建立此连接，否则确保它包含一个名为“article”的表。
I would recommend adding some more code examples, for instance the code that calls this class/method before the error is triggered.
我建议添加更多代码示例，例如在触发错误之前调用此类/方法的代码。
回答by Aris
Some error handling will help you avoid issues like this:
一些错误处理将帮助您避免这样的问题：
$q = $this->pdo->query('SELECT * FROM article');
//error case
if(!$q)
{
die("Execute query error, because: ". print_r($this->pdo->errorInfo(),true) );
}
//success case
else{
//continue flow
}
回答by Hossein
I'm not sure wheatear this is exactly the error I struggled with, but my error was due to my $con variable, I used a single $con for 2 SQL statements, for example:
我不确定 wheatear 这正是我遇到的错误，但我的错误是由于我的 $con 变量造成的，我对 2 个 SQL 语句使用了一个 $con，例如：
$con = new mysqli($host,$username,$password,$database);
$sql = "SELECT name FROM users WHERE email = '$email'";
$stm = $con->prepare($sql);
$stm->execute();
and
和
$sql1 = "INSERT INTO posts
VALUES('$email','$body')";
$stm1 = $con->prepare($sql1);
if ($stm1->execute()) {
I should have done:
我应该这样做：
$con = new mysqli($host,$username,$password,$database);
$sql = "SELECT name FROM users WHERE email = '$email'";
$stm = $con->prepare($sql);
$stm->execute();
and
和
$con1 = new mysqli($host,$username,$password,$database);
$sql1 = "INSERT INTO posts
VALUES('$email','$body')";
$stm1 = $con1->prepare($sql1);
$stm1->execute()
