print_r($con->_con);
echo "(SELECT * FROM fvwItems WHERE itemname LIKE :key)";
echo "PDO::errorInfo():n";
print_r($con->_con->errorInfo());// no error showing
$stmt = $con->_con->prepare("(SELECT * FROM fvwItems WHERE itemname LIKE :key)", array(PDO::ATTR_CURSOR => PDO::CURSOR_SCROLL));
echo $stmt;//not putting any output
echo "PDO::errorInfo():n";
print_r($con->_con->errorInfo());
我有上面的查询,我只得到
Resource id #10
(SELECT * FROM fvwItems WHERE itemname LIKE :key)PDO::errorInfo():
在Chrome的开发工具中。我也有 try/catch
中的查询,但是我没有错误或查询中有什么问题,因为它没有给我足够的信息来判断哪个有问题。
$stmt = $con->_con->prepare("(SELECT * FROM fvwItems WHERE itemname LIKE :key)", array(PDO::ATTR_CURSOR => PDO::CURSOR_SCROLL));
以外的任何内容均未显示错误或未显示
检查查询有什么问题是什么?
UPDATE
我的连接就是这样:
$connectionOptions = array(
"Database" => DB_NAME,
"Uid" => DB_USER,
"PWD" => DB_PASS
);
//Establishes the connection
$this->_connection = sqlsrv_connect($serverName, $connectionOptions);
我以为我需要它是 new PDO
,以便它运行?
喜欢
$this->_connection = new PDO("sqlsrv:Server=$serverName;Database=".DB_NAME.";",DB_USER,DB_PASS);
print_r($this->_connection);
导致
PDO Object
(
)
,但这不是连接的,并且正在获得错误
创建PDO对象设置此选项:
$this->db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);