我已按照zend 说明使用数据库表实现我的 Web 身份验证。
完全相同的代码,但是在渲染页面时,出现以下异常:
Zend\Authentication\Adapter\Exception\RuntimeException
File:
C:\xampp\htdocs\pfc\vendor\ZF2\library\Zend\Authentication\Adapter\DbTable.php
Mensaje:
The supplied parameters to DbTable failed to produce a valid sql statement, please
check table and column names for validity.
由另一个产生:
Zend\Db\Adapter\Exception\InvalidQueryException
File:
C:\xampp\htdocs\pfc\vendor\ZF2\library\Zend\Db\Adapter\Driver\Mysqli\Statement.php
Mensaje:
Statement couldn't be produced with sql: SELECT `users`.*, (CASE WHEN `password` = ?
THEN 1 ELSE 0 END) AS `zend_auth_credential_match` FROM `users` WHERE `mail` = ?
似乎是 Statement.php 无法执行上面的 sql,但我通过 phpmyadmin 发送 sql 替换 ? 对于字符串和工作正常。
我确信 $dbAdapter 也可以正常工作,因为我已经对其进行了测试,并且列名是“mail”和“password”。
这在我的代码中,我也放了 $dbAdapter 测试代码。
$dbAdapter = new DbAdapter(array( //This DbAdapter Work ok sure!!
'driver' => 'Mysqli',
'database' => 'securedraw',
'username' => 'root',
'password' => ''
));
$fp = function($name) use ($dbAdapter) { return $dbAdapter->driver->formatParameterName($name);};
$sql = 'SELECT * FROM ' . $qi('users') . ' WHERE id = ' . $fp('id');
$statement = $dbAdapter->query($sql);
$parameters = array('id' => 1);
$sqlResult = $statement->execute($parameters);
$row = $sqlResult->current();
$mail = $row['mail'];
$password = $row['password']; //until here test $dbAdapter exitly!!
//Start the auth proccess!!
$authAdapter = new AuthDbTableAdapter($dbAdapter);
$authAdapter->setTableName('users')
->setIdentityColumn('mail')
->setCredentialColumn('password');
$authAdapter->setIdentity('josep')
->setCredential('josep');
$authResult = $authAdapter->authenticate(); //This is the fail method!!!