在Laravel 5中,我正在尝试进行客户查询。我的代码是这样的:
$ params = array( 'criteria'=> $标准, 'criteria1'=> $标准 );
//Define the SQL
$sql = 'SELECT * FROM ' . $this -> _taskTableName .'
JOIN ' . $this -> _userTableName .' ON
' . $this -> _userTableName .'.id = ' . $this -> _taskTableName .' .client_id
WHERE notes LIKE \'%:criteria%\' OR name LIKE \'%:criteria1%\' ';
//Exeute the search
$tasks = DB::statement(DB::raw($sql),$params);
除非我不断收到此错误,即使我删除了DB :: raw,我也尝试过DB :: select。
SQLSTATE[HY093]: Invalid parameter number: :criteria (SQL: SELECT * FROM tasks
JOIN users ON
users.id = tasks .client_id
WHERE notes LIKE '%:criteria%' OR name LIKE '%:criteria1 %' )
in Connection.php (line 647)
at Connection->runQueryCallback(object(Expression), array('criteria' => 'Devin', 'criteria1' => 'Devin'), object(Closure))
in Connection.php (line 607)
at Connection->run(object(Expression), array('criteria' => 'Devin', 'criteria1' => 'Devin'), object(Closure))
in Connection.php (line 450)
有没有人知道为什么会发生这种情况以及如何解决?
答案 0 :(得分:0)
尝试将DB:select()
与$params
数组一起用作第二个参数,如下所示:
$params = array( 'criteria' => $criteria, 'criteria1' => $criteria );
//Define the SQL
$sql = 'SELECT * FROM ' . $this -> _taskTableName .'
JOIN ' . $this -> _userTableName .' ON
' . $this -> _userTableName .'.id = ' . $this -> _taskTableName .' .client_id
WHERE notes LIKE \'%:criteria%\' OR name LIKE \'%:criteria1%\' ';
//Exeute the search
$tasks = DB::select($sql, $params);
答案 1 :(得分:0)
你可以这样做
$sql = 'SELECT * FROM ' . $this -> _taskTableName .'
JOIN ' . $this -> _userTableName .' ON
' . $this -> _userTableName .'.id = ' . $this -> _taskTableName .' .client_id
WHERE notes LIKE \'%?%\' OR name LIKE \'%?%\' ';
DB::select($sql,array($criteria,$criteria));