Question

因为我的codeigniter返回语法错误？

        $handler_feedback = $this->CI->db
            ->select('
                feedback.id as feedback_id,
                feedback.titulo,
                relacion_feedback_usuario_principal.valor
            ')
            ->from('feedback')
            ->join(
                'relacion_feedback_usuario_principal', 
                '
                    relacion_feedback_usuario_principal.feedback_id = feedback.id AND
                    relacion_feedback_usuario_principal.usuario_principal_id = 20
                ', 
                'left'
            )
            ->get();

Error Number: 1064

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 3

SELECT `feedback`.`id` as feedback_id, `feedback`.`titulo`, `relacion_feedback_usuario_principal`.`valor` FROM (`feedback`) LEFT JOIN `relacion_feedback_usuario_principal` ON `relacion_feedback_usuario_principal`.`feedback_id` = `feedback`.`id` AND

Filename: /dir/file.php

Line Number: 289

Mi原始查询：

        SELECT 
            feedback.id as feedback_id,
            feedback.titulo,
            relacion_feedback_usuario_principal.valor
        FROM feedback
        LEFT JOIN relacion_feedback_usuario_principal ON (
            relacion_feedback_usuario_principal.feedback_id = feedback.id AND
            relacion_feedback_usuario_principal.usuario_principal_id = 20
        )

Answer 1

它应该按照您编写的方式工作它很简单，

$this->db->join('B', 'aCol = bCol AND bOtherCol = 0');
$this->db->get('A');

无论如何，您可以尝试其他方式（join带where子句），

$this->db->select('t1.id as feedback_id, t1.titulo, t2.valor');
$this->db->from('feedback as t1');
$this->db->join('relacion_feedback_usuario_principal as t2', 't1.id = t2.feedback_id', 'left');
$this->db->where('t2.usuario_principal_id', 20);

$query = $this->db->get();

Answer 2

relacion_feedback_usuario_principal.feedback_id = feedback.id AND
                    relacion_feedback_usuario_principal.usuario_principal_id = 20

AND和relaction_feedback_ususario_principal.usuario_principal_id=20之间是否有空格？

Answer 3

我得出的结论是我的代码还不错，这是codeigniter中的一个错误。

Codeigniter sql语法错误？

3 个答案: