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Set_relation_n_n杂货crud和condeigniter和MYSQL

[英]Set_relation_n_n grocery crud and condeigniter and MYSQL

我在杂货店使用set_relation_n_n时遇到问题

问题是mysql告诉我,检索数据的数据库有错误的MySQL 5.6.23语法的查询是我服务器的版本。 并且不知道在哪里可以找到不允许我编辑系统中的字段的解决方案

我把这两张桌子搞得一团糟

Docente_Escuela (Id_afiliado int(11)NOT NULL,
Id_escuela int(11)NOT NULL,优先级 int(11)NOT NULL,
PRIMARY KEY( id_afiliadoid_escuela ))
ENGINE = InnoDB DEFAULT CHARSET = utf8 COLLATE = utf8_swedish_ci;

另一个

Escuela (Id_escuela int(11)NOT NULL AUTO_INCREMENT,
NÚMERO varchar(10)NOT NULL utf8_spanish_ci COLLATE,
Name varchar(200)NOT NULL utf8_spanish_ci COLLATE,
Domicilio varchar(100)NOT NULL utf8_spanish_ci COLLATE,
Id_nivel int(11)NOT NULL,Id_modalidad int(11)NOT NULL,
Id_ciudad int(11)NOT NULL,
Cod_postal varchar(30)NOT NULL utf8_spanish_ci COLLATE,
Telefono varchar(50)NOT NULL utf8_spanish_ci COLLATE,
Celular varchar(50)NOT NULL utf8_spanish_ci COLLATE,
邮件 varchar(100)NOT NULL utf8_spanish_ci COLLATE,
Mail_alt varchar(100)NOT NULL utf8_spanish_ci COLLATE,
Mail_alt_2 varchar(100)NOT NULL utf8_spanish_ci COLLATE,
Observaciones int(11)NOT NULL,
Clave_cobro varchar(10)NOT NULL utf8_spanish_ci COLLATE,
Status tinyint(1)NOT NULL id_escuela KEY( id_escuela ))ENGINE = InnoDB DEFAULT CHARSET = utf8 COLLATE = utf8_spanish_ci AUTO_INCREMENT = 2;

和表会员

Afiliado (Id_afiliado int(11)NOT NULL AUTO_INCREMENT,
Name varchar(100)NOT NULL utf8_spanish2_ci COLLATE,
Apellido varchar(100)NOT NULL utf8_spanish2_ci COLLATE,
这份执行 varchar(11)NOT NULL utf8_spanish2_ci COLLATE,
Domicilio varchar(200)NOT NULL utf8_spanish2_ci COLLATE,
Id_situacion int(11)NOT NULL,Id_localidad int(11)NOT NULL,
Id_provincia int(11)NOT NULL,
Cod_postal varchar(30)NOT NULL utf8_spanish2_ci COLLATE,
Telefono_fijo varchar(30)NOT NULL utf8_spanish2_ci COLLATE,
Telefono_alt varchar(30)NOT NULL utf8_spanish2_ci COLLATE,
Celular varchar(30)NOT NULL utf8_spanish2_ci COLLATE,
邮件 varchar(100)NOT NULL utf8_spanish2_ci COLLATE,
Mail_alt varchar(100)NOT NULL utf8_spanish2_ci COLLATE,
Clave_cobro varchar(20)NOT NULL utf8_spanish2_ci COLLATE,
Valor_cuota varchar(20)NOT NULL utf8_spanish2_ci COLLATE,
Status tinyint(1)NOT NULL id_afiliado KEY( id_afiliado ))ENGINE = InnoDB DEFAULT CHARSET = utf8 COLLATE = utf8_spanish2_ci AUTO_INCREMENT = 6;

所以我打电话给系统

$ Crud-> set_relation_n_n('Schools','Docente_Escuela','School','id_afiliado''id_escuela','name','priority');

这是错误

错误号码:1064

您的SQL语法有错误; 查看与您的MySQL服务器版本对应的手册,以便在'WHERE``Escuela附近使用正确的语法. id_escuela id_afiliado ='5'在第3行按Docente_Escuela .`pri'排序

SELECT * FROM s7a675883 name as Docente_Escuela Escuela Docente_Escuela ON LEFT JOIN Docente_Escuela Id_escuela =``在WHERE Escuela . id_escuela id_afiliado ='5'按Docente_Escuela priority

文件名:models / Grocery_crud_model.php

行号:336

有人可以帮助我推进一下吗? 我被锁了

谢谢!! 布鲁诺

你的crud调用有语法错误。

检查'id_afiliado''id_escuela'之间$crud->set_relation缺少的逗号。

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