[英]Set_relation_n_n grocery crud and condeigniter and MYSQL
我在雜貨店使用set_relation_n_n時遇到問題
問題是mysql告訴我,檢索數據的數據庫有錯誤的MySQL 5.6.23語法的查詢是我服務器的版本。 並且不知道在哪里可以找到不允許我編輯系統中的字段的解決方案
我把這兩張桌子搞得一團糟
表
Docente_Escuela
(Id_afiliadoint(11)NOT NULL,
Id_escuelaint(11)NOT NULL,優先級
int(11)NOT NULL,
PRIMARY KEY(id_afiliado
,id_escuela
))
ENGINE = InnoDB DEFAULT CHARSET = utf8 COLLATE = utf8_swedish_ci;
另一個
表
Escuela
(Id_escuelaint(11)NOT NULL AUTO_INCREMENT,
NÚMEROvarchar(10)NOT NULL utf8_spanish_ci COLLATE,
Name
varchar(200)NOT NULL utf8_spanish_ci COLLATE,
Domiciliovarchar(100)NOT NULL utf8_spanish_ci COLLATE,
Id_nivelint(11)NOT NULL,Id_modalidad
int(11)NOT NULL,
Id_ciudadint(11)NOT NULL,
Cod_postalvarchar(30)NOT NULL utf8_spanish_ci COLLATE,
Telefonovarchar(50)NOT NULL utf8_spanish_ci COLLATE,
Celularvarchar(50)NOT NULL utf8_spanish_ci COLLATE,
郵件varchar(100)NOT NULL utf8_spanish_ci COLLATE,
Mail_altvarchar(100)NOT NULL utf8_spanish_ci COLLATE,
Mail_alt_2varchar(100)NOT NULL utf8_spanish_ci COLLATE,
Observacionesint(11)NOT NULL,
Clave_cobrovarchar(10)NOT NULL utf8_spanish_ci COLLATE,
Status
tinyint(1)NOT NULLid_escuela
KEY(id_escuela
))ENGINE = InnoDB DEFAULT CHARSET = utf8 COLLATE = utf8_spanish_ci AUTO_INCREMENT = 2;
和表會員
表
Afiliado
(Id_afiliadoint(11)NOT NULL AUTO_INCREMENT,
Name
varchar(100)NOT NULL utf8_spanish2_ci COLLATE,
Apellidovarchar(100)NOT NULL utf8_spanish2_ci COLLATE,
這份執行varchar(11)NOT NULL utf8_spanish2_ci COLLATE,
Domiciliovarchar(200)NOT NULL utf8_spanish2_ci COLLATE,
Id_situacionint(11)NOT NULL,Id_localidad
int(11)NOT NULL,
Id_provinciaint(11)NOT NULL,
Cod_postalvarchar(30)NOT NULL utf8_spanish2_ci COLLATE,
Telefono_fijovarchar(30)NOT NULL utf8_spanish2_ci COLLATE,
Telefono_altvarchar(30)NOT NULL utf8_spanish2_ci COLLATE,
Celularvarchar(30)NOT NULL utf8_spanish2_ci COLLATE,
郵件varchar(100)NOT NULL utf8_spanish2_ci COLLATE,
Mail_altvarchar(100)NOT NULL utf8_spanish2_ci COLLATE,
Clave_cobrovarchar(20)NOT NULL utf8_spanish2_ci COLLATE,
Valor_cuotavarchar(20)NOT NULL utf8_spanish2_ci COLLATE,
Status
tinyint(1)NOT NULLid_afiliado
KEY(id_afiliado
))ENGINE = InnoDB DEFAULT CHARSET = utf8 COLLATE = utf8_spanish2_ci AUTO_INCREMENT = 6;
所以我打電話給系統
$ Crud-> set_relation_n_n('Schools','Docente_Escuela','School','id_afiliado''id_escuela','name','priority');
這是錯誤
錯誤號碼:1064
您的SQL語法有錯誤; 查看與您的MySQL服務器版本對應的手冊,以便在'WHERE``Escuela附近使用正確的語法
.
id_escuelaid_afiliado
='5'在第3行按Docente_Escuela
.`pri'排序SELECT * FROM
s7a675883 name as Docente_Escuela
EscuelaDocente_Escuela
ON LEFT JOIN
Docente_Escuela
。 Id_escuela =``在WHERE
Escuela.
id_escuelaid_afiliado
='5'按Docente_Escuela
。priority
文件名:models / Grocery_crud_model.php
行號:336
有人可以幫助我推進一下嗎? 我被鎖了
謝謝!! 布魯諾
你的crud調用有語法錯誤。
檢查'id_afiliado'
和'id_escuela'
之間$crud->set_relation
缺少的逗號。
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