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ajaxCRUD.com多重关系_Ajax_Foreign Keys_Relational Database_Crud_Foreign Key Relationship - Fatal编程技术网

ajaxCRUD.com多重关系

ajaxCRUD.com多重关系,ajax,foreign-keys,relational-database,crud,foreign-key-relationship,Ajax,Foreign Keys,Relational Database,Crud,Foreign Key Relationship,我真的不确定我是否会找到人来帮我,但让我们试试看 我正在使用来自ajaxCRUD.com的CRUD脚本,我想在我的表之间创建多个关系。但是,当我尝试这样做时,只有第一个关系出现 我想将我的水果表与属性表链接: $tblDemo = new ajaxCRUD("Fruits", "my_fruits", "id", "../"); $tblDemo->defineRelationship("id", "properties", "id", "color"); $tblDemo->def

我真的不确定我是否会找到人来帮我,但让我们试试看

我正在使用来自ajaxCRUD.com的CRUD脚本,我想在我的表之间创建多个关系。但是,当我尝试这样做时,只有第一个关系出现

我想将我的
水果
表与
属性
表链接:

$tblDemo = new ajaxCRUD("Fruits", "my_fruits", "id", "../");
$tblDemo->defineRelationship("id", "properties", "id", "color");
$tblDemo->defineRelationship("id", "properties", "id", "taste");
如果我想从同一个“relationshipped”表中获得颜色和味道,则只会显示其中一个


关于如何在同一个表的两列(或更多列)之间建立关系的任何提示?

如果在数据库中创建一个视图来连接颜色和味道,例如:

CREATE VIEW v_properties AS
SELECT id, CONCAT(color, ' - ', taste) AS color_taste
FROM properties;
然后在代码中引用该视图:

$tblDemo->defineRelationship("id", "v_properties", "id", "color_taste");
这样行吗

或者,如果使用视图,则创建一个仅作为属性别名的视图:

CREATE VIEW properties2 AS
SELECT * FROM properties
并在第二个链接中使用属性2:

$tblDemo->defineRelationship("id", "properties", "id", "color");
$tblDemo->defineRelationship("id", "properties2", "id", "taste");
以我为例

$tblFriend->defineRelationship("idCliente", "clientes", "idClientes","nombreCliente");
$tblFriend->defineRelationship("idCategoria", "categorias", "idCategoria","nombreCategoria");
我认为两人的关系很完美。没有额外的代码或任何东西