Question

我有2张桌子，用户＆amp; country 注意：表名是单数

ADOdb_Active_Record::TableKeyBelongsTo(
    'user', // child table name
    'id', // child primary key
    'country', // parent table name
    'country_id', // child foreign key
    'id' // parent primary key
);

class User_Model extends ADOdb_Active_Record {
    var $_table = 'user';
    ...
}

ADOdb_Active_Record::TableKeyHasMany(
    'country', // parent table name
    'id', // parent primary key
    'user', // child table name
    'country_id' // child foreign key
);

class Country_Model extends ADOdb_Active_Record {
    var $_table = 'country';
    ...
}

$user = new User_Model();
$user->load('id=1');

但是出现了这个错误

1146: Table 'happylucky.countries' doesn't exist

                           ADOConnection._Execute(SHOW COLUMNS FROM `countries`, false) % line 1017, file: adodb.inc.php
                        ADOConnection.Execute(SHOW COLUMNS FROM `countries`) % line  410, file: adodb-mysql.inc.php
                     ADODB_mysql.MetaColumns(countries) % line  390, file: adodb-active-record.inc.php
                  ADODB_Active_Record.UpdateActiveTable() % line  239, file: adodb-active-record.inc.php
               ADODB_Active_Record.belongsTo(country, country_id, id, ADODB_Active_Record) % line  263, file: adodb-active-record.inc.php

ADODB_Active_Record::UpdateActiveTable: Invalid table name: countries
Fatal error: Class 'User' not found in /home/user/public_html/project/application/controllers/welcome.php on line 23

我的关系设置错了吗？或者表名必须是复数？

Answer 1

您可以尝试设置

ADODB_Active_Record::$_changeNames = FALSE;

使用活动记录之前

php ADOdb Active Record关系设置

1 个答案: