将不同树中的两个表加入 phpmyadmin

Question

我需要在两个表之间进行连接，其中一个在分组内。但是查询给了我一个错误，我不知道该怎么办。enter image description here

我必须加入 users_field_data.uid 和 user__field_apellidos.entity_id，但它给我错误。

我应该怎么做？

enter image description here

谢谢！

编辑： Table users_field_data:

    CREATE TABLE IF NOT EXISTS `user__field_apellidos` (
      `bundle` varchar(128) CHARACTER SET ascii NOT NULL DEFAULT '' COMMENT 'The field instance bundle to which this row belongs, used when deleting a field instance',
      `deleted` tinyint(4) NOT NULL DEFAULT '0' COMMENT 'A boolean indicating whether this data item has been deleted',
      `entity_id` int(10) unsigned NOT NULL COMMENT 'The entity id this data is attached to',
      `revision_id` int(10) unsigned NOT NULL COMMENT 'The entity revision id this data is attached to, which for an unversioned entity type is the same as the entity id',
      `langcode` varchar(32) CHARACTER SET ascii NOT NULL DEFAULT '' COMMENT 'The language code for this data item.',
      `delta` int(10) unsigned NOT NULL COMMENT 'The sequence number for this data item, used for multi-value fields',
      `field_apellidos_value` varchar(255) NOT NULL
    ) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COMMENT='Data storage for user field field_apellidos.';

ALTER TABLE `user__field_apellidos`
      ADD PRIMARY KEY (`entity_id`,`deleted`,`delta`,`langcode`),
      ADD KEY `bundle` (`bundle`),
      ADD KEY `revision_id` (`revision_id`);


CREATE TABLE IF NOT EXISTS `users_field_data` (
  `uid` int(10) unsigned NOT NULL,
  `langcode` varchar(12) CHARACTER SET ascii NOT NULL,
  `preferred_langcode` varchar(12) CHARACTER SET ascii DEFAULT NULL,
  `preferred_admin_langcode` varchar(12) CHARACTER SET ascii DEFAULT NULL,
  `name` varchar(60) NOT NULL,
  `pass` varchar(255) DEFAULT NULL,
  `mail` varchar(254) DEFAULT NULL,
  `timezone` varchar(32) DEFAULT NULL,
  `status` tinyint(4) DEFAULT NULL,
  `created` int(11) NOT NULL,
  `changed` int(11) DEFAULT NULL,
  `access` int(11) NOT NULL,
  `login` int(11) DEFAULT NULL,
  `init` varchar(254) DEFAULT NULL,
  `default_langcode` tinyint(4) NOT NULL,
  `content_translation_source` varchar(12) CHARACTER SET ascii DEFAULT NULL,
  `content_translation_outdated` tinyint(4) DEFAULT NULL,
  `content_translation_uid` int(10) unsigned DEFAULT NULL COMMENT 'The ID of the target entity.',
  `content_translation_status` tinyint(4) DEFAULT NULL,
  `content_translation_created` int(11) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COMMENT='The data table for user entities.';


ALTER TABLE `users_field_data`
  ADD PRIMARY KEY (`uid`,`langcode`),
  ADD UNIQUE KEY `user__name` (`name`,`langcode`),
  ADD KEY `user__id__default_langcode__langcode` (`uid`,`default_langcode`,`langcode`),
  ADD KEY `user_field__mail` (`mail`(191)),
  ADD KEY `user_field__created` (`created`),
  ADD KEY `user_field__access` (`access`),
  ADD KEY `user_field__content_translation_uid__target_id` (`content_translation_uid`);

Table user__field_apellidos:
CREATE TABLE IF NOT EXISTS `user__field_apellidos` (
  `bundle` varchar(128) CHARACTER SET ascii NOT NULL DEFAULT '' COMMENT 'The field instance bundle to which this row belongs, used when deleting a field instance',
  `deleted` tinyint(4) NOT NULL DEFAULT '0' COMMENT 'A boolean indicating whether this data item has been deleted',
  `entity_id` int(10) unsigned NOT NULL COMMENT 'The entity id this data is attached to',
  `revision_id` int(10) unsigned NOT NULL COMMENT 'The entity revision id this data is attached to, which for an unversioned entity type is the same as the entity id',
  `langcode` varchar(32) CHARACTER SET ascii NOT NULL DEFAULT '' COMMENT 'The language code for this data item.',
  `delta` int(10) unsigned NOT NULL COMMENT 'The sequence number for this data item, used for multi-value fields',
  `field_apellidos_value` varchar(255) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COMMENT='Data storage for user field field_apellidos.';


ALTER TABLE `user__field_apellidos`
  ADD PRIMARY KEY (`entity_id`,`deleted`,`delta`,`langcode`),
  ADD KEY `bundle` (`bundle`),
  ADD KEY `revision_id` (`revision_id`);

Answer 1

你错过了连接部分。尝试以下查询：

s`elect * from user__field_apellidos inner join users_field_data on entity_id = users_field_data.uid`

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