Php PDO异常错误-Drupal 7

创建图像内容类型并添加内容时，出现以下错误：

DOException: SQLSTATE[42S22]: Column not found: 1054 Unknown column 'base.12' in 'where clause': SELECT base.tid AS tid, base.vid AS vid, base.name AS name, base.description AS description, base.format AS format, base.weight AS weight, v.machine_name AS vocabulary_machine_name FROM {taxonomy_term_data} base INNER JOIN {taxonomy_vocabulary} v ON base.vid = v.vid WHERE (base.12 = :db_condition_placeholder_0) ; Array ( [:db_condition_placeholder_0] => 12 ) in DrupalDefaultEntityController->load() (line 196 of /Users/httdocs/includes/entity.inc).

但是我不使用分类术语，vocab等等。。。如何修复它？

PGSQL查询如下：-

 SELECT base.tid AS tid, base.vid AS vid, base.name AS name, base.description AS description, base.format AS format, base.weight AS weight, v.machine_name AS vocabulary_machine_name FROM taxonomy_term_data.base INNER JOIN taxonomy_vocabulary.v ON base.vid = v.vid WHERE (base.name = v.machine_name) ;

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在查询中，您通过了base.12，没有名称为12的列。因此它显示了错误

#1064-您的SQL语法有错误；请查看与MySQL服务器版本对应的手册，以了解可在第1行使用的正确语法：db_condition_placeholder_0）LIMIT 0，30

我不知道drupal。。。但该psql查询在where条件下出错。。。where条件，如where（base.name=v.machine\u name）