PDOException错误 - Drupal 7

时间:2012-12-11 10:28:13

标签: php mysql drupal-7

当我创建图像内容类型并添加内容时,会收到以下错误:

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).

但我不使用分类术语,词汇等...我该如何解决?

1 个答案:

答案 0 :(得分:0)

PGSQL查询如: -

 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) ;

在您的查询中,您传递了base.12,没有名称为12的列。因此它显示错误

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