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當我創建了一個形象的內容類型和添加內容,出現此錯誤:PDOException錯誤 - Drupal的7
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).
但我不使用分類術語,詞彙等...我怎麼能修理它?
'#1064 - 你必須在你的錯誤SQL語法;檢查與您的MySQL服務器版本相對應的手冊,以在':db_condition_placeholder_0'附近使用正確的語法。LIMIT 0,30'在第1行' –
我不知道drupal ...但是psql查詢在where條件處有錯誤。 ..在哪裏條件像WHERE(base.name = v.machine_name) – Daya