Django模型保存方法：錯誤：在「args」處或附近的語法錯誤

Question

我試圖保存我的模型窗體。 Django重新啓動InternalError: current transaction is aborted, commands ignored until end of transaction block

這是執行INSERT sql-query 時出現的一個問題，它由Django生成。它看起來像：

INSERT INTO "myapp_mymodel" ("title", ...) VALUES ("Test", ...) RETURNING "catalog_ad"."id"; args=("title", ...) 

我試圖在PgAdmin中執行此查詢。它返回了這個錯誤：

ERROR: syntax error at or near "args" 

什麼可能是錯的？

UPD：這裏是tracback爲InternalError該：

File "/Library/Python/2.7/site-packages/django/core/handlers/base.py" in get_response 
    111.       response = callback(request, *callback_args, **callback_kwargs) 
File "/<Path to my app>/views.py" in master 
    115.    ad = form.save() 
File "/Library/Python/2.7/site-packages/django/forms/models.py" in save 
    364.        fail_message, commit, construct=False) 
File "/Library/Python/2.7/site-packages/django/forms/models.py" in save_instance 
    86.   instance.save() 
File "<Path to my app>/models.py" in save 
    145.   super(Ad, self).save(*args, **kwargs) 
File "/Library/Python/2.7/site-packages/django/db/models/base.py" in save 
    463.   self.save_base(using=using, force_insert=force_insert, force_update=force_update) 
File "/Library/Python/2.7/site-packages/django/db/models/base.py" in save_base 
    551.     result = manager._insert([self], fields=fields, return_id=update_pk, using=using, raw=raw) 
File "/Library/Python/2.7/site-packages/django/db/models/manager.py" in _insert 
    203.   return insert_query(self.model, objs, fields, **kwargs) 
File "/Library/Python/2.7/site-packages/django/db/models/query.py" in insert_query 
    1576.  return query.get_compiler(using=using).execute_sql(return_id) 
File "/Library/Python/2.7/site-packages/django/db/models/sql/compiler.py" in execute_sql 
    910.    cursor.execute(sql, params) 

UPD2：

我的模型具有overwrited保存方法：

def save(self, *args, **kwargs): 
    self.title = self.model.brand.name + " " + self.model.name 
    super(Ad, self).save(*args, **kwargs) 

數據由視圖，它使用處理連線ModelForm（從POST獲取數據）。表單在保存之前已經過驗證（form.is_valid（））。

Answer 1

我會說你在你的代碼中有一個通過Django產生INSERT語句的錯誤。如果沒有看到代碼，很難說出問題所在。您發佈的SQL中的分號結束了一個看起來有效的SQL語句。

args=("title", ...)部分看起來不像SQL。檢查你的代碼，看看什麼可能會意外地附加到生成SQL的任何東西。