Python 重复错误

我正在使用下面的模块生成一个唯一的id。但是每天它都会抛出

IntegrityError:（1062，“对键'PRIMARY'重复输入'122830001'）

，用于第一个loanid

我找不到问题的原因。请帮忙

def generate_loanid(leadid='0', counter=0):
    """ Locks the customer_loanid table retrieves maximum loanid from it.
        Generates a new loanid and puts it back in the table.
        Returns the new loanid.
    """
    def gen_id(max_loanid=0, count=0):
        """
        Increments the counter if an id exists for the day.
        Else resets the counter and creates a fresh one.
        """
        timestamp_now = time.localtime()
        if max_loanid:
            logger.debug("""Current Maximum Loan id :
                        (Year(YY)+Julian Day(DDD)+Counter(CCCC) %d,
                        Current timestamp : %s""" %(max_loanid, timestamp_now))
            julian_day = (max_loanid/10000) % 1000
            if julian_day == timestamp_now[7]:
                count = max_loanid % 10000
        return (str(timestamp_now[0])[2:] +
                str(timestamp_now[7]).rjust(3, '0') +
                str(count+1).rjust(4, '0')
                )

    logger.debug("Leadid:%s - Counter:%s"%(leadid, counter))
    db_obj = dbLis(pooling=False)
    try:
        try:
            db_obj.query("lock tables customer_loanid write")
            max_loanid = db_obj.query("select max(loanid) as loanid from customer_loanid")
            curr_loanid = gen_id(max_loanid = max_loanid.__len__() and max_loanid[0].loanid)
            db_obj.insert('customer_loanid', loanid=curr_loanid)
        except (MySQLdb.IntegrityError,MySQLdb.OperationalError):
            logger.warning(traceback.format_exc())
            #There is no logical backing for this piece of code.
            if counter < 3:
                db_obj.query("unlock tables")
                return generate_loanid(counter=counter+1)
            else:
                raise
    finally:
        try:
            db_obj.query("unlock tables")
            logger.debug("Unlocked All Tables")
            db_obj.ctx.db.close()
        except MySQLdb.OperationalError:
            logger.info("Table unlocked already")
            logger.debug(traceback.format_exc())
    logger.info("Generated Loanid : %s"%(curr_loanid))
    return curr_loanid

---

您可以在一个查询中完成这一切：

timestamp_now = time.localtime()
day_part_of_id = str(timestamp_now[0])[2:] + str(timestamp_now[7]).rjust(3, '0')
min_id_of_day = day_part_of_id + '0001'

sql = "INSERT INTO customer_loanid (SELECT "+min_id_of_day+" + count(*) FROM customer_loanid WHERE loanid LIKE '"+day_part_of_id+"%'"

---

为什么要使用

max\u loanid.\uu len\uunid（）

而不是

len（max\u loanid）

？@shiva：我怀疑，但直接调用特殊方法通常不是一个好主意。@MartijnPieters谢谢。我将它改为len（）。你为什么需要如此复杂的方法来计算loanid呢？带有

AUTO_INCREMENT

的id不会起作用吗？@shiva似乎问题在于，计数器以“下一个可用id”的方式工作会产生“溢出”（在代码中由

count=max_loanid%10000处理）。这意味着，如果给定日期的ID数超过10000，则返回的ID将被剥离（计数10001将给出计数1），因此会生成冲突。min_ID_of_day=day_part_of_ID+“0001”将抛出重复错误，但在查询中，ID将添加到该前缀已被删除的ID数（视为一个数字）. 当表中已有三个ID具有相同前缀时，ID将为XXXXX 0004。
timestamp_now = time.localtime()
day_part_of_id = str(timestamp_now[0])[2:] + str(timestamp_now[7]).rjust(3, '0')
min_id_of_day = day_part_of_id + '0001'

sql = "INSERT INTO customer_loanid (SELECT "+min_id_of_day+" + count(*) FROM customer_loanid WHERE loanid LIKE '"+day_part_of_id+"%'"