Question

我正在尝试使用python访问和更新Oracle数据库。以下是我的代码：

import cx_Oracle
import pandas as pd
import datetime
import numpy
import math
#import pandasql

conn_str = 'sahil/sahil@52.20.141.126:1521/xe'
conn = cx_Oracle.connect(conn_str)
c = conn.cursor()

def update_output_table(customer_id_list,column_name,column_vlaue_list) :
    num_rows_to_add = len(customer_id_list)
    conn = cx_Oracle.connect(conn_str)
    c = conn.cursor()

    for i in range(0,num_rows_to_add,1) :
        c.execute("""UPDATE output SET """+column_name+""" = %s WHERE customer_id = %s""" %(column_vlaue_list[i],customer_id_list[i]))
    print "Completed updating " + column_name

    conn.commit()

total_transaction_df = pd.read_sql("""select distinct b.customer_id,count(a.transaction_id) as total_transaction from transaction_fact a,customer_dim b where a.customer_id = b.customer_id group by b.customer_id""",conn)

# Update this details to the output table
update_output_table(list(total_transaction_df['CUSTOMER_ID']),'TOTAL_TRANSACTION',list(total_transaction_df['TOTAL_TRANSACTION']))

english_movies_df = pd.read_sql("""select b.customer_id, count(a.product_id) as "ENGLISH_MOVIES" from transaction_fact a inner join customer_dim b on a.customer_id = b.customer_id where a.product_id like 'E%' group by b.customer_id""",conn)

# Update this details to the output table
update_output_table(list(english_movies_df['CUSTOMER_ID']),'ENGLISH_MOVIES',list(english_movies_df['ENGLISH_MOVIES']))

hindi_movies_df = pd.read_sql("""select b.customer_id, count(a.product_id) as "HINDI_MOVIES" from transaction_fact a inner join customer_dim b on     a.customer_id = b.customer_id where a.product_id like 'H%' group by b.customer_id""",conn)

# Update this details to the output table
update_output_table(list(hindi_movies_df['CUSTOMER_ID']),'HINDI_MOVIES',list(hindi_movies_df['HINDI_MOVIES']))

most_popular_genre_df = pd.read_sql("""select x.customer_id, x.genre as "MOST_POPULAR_GENRE",x.count1 from (select b.customer_id,c.genre,count(a.transaction_id) as count1, RANK() OVER (PARTITION BY b.customer_id order by count(a.transaction_id) desc,c.genre) as Rank1 from transaction_fact a inner join customer_dim b on a.customer_id = b.customer_id inner join product_dim c on a.product_id = c.product_id group by b.customer_id, c.genre)x where x.Rank1 = 1""",conn)

# Update this details to the output table
update_output_table(list(most_popular_genre_df['CUSTOMER_ID']),'MOST_POPULAR_GENRE',list(most_popular_genre_df['MOST_POPULAR_GENRE']))

conn.close()

整个脚本正在执行，但最后一次调用update_output_table的地方除外ORA-00904 : 'UNCONVENTIONAL' invalid identifier。

我已经传递了正确的列名。以下是我创建output表的代码：

create table output
(
    customer_id integer,
    total_transaction integer,
    english_movies integer,
    hindi_movies integer,
    most_popular_genre varchar2(30),
    last_transaction_date date,
    overall_score integer
)

UNCONVENTIONAL是我的事实表产品类别的第一个条目。是选择UNCONVENTIONAL作为我的列名而不是MOST_POPULAR_GENRE吗？如果是，那怎么回事？

注意：我是新手，所以如果这是一个愚蠢的怀疑，请光顾。

Answer 1

由于您的代码中没有提及字段UNCONVENTIONAL，因此很难指出问题所在。但错误信息告诉了很多事情。我鼓励您理解错误消息，以便将来帮助您。

Oracle / PLSQL：ORA-00904错误消息

<强>描述

如果遇到ORA-00904错误，将显示以下错误消息：

ORA-00904: invalid identifier

<强>原因

您尝试执行包含无效列名或缺少列名的SQL语句。当您在SELECT语句中引用无效别名时，通常会发生这种情况。

解决

重写SQL以包含有效的列名。要成为有效的列名，必须满足以下条件：

列名必须以字母开头。

列名称不能超过30个字符。

列名必须由字母数字字符或以下特殊字符组成：$, _, and #。如果列名使用任何其他字符，则必须用双引号括起来。

列名不能是保留字。

让我们看一个如何解决ORA-00904错误的例子。

SQL> SELECT contact_id AS "c_id", last_name, first_name 2 FROM contacts 3 ORDER BY "cid"; ORDER BY "cid" * ERROR at line 3: ORA-00904: "cid": invalid identifie

此错误是通过对列进行别名来创建的，但稍后会对别名进行错误输入。在这个例子中，我们为contact_id创建了名为“c_id”的别名，但后来在ORDER BY子句中将其称为“cid”。

要解决此错误，我们可以修改SELECT语句，以便在ORDER BY子句中使用正确的别名，如下所示：

SQL> SELECT contact_id AS "c_id", last_name, first_name 2 FROM contacts 3 ORDER BY "c_id"; 10 rows selected

参考：Oracle / PLSQL: ORA-00904 Error Message。

Oracle错误：无效的标识符

1 个答案: