Question

I m试图在数据库内一栏中更新记录。我目前的法典如下:

curs = conn.cursor()
statement= SELECT column FROM table 
curs.execute(statement)
curs.execute("INSERT INTO table VALUES (4,  Five )")

In my understanding, the 4th row of the column should be updated to Five . After I ran it, there is no error, but no update neither. There must be something wrong in my codes, or I m missing something important. What if I want to update the whole column s records? Thanks in advance for any clarification.

Answer 1

我用以下法典解决了我的问题:

conn = psycopg2.connect(conn_string)
curs = conn.cursor()
statement= UPDATE table SET column = false 
curs.execute(statement)
conn.commit()
conn.close()

你们的帮助

Answer 2

update `table_name` set `column_name` = value

虽然你想要使核查组相信,你会试图将新一栏的数值中的所有段落加以更新,否则,你需要增加。

where `unique_key` = unique_value

Answer 3

根据你使用MSSQL、Oracle、MySQL的版本,你需要不同的辛加税。你们喜欢再次使用劳动和社会保障部,因此,你希望从此开始:

UPDATE table SET column = "Five"

但你只能说第4行至X射线,没有像Excel电子表格这样固定的流量。你希望增加一栏,把它称为PK——表名,把它定为表上的主要关键。然后,你可以作类似发言,并总是更新正确的行文:

UPDATE table SET column = "Five" WHERE PK_tablename = 4

我建议阅读一下你帮助的“小学钥匙”。

Answer 4

<编码>insert/code>没有更新数据库中的数值,而是增加了记录。页: 1

Answer 5

页: 1 正如其他人所说的那样,这将添加记录而不是更新记录。

基本上有4项简单查询的基本业务:

SELECT: Use this to view data. JOIN tables, set a WHERE clause, etc. to manipulate the view of the data. This is the safest operation (though a bad SELECT can bring a server s performance to its knees).
INSERT: This will insert a row into the table. It s fairly safe, as it doesn t change any existing data.
UPDATE: This will update one or _more_ rows in the table. The main thing to worry about here is the WHERE clause. If you don t include one, you will update every row in the table. Generally you want to test your WHERE clause in a SELECT statement first to make absolutely sure you re updating only the rows you want to update.
DELETE: This is, naturally, the most dangerous. Again, without a WHERE clause it will delete all rows in the table. Test your WHERE clause to make sure you re deleting only the rows you want to delete.

根据您的<代码>INSERT说明,你似乎重新尝试根据钥匙(号码4)更新一行。由于缺乏错误,这似乎实际上不是一个关键因素。如果该表是关键,它本会再回一个错误,说你可以插入一条带有重复钥匙的行文。

seems (根据有关信息的有限,自然)你想要的是:

UPDATE table SET column =  five  WHERE id = 4

详情见here。

友情链接