langchain sql agent 案例
原文：SQL 数据库 |️ 朗链 (langchain.com)langchainSQL 数据库 |️ 朗链 (langchain.com)
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SQL 数据库
此笔记本展示了旨在与数据库交互的代理。 该代理基于 SQLDatabaseChain 构建，旨在回答有关数据库的更多常见问题，以及从错误中恢复。SQL
请注意，由于此代理处于积极开发阶段，因此所有答案可能都不正确。此外，对于某些问题，不能保证代理不会对您的数据库执行 DML 语句。小心在敏感数据上运行它！
这将使用示例数据库。要按照 2 Sample Databases for SQLite 上的说明进行设置，将.db文件放在此存储库根目录下的笔记本文件夹中。Chinook
初始化
from langchain.agents import create_sql_agent
from langchain.agents.agent_toolkits import SQLDatabaseToolkit
from langchain.sql_database import SQLDatabase
from langchain.llms.openai import OpenAI
from langchain.agents import AgentExecutor
from langchain.agents.agent_types import AgentType
from langchain.chat_models import ChatOpenAI
db = SQLDatabase.from_uri("sqlite:///../../../../../notebooks/Chinook.db")
toolkit = SQLDatabaseToolkit(db=db, llm=OpenAI(temperature=0))
用 ZERO_SHOT_REACT_DESCRIPTION
这显示了如何使用ZERO_SHOT_REACT_DESCRIPTION代理类型初始化代理。
agent_executor = create_sql_agent(
llm=OpenAI(temperature=0),
toolkit=toolkit,
verbose=True,
agent_type=AgentType.ZERO_SHOT_REACT_DESCRIPTION,
)
使用 OpenAI 函数
这显示了如何使用OPENAI_FUNCTIONS代理类型初始化代理。请注意，这是上述方法的替代方法。
# agent_executor = create_sql_agent(
# llm=ChatOpenAI(temperature=0, model="gpt-3.5-turbo-0613"),
# toolkit=toolkit,
# verbose=True,
# agent_type=AgentType.OPENAI_FUNCTIONS
# )
免責聲明 ⚠️
查询链可能会生成插入/更新/删除查询。如果不希望出现这种情况，请使用自定义提示符或创建没有写入权限的 SQL 用户。
最终用户可能会通过询问一个简单的问题（例如“运行可能的最大查询”）来重载 SQL 数据库。生成的查询可能如下所示：
SELECT * FROM "public"."users"
JOIN "public"."user_permissions" ON "public"."users".id = "public"."user_permissions".user_id
JOIN "public"."projects" ON "public"."users".id = "public"."projects".user_id
JOIN "public"."events" ON "public"."projects".id = "public"."events".project_id;
对于事务性 SQL 数据库，如果上表之一包含数百万行，则查询可能会给使用同一数据库的其他应用程序带来麻烦。
大多数面向数据仓库的数据库都支持用户级配额，以限制资源使用。
示例：描述表
agent_executor.run("Describe the playlisttrack table")
> Entering new chain...
Invoking: `list_tables_sql_db` with `{}`
Album, Artist, Track, PlaylistTrack, InvoiceLine, sales_table, Playlist, Genre, Employee, Customer, Invoice, MediaType
Invoking: `schema_sql_db` with `PlaylistTrack`
CREATE TABLE "PlaylistTrack" (
"PlaylistId" INTEGER NOT NULL,
"TrackId" INTEGER NOT NULL,
PRIMARY KEY ("PlaylistId", "TrackId"),
FOREIGN KEY("TrackId") REFERENCES "Track" ("TrackId"),
FOREIGN KEY("PlaylistId") REFERENCES "Playlist" ("PlaylistId")
)
/*
3 rows from PlaylistTrack table:
PlaylistId TrackId
1 3402
1 3389
1 3390
*/The `PlaylistTrack` table has two columns: `PlaylistId` and `TrackId`. It is a junction table that represents the relationship between playlists and tracks.
Here is the schema of the `PlaylistTrack` table:
```
CREATE TABLE "PlaylistTrack" (
"PlaylistId" INTEGER NOT NULL,
"TrackId" INTEGER NOT NULL,
PRIMARY KEY ("PlaylistId", "TrackId"),
FOREIGN KEY("TrackId") REFERENCES "Track" ("TrackId"),
FOREIGN KEY("PlaylistId") REFERENCES "Playlist" ("PlaylistId")
)
```
Here are three sample rows from the `PlaylistTrack` table:
```
PlaylistId TrackId
1 3402
1 3389
1 3390
```
Please let me know if there is anything else I can help you with.
> Finished chain.
'The `PlaylistTrack` table has two columns: `PlaylistId` and `TrackId`. It is a junction table that represents the relationship between playlists and tracks. \n\nHere is the schema of the `PlaylistTrack` table:\n\n```\nCREATE TABLE "PlaylistTrack" (\n\t"PlaylistId" INTEGER NOT NULL, \n\t"TrackId" INTEGER NOT NULL, \n\tPRIMARY KEY ("PlaylistId", "TrackId"), \n\tFOREIGN KEY("TrackId") REFERENCES "Track" ("TrackId"), \n\tFOREIGN KEY("PlaylistId") REFERENCES "Playlist" ("PlaylistId")\n)\n```\n\nHere are three sample rows from the `PlaylistTrack` table:\n\n```\nPlaylistId TrackId\n1 3402\n1 3389\n1 3390\n```\n\nPlease let me know if there is anything else I can help you with.'
示例：描述表，从错误中恢复
在此示例中，代理尝试搜索不存在的表，但找到下一个最佳结果
agent_executor.run("Describe the playlistsong table")
> Entering new AgentExecutor chain...
Action: list_tables_sql_db
Action Input: ""
Observation: Genre, PlaylistTrack, MediaType, Invoice, InvoiceLine, Track, Playlist, Customer, Album, Employee, Artist
Thought: I should look at the schema of the PlaylistSong table
Action: schema_sql_db
Action Input: "PlaylistSong"
Observation: Error: table_names {'PlaylistSong'} not found in database
Thought: I should check the spelling of the table
Action: list_tables_sql_db
Action Input: ""
Observation: Genre, PlaylistTrack, MediaType, Invoice, InvoiceLine, Track, Playlist, Customer, Album, Employee, Artist
Thought: The table is called PlaylistTrack
Action: schema_sql_db
Action Input: "PlaylistTrack"
Observation:
CREATE TABLE "PlaylistTrack" (
"PlaylistId" INTEGER NOT NULL,
"TrackId" INTEGER NOT NULL,
PRIMARY KEY ("PlaylistId", "TrackId"),
FOREIGN KEY("TrackId") REFERENCES "Track" ("TrackId"),
FOREIGN KEY("PlaylistId") REFERENCES "Playlist" ("PlaylistId")
)
SELECT * FROM 'PlaylistTrack' LIMIT 3;
PlaylistId TrackId
1 3402
1 3389
1 3390
Thought: I now know the final answer
Final Answer: The PlaylistTrack table contains two columns, PlaylistId and TrackId, which are both integers and are used to link Playlist and Track tables.
> Finished chain.
'The PlaylistTrack table contains two columns, PlaylistId and TrackId, which are both integers and are used to link Playlist and Track tables.'
示例：运行查询
agent_executor.run(
"List the total sales per country. Which country's customers spent the most?"
)
> Entering new AgentExecutor chain...
Action: list_tables_sql_db
Action Input: ""
Observation: Invoice, MediaType, Artist, InvoiceLine, Genre, Playlist, Employee, Album, PlaylistTrack, Track, Customer
Thought: I should look at the schema of the relevant tables to see what columns I can use.
Action: schema_sql_db
Action Input: "Invoice, Customer"
Observation:
CREATE TABLE "Customer" (
"CustomerId" INTEGER NOT NULL,
"FirstName" NVARCHAR(40) NOT NULL,
"LastName" NVARCHAR(20) NOT NULL,
"Company" NVARCHAR(80),
"Address" NVARCHAR(70),
"City" NVARCHAR(40),
"State" NVARCHAR(40),
"Country" NVARCHAR(40),
"PostalCode" NVARCHAR(10),
"Phone" NVARCHAR(24),
"Fax" NVARCHAR(24),
"Email" NVARCHAR(60) NOT NULL,
"SupportRepId" INTEGER,
PRIMARY KEY ("CustomerId"),
FOREIGN KEY("SupportRepId") REFERENCES "Employee" ("EmployeeId")
)
SELECT * FROM 'Customer' LIMIT 3;
CustomerId FirstName LastName Company Address City State Country PostalCode Phone Fax Email SupportRepId
1 Luís Gonçalves Embraer - Empresa Brasileira de Aeronáutica S.A. Av. Brigadeiro Faria Lima, 2170 São José dos Campos SP Brazil 12227-000 +55 (12) 3923-5555 +55 (12) 3923-5566 [email protected] 3
2 Leonie Köhler None Theodor-Heuss-Straße 34 Stuttgart None Germany 70174 +49 0711 2842222 None [email protected] 5
3 François Tremblay None 1498 rue Bélanger Montréal QC Canada H2G 1A7 +1 (514) 721-4711 None [email protected] 3
CREATE TABLE "Invoice" (
"InvoiceId" INTEGER NOT NULL,
"CustomerId" INTEGER NOT NULL,
"InvoiceDate" DATETIME NOT NULL,
"BillingAddress" NVARCHAR(70),
"BillingCity" NVARCHAR(40),
"BillingState" NVARCHAR(40),
"BillingCountry" NVARCHAR(40),
"BillingPostalCode" NVARCHAR(10),
"Total" NUMERIC(10, 2) NOT NULL,
PRIMARY KEY ("InvoiceId"),
FOREIGN KEY("CustomerId") REFERENCES "Customer" ("CustomerId")
)
SELECT * FROM 'Invoice' LIMIT 3;
InvoiceId CustomerId InvoiceDate BillingAddress BillingCity BillingState BillingCountry BillingPostalCode Total
1 2 2009-01-01 00:00:00 Theodor-Heuss-Straße 34 Stuttgart None Germany 70174 1.98
2 4 2009-01-02 00:00:00 Ullevålsveien 14 Oslo None Norway 0171 3.96
3 8 2009-01-03 00:00:00 Grétrystraat 63 Brussels None Belgium 1000 5.94
Thought: I should query the Invoice and Customer tables to get the total sales per country.
Action: query_sql_db
Action Input: SELECT c.Country, SUM(i.Total) AS TotalSales FROM Invoice i INNER JOIN Customer c ON i.CustomerId = c.CustomerId GROUP BY c.Country ORDER BY TotalSales DESC LIMIT 10
Observation: [('USA', 523.0600000000003), ('Canada', 303.9599999999999), ('France', 195.09999999999994), ('Brazil', 190.09999999999997), ('Germany', 156.48), ('United Kingdom', 112.85999999999999), ('Czech Republic', 90.24000000000001), ('Portugal', 77.23999999999998), ('India', 75.25999999999999), ('Chile', 46.62)]
Thought: I now know the final answer
Final Answer: The customers from the USA spent the most, with a total of $523.06.
> Finished chain.
'The customers from the USA spent the most, with a total of $523.06.'
agent_executor.run(
"Show the total number of tracks in each playlist. The Playlist name should be included in the result."
)
> Entering new AgentExecutor chain...
Action: list_tables_sql_db
Action Input: ""
Observation: Invoice, MediaType, Artist, InvoiceLine, Genre, Playlist, Employee, Album, PlaylistTrack, Track, Customer
Thought: I should look at the schema of the Playlist and PlaylistTrack tables to see what columns I can use.
Action: schema_sql_db
Action Input: "Playlist, PlaylistTrack"
Observation:
CREATE TABLE "Playlist" (
"PlaylistId" INTEGER NOT NULL,
"Name" NVARCHAR(120),
PRIMARY KEY ("PlaylistId")
)
SELECT * FROM 'Playlist' LIMIT 3;
PlaylistId Name
1 Music
2 Movies
3 TV Shows
CREATE TABLE "PlaylistTrack" (
"PlaylistId" INTEGER NOT NULL,
"TrackId" INTEGER NOT NULL,
PRIMARY KEY ("PlaylistId", "TrackId"),
FOREIGN KEY("TrackId") REFERENCES "Track" ("TrackId"),
FOREIGN KEY("PlaylistId") REFERENCES "Playlist" ("PlaylistId")
)
SELECT * FROM 'PlaylistTrack' LIMIT 3;
PlaylistId TrackId
1 3402
1 3389
1 3390
Thought: I can use a SELECT statement to get the total number of tracks in each playlist.
Action: query_checker_sql_db
Action Input: SELECT Playlist.Name, COUNT(PlaylistTrack.TrackId) AS TotalTracks FROM Playlist INNER JOIN PlaylistTrack ON Playlist.PlaylistId = PlaylistTrack.PlaylistId GROUP BY Playlist.Name
Observation:
SELECT Playlist.Name, COUNT(PlaylistTrack.TrackId) AS TotalTracks FROM Playlist INNER JOIN PlaylistTrack ON Playlist.PlaylistId = PlaylistTrack.PlaylistId GROUP BY Playlist.Name
Thought: The query looks correct, I can now execute it.
Action: query_sql_db
Action Input: SELECT Playlist.Name, COUNT(PlaylistTrack.TrackId) AS TotalTracks FROM Playlist INNER JOIN PlaylistTrack ON Playlist.PlaylistId = PlaylistTrack.PlaylistId GROUP BY Playlist.Name LIMIT 10
Observation: [('90’s Music', 1477), ('Brazilian Music', 39), ('Classical', 75), ('Classical 101 - Deep Cuts', 25), ('Classical 101 - Next Steps', 25), ('Classical 101 - The Basics', 25), ('Grunge', 15), ('Heavy Metal Classic', 26), ('Music', 6580), ('Music Videos', 1)]
Thought: I now know the final answer.
Final Answer: The total number of tracks in each playlist are: '90’s Music' (1477), 'Brazilian Music' (39), 'Classical' (75), 'Classical 101 - Deep Cuts' (25), 'Classical 101 - Next Steps' (25), 'Classical 101 - The Basics' (25), 'Grunge' (15), 'Heavy Metal Classic' (26), 'Music' (6580), 'Music Videos' (1).
> Finished chain.
"The total number of tracks in each playlist are: '90’s Music' (1477), 'Brazilian Music' (39), 'Classical' (75), 'Classical 101 - Deep Cuts' (25), 'Classical 101 - Next Steps' (25), 'Classical 101 - The Basics' (25), 'Grunge' (15), 'Heavy Metal Classic' (26), 'Music' (6580), 'Music Videos' (1)."
从错误中恢复
在此示例中，代理能够在最初尝试访问不存在的属性 （） 后从错误中恢复。Track.ArtistId
agent_executor.run("Who are the top 3 best selling artists?")
> Entering new AgentExecutor chain...
Action: list_tables_sql_db
Action Input: ""
Observation: MediaType, Track, Invoice, Album, Playlist, Customer, Employee, InvoiceLine, PlaylistTrack, Genre, Artist
Thought: I should look at the schema of the Artist, InvoiceLine, and Track tables to see what columns I can use.
Action: schema_sql_db
Action Input: "Artist, InvoiceLine, Track"
Observation:
CREATE TABLE "Artist" (
"ArtistId" INTEGER NOT NULL,
"Name" NVARCHAR(120),
PRIMARY KEY ("ArtistId")
)
SELECT * FROM 'Artist' LIMIT 3;
ArtistId Name
1 AC/DC
2 Accept
3 Aerosmith
CREATE TABLE "Track" (
"TrackId" INTEGER NOT NULL,
"Name" NVARCHAR(200) NOT NULL,
"AlbumId" INTEGER,
"MediaTypeId" INTEGER NOT NULL,
"GenreId" INTEGER,
"Composer" NVARCHAR(220),
"Milliseconds" INTEGER NOT NULL,
"Bytes" INTEGER,
"UnitPrice" NUMERIC(10, 2) NOT NULL,
PRIMARY KEY ("TrackId"),
FOREIGN KEY("MediaTypeId") REFERENCES "MediaType" ("MediaTypeId"),
FOREIGN KEY("GenreId") REFERENCES "Genre" ("GenreId"),
FOREIGN KEY("AlbumId") REFERENCES "Album" ("AlbumId")
)
SELECT * FROM 'Track' LIMIT 3;
TrackId Name AlbumId MediaTypeId GenreId Composer Milliseconds Bytes UnitPrice
1 For Those About To Rock (We Salute You) 1 1 1 Angus Young, Malcolm Young, Brian Johnson 343719 11170334 0.99
2 Balls to the Wall 2 2 1 None 342562 5510424 0.99
3 Fast As a Shark 3 2 1 F. Baltes, S. Kaufman, U. Dirkscneider & W. Hoffman 230619 3990994 0.99
CREATE TABLE "InvoiceLine" (
"InvoiceLineId" INTEGER NOT NULL,
"InvoiceId" INTEGER NOT NULL,
"TrackId" INTEGER NOT NULL,
"UnitPrice" NUMERIC(10, 2) NOT NULL,
"Quantity" INTEGER NOT NULL,
PRIMARY KEY ("InvoiceLineId"),
FOREIGN KEY("TrackId") REFERENCES "Track" ("TrackId"),
FOREIGN KEY("InvoiceId") REFERENCES "Invoice" ("InvoiceId")
)
SELECT * FROM 'InvoiceLine' LIMIT 3;
InvoiceLineId InvoiceId TrackId UnitPrice Quantity
1 1 2 0.99 1
2 1 4 0.99 1
3 2 6 0.99 1
Thought: I should query the database to get the top 3 best selling artists.
Action: query_sql_db
Action Input: SELECT Artist.Name, SUM(InvoiceLine.Quantity) AS TotalQuantity FROM Artist INNER JOIN Track ON Artist.ArtistId = Track.ArtistId INNER JOIN InvoiceLine ON Track.TrackId = InvoiceLine.TrackId GROUP BY Artist.Name ORDER BY TotalQuantity DESC LIMIT 3
Observation: Error: (sqlite3.OperationalError) no such column: Track.ArtistId
[SQL: SELECT Artist.Name, SUM(InvoiceLine.Quantity) AS TotalQuantity FROM Artist INNER JOIN Track ON Artist.ArtistId = Track.ArtistId INNER JOIN InvoiceLine ON Track.TrackId = InvoiceLine.TrackId GROUP BY Artist.Name ORDER BY TotalQuantity DESC LIMIT 3]
(Background on this error at: https://sqlalche.me/e/14/e3q8)
Thought: I should double check my query before executing it.
Action: query_checker_sql_db
Action Input: SELECT Artist.Name, SUM(InvoiceLine.Quantity) AS TotalQuantity FROM Artist INNER JOIN Track ON Artist.ArtistId = Track.ArtistId INNER JOIN InvoiceLine ON Track.TrackId = InvoiceLine.TrackId GROUP BY Artist.Name ORDER BY TotalQuantity DESC LIMIT 3
Observation:
SELECT Artist.Name, SUM(InvoiceLine.Quantity) AS TotalQuantity
FROM Artist
INNER JOIN Track ON Artist.ArtistId = Track.ArtistId
INNER JOIN InvoiceLine ON Track.TrackId = InvoiceLine.TrackId
GROUP BY Artist.Name
ORDER BY TotalQuantity DESC
LIMIT 3;
Thought: I now know the final answer.
Action: query_sql_db
Action Input: SELECT Artist.Name, SUM(InvoiceLine.Quantity) AS TotalQuantity FROM Artist INNER JOIN Album ON Artist.ArtistId = Album.ArtistId INNER JOIN Track ON Album.AlbumId = Track.AlbumId INNER JOIN InvoiceLine ON Track.TrackId = InvoiceLine.TrackId GROUP BY Artist.Name ORDER BY TotalQuantity DESC LIMIT 3
Observation: [('Iron Maiden', 140), ('U2', 107), ('Metallica', 91)]
Thought: I now know the final answer.
Final Answer: The top 3 best selling artists are Iron Maiden, U2, and Metallica.
> Finished chain.
'The top 3 best selling artists are Iron Maiden, U2, and Metallica.'