Question

我使用这些库：

sqljdbc4.jar
INTEGER

在SQLite中，列类型为long tsMills = 123; preparedStatement.setLong(parameterIndex, tsMills);。

我的Java代码：

long

这会正确地将数据插入数据库。

但是，如果我使用long tsMills = 1522746908000l;值，如

{{1}}

此值不已插入数据库。

我如何解决这个问题？

这里是SQLite中所有可能的数据类型：

https://github.com/nltk/nltk/blob/develop/nltk/classify/naivebayes.py#L201

Answer 1

SQLite does not care much about types。使用Java插入长值非常有可能。

使用此库

<dependency>
    <groupId>org.xerial</groupId>
    <artifactId>sqlite-jdbc</artifactId>
    <version>3.21.0.1</version>
</dependency>

和一个包含一个表的数据库test.db

create table warehouses(name string, capacity integer);

此Java代码按预期工作：

import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.PreparedStatement;
import java.sql.SQLException;

public class InsertApp {

    private Connection connect() {
        String url = "jdbc:sqlite:/tmp/test.db";
        Connection conn = null;
        try {
            conn = DriverManager.getConnection(url);
        } catch (SQLException e) {
            System.out.println(e.getMessage());
        }
        return conn;
    }

    private void insert(String name, long capacity) {
        String sql = "INSERT INTO warehouses(name,capacity) VALUES(?,?)";

        try (Connection conn = this.connect();
                PreparedStatement pstmt = conn.prepareStatement(sql)) {
            pstmt.setString(1, name);
            pstmt.setLong(2, capacity);
            pstmt.executeUpdate();
        } catch (SQLException e) {
            System.out.println(e.getMessage());
        }
    }

    public static void main(String[] args) {

        InsertApp app = new InsertApp();
        app.insert("Raw Materials", 3000000000L);
        app.insert("Semifinished Goods", 4000000000L);
        app.insert("Finished Goods", 5000000000L);
    }

}

执行此代码后，该表有三行，其中包含预期值：

$ sqlite3 test.db .dump
PRAGMA foreign_keys=OFF;
BEGIN TRANSACTION;
CREATE TABLE warehouses(name string, capacity integer);
INSERT INTO "warehouses" VALUES('Raw Materials',3000000000);
INSERT INTO "warehouses" VALUES('Semifinished Goods',4000000000);
INSERT INTO "warehouses" VALUES('Finished Goods',5000000000);
COMMIT;

无法长期插入SQLite

1 个答案: