Jooq Java代码处理超过20万条记录的表

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英文:

Jooq Java Code to handle more than 200k records from a table

问题

我的翻译如下:

我的数据库中的表有超过200,000条记录,我不想一次性从resultSet中获取所有记录进行处理。我正在尝试编写一个服务中的函数,调用另一个持久化类的函数,只获取1000条记录进行处理。
一旦处理完这1000条记录,它应该获取接下来的1000条记录。我在我的Java代码中使用了JOOQ数据库库。我只是分享了我脑海中的一个示例代码框架。

class RecordService {
    RecordPersistence recordPersistence = new RecordPersistence();

    public void processRecords() {

        List<Record> records = recordPersistence.fetchRecords();
        // 处理记录的代码
    }
}


class RecordPersistence {
    public List<Record> fetchRecords(){

        Cursor<Record> records = dsl.select...fetchLazy();

        while (records.hasNext()) {
           records.fetchNext(1000);
        }
        return records;
    }
}

如何从fetchRecords()函数中仅返回少量记录?我应该将其编写为异步函数吗?这是处理结果的正确方式吗?还是有更好的处理方式来处理我的情况?

英文:

The table in my database has more than 200k records and I don't want to fetch all of them at once in the resultSet to process. I am trying to write a function in a service that calls another persistence class function to bring only 1000 records to process.
Once it processes the 1000 records, it should fetch the following thousand records. I am using the JOOQ database library in my Java code. I am just sharing a sample skeleton code that I had in my mind.

class RecordService {
    RecordPersistence recordPersistence = new RecordPersistence();

	public void processRecords() {

		List&lt;Record&gt; records = recordPersistence.fetchRecords();
		// Code to process them
	 }
} 


class RecordPersistence {
	public List&lt;Record&gt;  fetchRecords(){
		
		Cursor&lt;Record&gt; records = dsl.select...fetchLazy();

		while (records.hasNext()) {
		   records.fetchNext(1000);
		}
		return records
	}
}

How to return only a few records from the fetchRecords() function? Should I write this is an async function? Is this the correct way to process result? or is there a better way to handle my case?

答案1

得分: 0

如评论中所提到的,问题在于你将处理逻辑放在了获取逻辑之外。只需将其移动到内部,例如:

class RecordService {
    RecordPersistence recordPersistence = new RecordPersistence();

    public void processRecords() {

        recordPersistence.fetchRecords(
            (List<Record> records) -> {
                // 处理记录的代码
            }
        );
     }
} 

class RecordPersistence {
    public List<Record> fetchRecords(Consumer<? super List<Record>> consumer) {
        try (Cursor<Record> cursor = dsl.select...fetchLazy()) {
            while (cursor.hasNext()) {
                consumer.accept(cursor.fetchNext(1000));
            }
        }
    }
}

对于这种类型的流式处理,你不需要进行任何范式转换(例如异步、响应式等)。只需从外部迭代切换到内部迭代即可。

英文:

As mentioned in the comments, the problem is that you're keeping the processing logic outside of the fetch logic. Just move it inside, e.g. as follows:

class RecordService {
    RecordPersistence recordPersistence = new RecordPersistence();

    public void processRecords() {

        recordPersistence.fetchRecords(
            (List&lt;Record&gt; records) -&gt; {
                // Code to process them
            }
        );
     }
} 

class RecordPersistence {
    public List&lt;Record&gt; fetchRecords(Consumer&lt;? super List&lt;Record&gt;&gt; consumer) {
        try (Cursor&lt;Record&gt; cursor = dsl.select...fetchLazy()) {
            while (cursor.hasNext()) {
                consumer.accept(cursor.fetchNext(1000));
            }
        }
    }
}

You don't need any paradigm shift (e.g. async, reactive, etc.) for this type of streaming to work. Just switch from external to internal iteration

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  • 本文由 发表于 2023年8月9日 09:19:17
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