Android的更快数据库插入

本文介绍了Android的更快数据库插入的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我解析，其中包含近20000标记的大型XML，我只要我解析这些数据保存在XML中的所有条目进入我的SQLite数据库。

但显然XML是越来越解析pretty没有迅速插入操作，但是当我尝试插入每个值需要花费很长的时间。（约10分钟）。

数据库插入code：

 的for（int i = 0; I＆LT; tracksList.size（）;我++）{
    dataSource.addTracks（tracksList.get（I））;
}众长addTracks（曲目的曲目）{    长insertId = 0;    ContentValues值=新ContentValues（）;
    values.put（TRACKS_ID，tracks.getStrId（））;
    values.put（TRACKS_ARTISTID，tracks.getStrArtistId（））;
    values.put（TRACKS_ARTISTNAME，tracks.getStrArtistName（））;
    values.put（TRACKS_ALBUMNAME，tracks.getStrAlbumName（））;
    values.put（TRACKS_FILENAME，tracks.getStrFileName（））;
    values.put（TRACKS_TRACKNAME，tracks.getStrTrackName（））;    insertId = database.insert（TRACKS_TABLE，空，价值）;    返回insertId;
}

有没有替代从我的ArrayList中获取的每个元素，并将它们保存有更好的速度。

编辑：得到它的工作，所有的日Thnx ..

  database.beginTransaction（）;
    尝试{
        //标准的SQL INSERT语句，可重复使用
        SQLiteStatement插入=
                database.compileStatement（插入+ TRACKS_TABLE
                    +（+ TRACKS_ID +，+ TRACKS_ARTISTID
                    +，+ TRACKS_ARTISTNAME
                    +，+ TRACKS_ALBUMNAME
                    +，+ TRACKS_FILENAME
                    +，+ TRACKS_TRACKNAME +）
                    +值+（，，，，，？？？？？？））;        对于（曲目值：tracksList）{
            //绑定1索引？s到指定的值            的System.out.println（value.getStrId（））;            insert.bindLong（1，value.getStrId（））;
            insert.bindString（2，value.getStrArtistId（））;
            insert.bindString（3，value.getStrArtistName（））;
            insert.bindString（4，value.getStrAlbumName（））;
            insert.bindString（5，value.getStrFileName（））;
            insert.bindString（6，value.getStrTrackName（））;
            insert.execute（）;
        }
        database.setTransactionSuccessful（）;    } {最后
        database.endTransaction（）;
    }

解决方案

使用一个交易的所有插入，否则你得到的存储同步开销为每一个INSERT语句：

  database.beginTransaction（）;
尝试{
    的for（int i = 0; I＆LT; tracksList.size（）;我++）
        dataSource.addTracks（tracksList.get（I））;    database.setTransactionSuccessful（）;
} {最后
    database.endTransaction（）;
}

I am parsing a HUGE xml which contains nearly 20000 tags, I am saving all the entries from the xml into my sqlite database as soon as i am parsing them.

But apparently xml is getting parsed pretty quickly without insert operation, but when i try to insert each values it takes very long time.( Around 10 mins).

Database insert code :

for (int i = 0; i < tracksList.size(); i++) {
    dataSource.addTracks(tracksList.get(i));                        
}



public long addTracks(Tracks tracks) {

    long insertId = 0;

    ContentValues values = new ContentValues();
    values.put(TRACKS_ID, tracks.getStrId());
    values.put(TRACKS_ARTISTID, tracks.getStrArtistId());
    values.put(TRACKS_ARTISTNAME, tracks.getStrArtistName());
    values.put(TRACKS_ALBUMNAME, tracks.getStrAlbumName());
    values.put(TRACKS_FILENAME, tracks.getStrFileName());
    values.put(TRACKS_TRACKNAME, tracks.getStrTrackName());

    insertId = database.insert(TRACKS_TABLE, null, values);

    return insertId;
}

Is there any alternative to get each element from my arraylist and save them with much better speed.

EDIT : Got it working, thnx all..

    database.beginTransaction();
    try {
        //standard SQL insert statement, that can be reused
        SQLiteStatement insert = 
                database.compileStatement("insert into " + TRACKS_TABLE
                    + "(" + TRACKS_ID + "," + TRACKS_ARTISTID
                    + "," + TRACKS_ARTISTNAME 
                    + "," + TRACKS_ALBUMNAME 
                    + "," + TRACKS_FILENAME 
                    + "," + TRACKS_TRACKNAME + ")"
                    +" values " + "(?,?,?,?,?,?)");

        for (Tracks value : tracksList){
            //bind the 1-indexed ?'s to the values specified

            System.out.println(value.getStrId());

            insert.bindLong(1, value.getStrId());
            insert.bindString(2, value.getStrArtistId());
            insert.bindString(3, value.getStrArtistName());
            insert.bindString(4, value.getStrAlbumName());
            insert.bindString(5, value.getStrFileName());
            insert.bindString(6, value.getStrTrackName());
            insert.execute();
        }
        database.setTransactionSuccessful();

    } finally {
        database.endTransaction();
    }

解决方案

Use one transaction for all inserts, otherwise you get the storage synchronization overhead for every single insert:

database.beginTransaction();
try {
    for (int i = 0; i < tracksList.size(); i++)
        dataSource.addTracks(tracksList.get(i));                        

    database.setTransactionSuccessful();
} finally {
    database.endTransaction();
}

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