Question

首先，我知道我不应该使用SQLite数据库存储图像，但我只存储非常小的网站图标。

我的问题是我尝试将这些图标插入数据库（似乎工作），我使用NSData -tiffrepresentation方法将图标转换为NSimage，然后插入到我的数据库进入blob列：

 NSImage *favico = [webview mainFrameIcon];
[appDelegate insertBookmark:[titleField stringValue] url:[urlfield stringValue] data:[favico TIFFRepresentation]]

SQLite方法如下所示：

-(void)insertBookmark:(NSString *)title url:(NSString *)url data:(NSData *)data
{
    NSData *imagedata = [[NSData alloc]initWithData:data]; 
    sqlite3 *database;
    if(sqlite3_open([databasePath UTF8String], &database) == SQLITE_OK) {
        NSString *query = [NSString stringWithFormat:@"INSERT INTO Bookmarks (title, url, image) VALUES ('%@', '%@', '%@')",title, url, data];
        const char *querychar = [query UTF8String]; 
        sqlite3_stmt *statement; 
        if (sqlite3_prepare_v2(database, querychar, -1, &statement, NULL) == SQLITE_OK)
        {
            int row =  3;
            sqlite3_bind_blob(statement, row, [imagedata bytes], [imagedata length], NULL);
            sqlite3_step(statement);
            sqlite3_finalize(statement);

        }
        else
        {
            NSLog(@"Error"); 
        }

        [databasePath retain]; 
        [databaseName retain]; 
    }
    sqlite3_close(database);
    [imagedata release]; 

}

当我查看数据库时，我在<data>之间的图像列中有值（正常？）

现在，当我从数据库中提取blob并尝试将其放入我的对象时，我在NSimage中得到了null：

-(void) readDatabase {
// Setup the database object
sqlite3 *database;

bookmarks = [[NSMutableArray alloc] init];

// Open the database from the users filessytem
if(sqlite3_open([databasePath UTF8String], &database) == SQLITE_OK) {
    // Setup the SQL Statement and compile it for faster access
    const char *sqlStatement = "SELECT * FROM Bookmarks ORDER BY title ASC";
    sqlite3_stmt *compiledStatement;
    if(sqlite3_prepare_v2(database, sqlStatement, -1, &compiledStatement, NULL) == SQLITE_OK) {
        // Loop through the results and add them to the feeds array
        while(sqlite3_step(compiledStatement) == SQLITE_ROW) {
            // Read the data from the result row
            NSString *aTitle = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 1)];
            NSString *aUrl = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 2)];
            NSData *data = [[NSData alloc] initWithBytes:sqlite3_column_blob(compiledStatement, 3) length:sqlite3_column_bytes(compiledStatement, 3)];
            NSImage *image = [[NSImage alloc]initWithData:data];
            bookmarkObject *bookmark = [[bookmarkObject alloc] initWithName:aTitle url:aUrl favico:image]; 

            [bookmarks addObject:bookmark];
            [bookmark release];
            [data release];
            [image release]; 



        }
    }
    // Release the compiled statement from memory
    sqlite3_finalize(compiledStatement);
    sqlite3_close(database); 
    [databasePath retain]; 
    [databaseName retain]; 


}

}

提前致谢

Answer 1

实际数据周围的<和>来自NSData。通过在格式字符串中使用%@并提供data，NSString会将description发送到data。 [NSData description]将内容封装在<和>之间。

您的代码的另一个问题是，您似乎混合了无参数和＆amp;参数预处理语句：

使用stringWithFormat:创建完整的查询语句
或使用INSERT INTO Bookmarks (title ...) VALUES (? ...)与sqlite3_bind_blob

SQLite支持的参数语法可以在这里找到： http://www.sqlite.org/c3ref/bind_blob.html

对NSImage的blob和blob的NSImage（SQLite，NSData）

1 个答案: