Question

-(void)initializeTableData
{
    sqlite3 *db=[DatabaseTestAppDelegate getNewDBConnection];
    sqlite3_stmt *statement=nil;
    const char *sql="select * from WhereTo";
    if (sqlite3_prepare_v2(db, sql, -1, &statement, NULL)!=SQLITE_OK)
        NSAssert1(0,@"error in preparing staement",sqlite3_errmsg(db));
    else {
        while(sqlite3_step(statement)==SQLITE_ROW)
            [tableData addObject:[NSString stringWithFormat:@"%s",(char*)sqlite3_column_text(statement,1)]];
    }
    sqlite3_finalize(statement);
}

在sqlite3 *db=[DatabaseTestAppDelegate getNewDBConnection];＆lt; ---它说，DatabaseTestAppDelegate可能无法响应'+ getNewDbConnection'

这是我的getNewDbConnection

+(sqlite3 *) getNewDBConnection{
    sqlite3 *newDBconnection;
    NSArray *paths = NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES);
    NSString *documentsDirectory = [paths objectAtIndex:0];
    NSString *path = [documentsDirectory stringByAppendingPathComponent:@"Malacca-lah.sqlite"];
    // Open the database. The database was prepared outside the application.
    if (sqlite3_open([path UTF8String], &newDBconnection) == SQLITE_OK) {
        NSLog(@"Database Successfully Opened :)");
    }
    else {
        NSLog(@"Error in opening database :(");
    }
    return newDBconnection;
}

我是XCode和SQLite的新手......过去几周一直在学习这个，试图抓住它...无论如何，请帮我解决这个问题。我理解整个代码，但我不明白为什么继承有问题。

提前致谢

Answer 1

如果它说某个类可能没有响应选择器，则意味着它无法找到选择器的方法声明。您是否在+(sqlite3 *)getNewDBConnection的标题（DatabaseTestAppDelegate）文件中声明了方法".h"？

从不兼容的指针类型初始化

1 个答案: