Question

我是iphone应用程序开发人员的新手，我正在使用sqlite3数据库，在应用程序委托中，我正在跟踪代码并运行正常我们也在数据库中找到了我的应用程序中的值，

但是应用程序会立即讨论为什么会发生这种情况我不明白。

代码如下所示

        -(void)Data
      {
   databaseName = @"dataa.sqlite";

        NSArray *documentPaths =   NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES);
      NSString *documentsDir = [documentPaths objectAtIndex:0];
     databasePath =[documentsDir stringByAppendingPathComponent:databaseName];

    [self checkAndCreateDatabase];

    list1 = [[NSMutableArray alloc] init];  

    sqlite3 *database;
    if (sqlite3_open([databasePath UTF8String], &database) == SQLITE_OK)
    {
       if(detailStmt == nil)
        {
                    const char *sql = "Select * from Dataa";            
             if(sqlite3_prepare_v2(database, sql, -1, &detailStmt, NULL) ==    SQLITE_OK)
            {               
                    //NSLog(@"Hiiiiiii");
                //sqlite3_bind_text(detailStmt, 1, [t1 UTF8String], -1,    SQLITE_TRANSIENT);
                //sqlite3_bind_text(detailStmt, 2, [t2 UTF8String], -2, SQLITE_TRANSIENT);
                  //sqlite3_bind_int(detailStmt, 3, t3);

                   while(sqlite3_step(detailStmt) == SQLITE_ROW) 
                 {
                    //NSLog(@"Helllloooooo");




                     NSString *item= [NSString stringWithUTF8String:(char *)sqlite3_column_text(detailStmt, 0)];
                //NSString *fame= [NSString stringWithUTF8String:(char *)sqlite3_column_text(detailStmt, 1)];
                //NSString *cinemax = [NSString stringWithUTF8String:(char *)sqlite3_column_text(detailStmt, 2)];
                //NSString *big= [NSString stringWithUTF8String:(char *)sqlite3_column_text(detailStmt, 3)];

                //pvr1 = pvr;
                item1=item;
                //NSLog(@"%@",item1);

                data = [[NSMutableArray alloc] init];   


                list *animal=[[list alloc] initWithName:item1];

                // Add the animal object to the animals Array
                [list1 addObject:animal];
                //[list1 addObject:item];


            }
            sqlite3_reset(detailStmt);
        }
        sqlite3_finalize(detailStmt);
        //  sqlite3_clear_bindings(detailStmt);
    }
}
detailStmt = nil;
sqlite3_close(database);

}

当我们看到控制台时，他们会显示以下错误，并给出了吼叫声

    2010-03-09 10:02:40.262 SanjeevKapoor[430:20b] *** Terminating app due to uncaught     exception 'NSInvalidArgumentException', reason: '*** +[NSString stringWithUTF8String:]:    NULL cString'

当我们看到调试器时，它们在后续行中显示错误

          NSString *item= [NSString stringWithUTF8String:(char *)sqlite3_column_text(detailStmt, 0)];

我无法解决问题请帮助我。

Answer 1

您可以使用NSString stringWithFormat:@"%s"。这也可以优雅地处理NULL值。

Answer 2

Sqlite可以存储NULL对象，这些对象与空字符串不同。在您的情况下，您从base中提取了值，结果为NULL。但[NSString stringWithUTF8String：]方法不能使用NULL作为参数，因此会引发错误。你应该检查提取的值：

char * str = (char*)sqlite3_column_text(statement, i);
if (str){
    item = [NSString stringWithUTF8String:str];
}
else{
    // handle case when object is NULL, for example set result to empty string:  
    item = @"";
}

Answer 3

您可以使用：

((char *)sqlite3_column_text(sqlStatement, 11)) ? [NSString stringWithUTF8String:(char *)sqlite3_column_text(sqlStatement, 11)] : nil;

如何解决运行时错误NSString，sqlite3_column_text NULL问题？

3 个答案: