查询未从表中检索数据

Question

我试图将数据库集成到我的项目中。这是我第一次实现数据库项目。我将数据库集成到我的项目中。数据库有3列名为rowid，col1，col2的列。下面的代码用于从数据库访问数据。但这并不是在“ if循环”中迷惑，在这种情况下，谁能帮助我即时消息。事先感谢。！[在此处输入图片描述] [1]

 /*   -(void)print
    {
        sqlite3_stmt *statement;
        qsql=[NSString stringWithFormat:@"SELECT * from light where rowid = '%d'",1];

        if(sqlite3_prepare_v2(database, [qsql UTF8String], -1, &statement, NULL) == SQLITE_OK) {


             NSLog(@"hhhhhhhhh");
            while (sqlite3_step(statement) == (SQLITE_ROW)){

                   NSString *Q_NO = [[NSString alloc] initWithUTF8String:(char *)sqlite3_column_text(statement, 0)];


               // tv.text=qsql;



            }

        }}*/
    -(void)print
    {
        sqlite3_stmt *statement;
        //qsql=[NSString stringWithFormat:@"SELECT * from light where rowid = %d",1];
        qsql=[NSString stringWithFormat:@"SELECT * from light where rowid = %i",1];

        //const char *sqlStatement = [qsql cStringUsingEncoding:NSUTF8StringEncoding];
       const char *qsql_stament = [qsql cStringUsingEncoding:NSUTF8StringEncoding];
        if(sqlite3_prepare_v2(database, qsql_stament, -1, &statement, NULL) == SQLITE_OK) {


            NSLog(@"hhhhhhhhh");
            while (sqlite3_step(statement) == (SQLITE_ROW)){
              //  NSString *Q_NO = [[NSString alloc] initWithUTF8String:(char *)sqlite3_column_text(statement, 0)];
                NSString *Q_NO = [[NSString alloc] initWithString:[NSString stringWithFormat:@"%i",sqlite3_column_int(statement, 0)]];

                // tv.text=qsql;
            }

        }            

            }

    -(void) checkAndCreateDatabase{
        BOOL success;
        NSFileManager *fileManager = [NSFileManager defaultManager];
        success = [fileManager fileExistsAtPath:databasePath];
        if(success) return;
        NSString *databasePathFromApp = [[[NSBundle mainBundle] resourcePath] stringByAppendingPathComponent:databaseName];
        [fileManager copyItemAtPath:databasePathFromApp toPath:databasePath error:nil];
        [fileManager release];
    }

    -(void) openDB {

        databaseName = @"Lighthr.sqlite";
        NSArray *documentsPaths = NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES);
        NSString *documentDir = [documentsPaths objectAtIndex:0];   
        databasePath = [documentDir stringByAppendingPathComponent:databaseName];
        [self checkAndCreateDatabase];  


        if(sqlite3_open([databasePath UTF8String],&database) == SQLITE_OK){
            NSLog(@"coming here???");
        }
    }


  [1]: http://i.stack.imgur.com/lm1ZV.png

Answer 1

因为rowid始终是整数，并且您尝试使用字符串进行访问。
我想你必须尝试一下。

编辑

if(sqlite3_open([databasePath UTF8String], &database) == SQLITE_OK) {
    qsql=[NSString stringWithFormat:@"SELECT * from light where rowid = %i",1];
    const char *sqlStatement = [qsql cStringUsingEncoding:NSUTF8StringEncoding];
    if(sqlite3_prepare_v2(database, sqlStatement, -1, &statement, NULL) == SQLITE_OK) {


        NSLog(@"hhhhhhhhh");
        while (sqlite3_step(statement) == (SQLITE_ROW)){
            NSString *Q_NO = [[NSString alloc] initWithUTF8String:(char *)sqlite3_column_text(statement, 0)];
           // tv.text=qsql;
        }

    }
}

Answer 2

您是否要在没有“ where”子句的情况下进行测试？

sqlite3_prepare_v2返回的错误代码是什么？

int sql_answer = sqlite3_prepare_v2(database, qsql_stament, -1, &statement, NULL);
NSLog(@" SQL answer : %d",sql_answer );
if(sql_answer == SQLITE_OK) {
  ...

Answer 3

NSString *Q_NO = [[NSString alloc] initWithUTF8String:(char *)sqlite3_column_text(statement, 0)];

雄蕊以上用以下陈述代替

NSString *Q_NO = [[NSString alloc] initWithString:[NSString stringWithFormat:@"%i",sqlite3_column_int(statement, 0)]];

和

const char *qsql_stament = [qsql cStringUsingEncoding:NSUTF8StringEncoding];

现在将[qsql UTF8String]替换为qsql_stament 。 它将正常工作。

Answer 4

我认为你应该这样。

-（无效）打印{

sqlite3_stmt *声明； qsql = [NSString stringWithFormat：@“ SELECT * from light where rowid =％d”，1];

const char *qsql_stament = [qsql cStringUsingEncoding:NSUTF8StringEncoding];
if(sqlite3_prepare_v2(database, qsql_stament, -1, &statement, NULL) == SQLITE_OK) {


    NSLog(@"hhhhhhhhh");
    while (sqlite3_step(statement) == (SQLITE_ROW)){            
        NSString *Q_NO =[NSString stringWithUTF8String:(char *)sqlite3_column_text(statement, 0)]



    }