如何在Objective-C中將字符串插入sqlite？

Question

我無法將NSMutableString插入sqlite。 我從Web服務器成功獲取了信息，並成功創建了sqlite文件，但是無法將Web中的數據插入sqlite。

我認為問題就在這里。

NSString *insertSQL = @"INSERT INTO Questions VALUES (result)";

但我不確定，無法解決此問題。 有人可以幫我嗎？

- (void)getInforamtionFromWeb {
    NSURL *url = [NSURL URLWithString:kGetUrl];
    data = [NSData dataWithContentsOfURL:url];
    NSError *error;
    json = [NSJSONSerialization JSONObjectWithData:data options:kNilOptions error:&error];
    result = [[NSMutableString alloc] init];
    for (NSObject * obj in json)
    {
        [result appendString:[obj description]];
    }
}

-(void)initiatSqlite{
    // Get the documents directory
    dirPaths = NSSearchPathForDirectoriesInDomains(NSDocumentDirectory, NSUserDomainMask, YES);
    docsDir = dirPaths[0];
    // Build the path to the database file
    _databasePath = [[NSString alloc]
                     initWithString: [docsDir stringByAppendingPathComponent:
                                      @"Questions.db"]];
    NSFileManager *filemgr = [NSFileManager defaultManager];
    if ([filemgr fileExistsAtPath: _databasePath ] == NO)
    {
        const char *dbpath = [_databasePath UTF8String];
        if (sqlite3_open(dbpath, &_contactDB) == SQLITE_OK)
        {
            char *errMsg;
            const char *sql_stmt =
            "CREATE TABLE IF NOT EXISTS Questions (ID INTEGER PRIMARY KEY AUTOINCREMENT, Question Text, AnswerA Text, AnswerB Text, AnswerC Text, AnswerD Text, CorrectAnswer Text, Explanation Text)";
            if (sqlite3_exec(_contactDB, sql_stmt, NULL, NULL, &errMsg) != SQLITE_OK)
            {
                [self Worningtitle:@"Error" Message:@"Failed to create table"];
            }
            sqlite3_close(_contactDB);
        } else {
            [self Worningtitle:@"Error" Message:@"Failed to open/create database"];

        }
    }

}
- (void) insertData
{
    sqlite3_stmt    *statement;
    const char *dbpath = [_databasePath UTF8String];


    if (sqlite3_open(dbpath, &_contactDB) == SQLITE_OK)
    {
        NSString *insertSQL = @"INSERT INTO Questions VALUES (result)";

        const char *insert_stmt = [insertSQL UTF8String];
        NSLog(@"%s",insert_stmt);
        sqlite3_prepare_v2(_contactDB, insert_stmt, -1, &statement, NULL);
        sqlite3_bind_text(statement, 1, [result UTF8String], -1, SQLITE_TRANSIENT);

        if (sqlite3_step(statement) == SQLITE_DONE)
        {
            NSLog(@"Product added");
        } else {
            NSLog(@"Failed to add Product");
        }

        sqlite3_finalize(statement);
        sqlite3_close(_contactDB);
    }
}

Answer 1

你是對的。 您的INSERT語句不正確。 它必須是：

NSString *insertSQL = @"INSERT INTO Questions VALUES (?)";

? 是將應用sqlite3_bind_xxx 。

請記住，您的Questions表有很多列，但是您只插入一個值。 您應指定該值應插入到哪一列。 例：

NSString *insertSQL = @"INSERT INTO Questions (Question) VALUES (?)";

用要輸入result值的正確列名替換“ Question 。

順便說一句-您的代碼需要做很多工作才能使其變得更好。 您需要添加更多錯誤檢查，尤其是使用sqlite3_prepare_v2 。

基於要插入更多數據，您可以執行以下操作：

- (BOOL)insertData {
    BOOL result = NO; // failed
    const char *dbpath = [_databasePath UTF8String];
    if (sqlite3_open(dbpath, &_contactDB) == SQLITE_OK) {
        const char *insert_stmt = "INSERT INTO Questions(AnswerA,AnswerB,AnswerC,AnswerD,CorrectAnswer,Question) VALUES (?,?,?,?,?,?)";
        sqlite3_stmt *statement;

        if (sqlite3_prepare_v2(_contactDB, insert_stmt, -1, &statement, NULL) == SQLITE_OK) {
            sqlite3_bind_text(statement, 1, [answerA UTF8String], -1, SQLITE_TRANSIENT);
            sqlite3_bind_text(statement, 2, [answerB UTF8String], -1, SQLITE_TRANSIENT);
            sqlite3_bind_text(statement, 3, [answerC UTF8String], -1, SQLITE_TRANSIENT);
            sqlite3_bind_text(statement, 4, [answerD UTF8String], -1, SQLITE_TRANSIENT);
            sqlite3_bind_text(statement, 5, [correctAnswer UTF8String], -1, SQLITE_TRANSIENT);
            sqlite3_bind_text(statement, 6, [question UTF8String], -1, SQLITE_TRANSIENT);

            if (sqlite3_step(statement) == SQLITE_DONE) {
                NSLog(@"Product added");
                result = YES;
            } else {
                NSLog(@"Failed to add Product: %s", sqlite3_errmsg(_contactDB));
            }

            sqlite3_finalize(statement);
        } else {
            NSLog(@"Unable to prepare statement: %s", sqlite3_errmsg(_contactDB);
        }

        sqlite3_close(_contactDB);
    } else {
        NSLog(@"Unable to open database: %@", sqlite3_errmsg(_contactDB));
    }

    return result;
}

請注意，您需要使用對包含每個列數據的實際變量/值的引用來更新每個sqlite3_bind_text調用。

您也不需要綁定ID值，因為它已設置為自動填充。

注意所有正確的錯誤檢查。 我也這樣做了，所以insertData返回一個BOOL指示是否成功添加了數據。

Answer 2

試聽所有表的通用方法：

+(BOOL)insertWithReturnConfirmInTable:(NSString *)tableName Param:(NSDictionary *)params
{
        NSArray *keys = [[NSArray alloc] initWithArray:[self GetColumnsForTable:tableName]];

        //insert statement
        NSString *query = [NSString stringWithFormat:@"INSERT INTO %@(",tableName];

        //set keys
        for (int i=0; i<[keys count]; i++) {
            if([keys objectAtIndex:i] == [keys lastObject])
            {
                query = [query stringByAppendingString:[NSString stringWithFormat:@"%@) VALUES (",[keys objectAtIndex:i]]];
                for (int m=0; m<[keys count]; m++) {

                    if([keys objectAtIndex:m] == [keys lastObject])
                        query = [query stringByAppendingString:@"?)"];
                    else
                        query = [query stringByAppendingString:@"?,"];
                }
            }
            else
            {
                query = [query stringByAppendingString:[NSString stringWithFormat:@"%@,",[keys objectAtIndex:i]]];
            }
        }
        NSLog(@"Query : %@",query);

        if (sqlite3_open([[self getDBPath] UTF8String], &dbObj) == SQLITE_OK) {

            sqlite3_stmt *addStmt = nil;

            if (sqlite3_prepare_v2(dbObj, [query UTF8String], -1, &addStmt, NULL) != SQLITE_OK)
                NSAssert1(0, @"Error while creating insert statement. '%s'",sqlite3_errmsg(dbObj));

            //for Values
            for (int i=0; i<[keys count]; i++) {
                sqlite3_bind_text(addStmt, i+1, [[NSString stringWithFormat:@"%@",[params objectForKey:[keys objectAtIndex:i]]] UTF8String], -1, SQLITE_TRANSIENT);
            }

            if (SQLITE_DONE != sqlite3_step(addStmt)) {
                NSLog(@"Error while inserting data. '%s'", sqlite3_errmsg(dbObj));
                sqlite3_close(dbObj);
                return NO;
            }
            else {
                sqlite3_finalize(addStmt);
                NSLog(@"data inserted in table : %@",tableName);
                sqlite3_close(dbObj);
                return YES;
            }


        } else {
            NSLog(@"Error in opening database.");
            sqlite3_close(dbObj);
            return NO;
        }

    }
}

---

調用方法：將所有字段名稱作為鍵，將所有值作為字典中的值。 並傳遞表名。

NSDictionary *param = [[NSDictionary alloc]initWithObjectsAndKeys:@"1",@"S_Id",@"abc",@"Name", nil];
[database insertInTable:@"Child" Param:param];