Sqlite插入錯誤的目標c

我正在使用sqlite，我試圖插入一些數據到sqlite數據庫。它不工作。我有一個表，它的名字是LabUpdate但我有一些象這樣的錯誤：Sqlite插入錯誤的目標c

2012-04-02 08:49:58.158 SqliteDeneme[361:207] Failed from sqlite3_prepare_v2. Error is: no such column: deneme 
2012-04-02 08:49:58.159 SqliteDeneme[361:207] Compiled Statement has error code:1:INSERT INTO LabUpdate (IsSuccess, ProducerId, Latitude, Longitude, Altitude, Slope, SampleDate, PackageNo, Status, Description) VALUES(1,1,0.100000,0.100000,0.100000,0.100000,deneme,1,1,deneme) 
2012-04-02 08:49:58.160 SqliteDeneme[361:207] ExecuteNonQuery has error 
2012-04-02 08:49:58.160 SqliteDeneme[361:207] Failed from sqlite3_step. Error is: library routine called out of sequence

我這樣的代碼：

- (IBAction)buttonClick:(id)sender { 

NSFileManager *fileMgr = [NSFileManager defaultManager]; 
NSString *dbPath = [[[NSBundle mainBundle] resourcePath ]stringByAppendingPathComponent:@"SqliteTestDb.sqlite"]; 
BOOL success = [fileMgr fileExistsAtPath:dbPath]; 
if(!success) 
{ 
    NSLog(@"Cannot locate database file '%@'.", dbPath); 
} 
if(!(sqlite3_open([dbPath UTF8String], &cruddb) == SQLITE_OK)) 
{ 
    NSLog(@"An error has occured."); 
} 

if(sqlite3_open([dbPath UTF8String], &cruddb) ==SQLITE_OK){ 

int str1 =1; 
int str2 =1; 
float str3 =0.1; 
float str4 =0.1; 
float str5 =0.1; 
float str6 =0.1; 
NSString *str7 [email protected]"deneme"; 
int str8 =1; 
int str9 =1; 
NSString *[email protected]"deneme"; 

NSString* SQL = [NSString stringWithFormat:@"INSERT INTO LabUpdate (IsSuccess, ProducerId, Latitude, Longitude, Altitude, Slope, SampleDate, PackageNo, Status, Description) VALUES(%i,%i,%f,%f,%f,%f,%@,%i,%i,%@)",str1,str2,str3,str4,str5,str6,str7,str8,str9,str10]; 

stmt = [self prepare:SQL]; 

    sqlite3_step(stmt); 
    sqlite3_finalize(stmt); 
    sqlite3_close(cruddb); 

if (sqlite3_step(stmt) != SQLITE_DONE) 
{ 
    NSLog(@"ExecuteNonQuery has error"); 
    NSLog(@"Failed from sqlite3_step. Error is: %s", sqlite3_errmsg(cruddb)); 

} 
else 
{ 
    int rowsaffected = sqlite3_changes(cruddb); 
    NSLog(@" rowsaffected %i",rowsaffected); 
} 

} 
} 

-(sqlite3_stmt*)prepare:(NSString*)query 
{ 
sqlite3_stmt *queryHandle; 


const char *sqlStatement = (const char *) [query UTF8String]; 

if(sqlite3_prepare_v2(cruddb, sqlStatement, -1, &queryHandle, NULL) != SQLITE_OK) 
{ 
    int error = sqlite3_prepare_v2(cruddb, sqlStatement, -1, &queryHandle, NULL); 

    NSLog(@"Failed from sqlite3_prepare_v2. Error is: %s", sqlite3_errmsg(cruddb)); 

    NSLog(@"Compiled Statement has error code:%i:%@",error,query); 
} 

return queryHandle; 
}

是什麼問題？我該如何解決這個問題？感謝您的回覆。

來源

2012-04-02 Hacer Akac

這並不是說它真的可以解決您的問題，但我完全推薦使用像FMDB或PLDatabase這樣的Objective-C sqlite包裝器 - 它會讓您的生活變得更加輕鬆。 – 2012-04-02 06:32:16

你可以發佈NSLog（@「SQL：％@」，SQL）嗎？但你應該引用你的字符串！例如：NSString * SQL = [NSString stringWithFormat：@「INSERT INTO LabUpdate（IsSuccess，ProducerId，Latitude，Longitude，Altitude，Slope，SampleDate，PackageNo，Status，Description）VALUES（％i，％i，％f，％f，％女，％F， '％@'，％I，％I， '％@'）」，STR1，STR2，STR3，STR4，STR5，STR6，STR7，STR8，STR9，str10]; – CarlJ 2012-04-02 07:43:54

出於某種原因，您正在執行'sqlite3_prepare_v2（）'，如果失敗，您再次執行它**以獲取錯誤代碼。這是無稽之談;一步完成。 – trojanfoe 2012-04-02 09:18:48

當您嘗試運行SQL語句時，所有不是數字的內容都需要引用，例如

INSERT INTO foo (column1, column2) VALUES ('bar' , 'baz'); 
             // ^^^^^ ^^^^^ string values quoted.

不過，我強烈建議使用sqlite3_bind，而不是格式值的直接到你的SQL語句。

來源

2012-04-02 07:59:09 JeremyP

Sqlite插入錯誤的目標c

回答

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