我不能在目标C插入数据到SQLite数据库

问题描述：

可能重复：
Assertion failure error in objective c 我不能在目标C插入数据到SQLite数据库

enter image description here 我想尝试到SQLite数据库表中插入一些数据，但我得到了一个错误像这样：

***Assertion failure in -[ViewController buttonClick:],/Users/ds/Desktop/SqliteDeneme/SqliteDeneme/ViewController.m:57

我代码在这里：

- (IBAction)buttonClick:(id)sender { 

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

NSArray *pathsArray=NSSearchPathForDirectoriesInDomains(NSDocumentDirectory,NSUserDomainMask,YES); 
NSString *doumentDirectoryPath=[pathsArray objectAtIndex:0]; 
destinationPath=[doumentDirectoryPath stringByAppendingPathComponent:@"SqliteTestDb.sqlite"]; 
NSLog(@"database path %@",destinationPath); 

if (sqlite3_open([destinationPath UTF8String], &cruddb)==SQLITE_OK) 
{ 
    NSLog(@"dataBaseOpen"); 
    // leak happens here, do stuff then call sqlite3_close(database), or move it out of the if/else block. 
    if(stmt == nil) { 
     const char *sql = "INSERT INTO LabUpdate (IsSuccess, ProducerId, Latitude, Longitude, Altitude, Slope, SampleDate, PackageNo, Status, Description) VALUES (?,?,?,?,?,?,?,?,?,?)";    
     if(sqlite3_prepare_v2(cruddb, sql, -1, &stmt, NULL) == SQLITE_OK){ 
      //sqlite3_prepare_v2(cruddb, sql, 1, &stmt, NULL); 
      sqlite3_bind_int(stmt, 1, [str1 integerValue]); 
      sqlite3_bind_int(stmt, 2, [str2 integerValue]); 
      sqlite3_bind_double(stmt, 3, [str3 floatValue]); 
      sqlite3_bind_double(stmt, 4, [str4 floatValue]); 
      sqlite3_bind_double(stmt, 5, [str5 floatValue]); 
      sqlite3_bind_double(stmt, 6, [str6 floatValue]); 
      sqlite3_bind_text(stmt, 7, [str7 UTF8String], -1, SQLITE_TRANSIENT); 
      sqlite3_bind_int(stmt, 8, [str8 integerValue]); 
      sqlite3_bind_int(stmt, 9, [str9 integerValue]); 
      sqlite3_bind_text(stmt, 10, [str10 UTF8String], -1, SQLITE_TRANSIENT); 
     } 
     else 
      NSAssert1(0, @"Error while creating add statement. '%s'", sqlite3_errmsg(cruddb)); 

    } 

    if(SQLITE_DONE != sqlite3_step(stmt)) 
     NSAssert1(0, @"Error while inserting data. '%s'", sqlite3_errmsg(cruddb)); 
    else 
     //SQLite provides a method to get the last primary key inserted by using sqlite3_last_insert_rowid 
     recordID = sqlite3_last_insert_rowid(cruddb); 

    //Reset the add statement. 
    sqlite3_reset(stmt); 

} 
else { 
    sqlite3_close(cruddb); 
    NSLog(@"dataBaseNotOpen"); 
    NSAssert1(0, @"Error while opening database '%s'", sqlite3_errmsg(cruddb)); 

} 

}

我该如何解决这个问题？我把一个断点，我看到了这条线后，没有进入：

if(sqlite3_prepare_v2(cruddb, sql, -1, &stmt, NULL) == SQLITE_OK){

这是我的数据库表和colums：

但代码是不同的！ – 2012-03-30 10:04:59

所以你应该编辑你的原始问题，而不是发布一个新的问题。 – borrrden 2012-03-30 10:09:48

sqlite3_prepare_v2不等于SQLITE_OK！请帮助我如何解决？ – 2012-03-30 10:13:42

答

可能是它的工作原理

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

    sqlite3_prepare_v2(cruddb, "BEGIN TRANSACTION", -1, &compiledStmt, NULL); 
    sqlite3_step(compiledStmt); 
    sqlite3_finalize(compiledStmt); 

    const char *sql = "INSERT INTO LabUpdate (IsSuccess, ProducerId, Latitude, Longitude, Altitude, Slope, SampleDate, PackageNo, Status, Description) VALUES (?,?,?,?,?,?,?,?,?,?)"; 
    if(sqlite3_prepare_v2(cruddb, sql, -1, &stmt, NULL) == SQLITE_OK){ 

     sqlite3_bind_int(stmt, 1, [str1 integerValue]); 
     sqlite3_bind_int(stmt, 2, [str2 integerValue]); 
     sqlite3_bind_double(stmt, 3, [str3 floatValue]); 
     sqlite3_bind_double(stmt, 4, [str4 floatValue]); 
     sqlite3_bind_double(stmt, 5, [str5 floatValue]); 
     sqlite3_bind_double(stmt, 6, [str6 floatValue]); 
     sqlite3_bind_text(stmt, 7, [str7 UTF8String], -1, SQLITE_TRANSIENT); 
     sqlite3_bind_int(stmt, 8, [str8 integerValue]); 
     sqlite3_bind_int(stmt, 9, [str9 integerValue]); 
     sqlite3_bind_text(stmt, 10, [str10 UTF8String], -1, SQLITE_TRANSIENT); 

      NSUInteger err = sqlite3_step(compiledStmt); 
      if (err != SQLITE_DONE){ 
       NSLog(@"error while binding %d %s",err, sqlite3_errmsg(database)); 
      } 
      sqlite3_reset(compiledStmt); 
     sqlite3_finalize(compiledStmt);  
    } else { 
     NSLog(@"Invalid Query"); 
    } 

    sqlite3_prepare_v2(cruddb, "END TRANSACTION", -1, &compiledStmt, NULL); 
    sqlite3_step(compiledStmt); 
    sqlite3_finalize(compiledStmt); 
    sqlite3_close(cruddb);

我得到thie消息：2012-03-30 13:57:22688 SqliteDeneme [1167：207]无效的查询。我该如何解决查询问题？ – 2012-03-30 10:58:04

我错误地设置了其他查询现在检查答案 – Hiren 2012-03-30 11:05:15

它给仍然无效的查询错误：S – 2012-03-30 11:07:55

答

我可以建议你使用绑定列的不同技术：

NSString* SQL = [NSString stringWithFormat:@"INSERT INTO table1(col1,col2,col3) VALUES(%i,'%@',%i)",number,Surname,age];

其中数字和年龄是int的，姓是NSSt ring *（注意格式中字符串的引号）。

你可以像这样的代码执行以下命令：

sqlite3_stmt *queryHandle = [self prepare:SQL]; 

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

} 
else 
{ 
    int rowsaffected = sqlite3_changes(database); 
}

一般准备功能应该是这样的：

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


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

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

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

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

return queryHandle;

}

这是我使用的方式，是快速，可以推广到子功能，如准备。

希望这会有所帮助。

我不能在目标C插入数据到SQLite数据库

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