Question

我需要从sqlite数据库中动态获取表名。

我的代码

-(void) readItemsFromDatabaseforTable:(NSString *)tableName {
    // Setup the database object
    sqlite3 *database;

    // Init the animals Array
    itemsList = [[NSMutableArray alloc] init];

    // Open the database from the users filessytem
    if(sqlite3_open([databasePath UTF8String], &database) == SQLITE_OK) {
        // Setup the SQL Statement and compile it for faster access
        const char *sqlStatement = "select * from %@",tableName ;
        sqlite3_stmt *compiledStatement;
        if(sqlite3_prepare_v2(database, sqlStatement, -1, &compiledStatement, NULL) == SQLITE_OK) {
            // Loop through the results and add them to the feeds array
            while(sqlite3_step(compiledStatement) == SQLITE_ROW) {
                // Read the data from the result row
                NSString *aName = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 1)];
                NSInteger aDescription =(compiledStatement, 2);
                //  NSString *aImageUrl = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 3)];

                // Create a new animal object with the data from the database
                Category *item = [[Category alloc] initWithName:aName Quantity:aDescription];

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

                [item release];
            }
        }
        // Release the compiled statement from memory
        sqlite3_finalize(compiledStatement);

    }
    sqlite3_close(database);


}

但我没有得到。

我收到一个警告未使用的变量tableName。

实际字符串是const char * sqlStatement =“select * from allcategories”;

如何在该类别中动态传递该表名。

任何人都可以帮助我。

提前感谢你。

Answer 1

试试这个，

NSString *sql_str = [NSString stringWithFormat:@"select * from %@", tableName];
const char *sqlStatement = (char *)[sql_str UTF8String];

Answer 2

const char *sqlStatement = "select * from %@",tableName ;

将此行更改为：

const char *sqlStatement = [NSString stringWithFormat: @"select * from %@",tableName];

更好的选择是将此sqlStatement设为字符串，因为我不知道char是否可以存储NSString对象...

尝试：

NSString *sqlStatement = [NSString stringWithFormat:@"select * from %@",tableName];

这肯定会有效，我自己尝试过（如果你的tableName是正确的话）。

NSLog tableName以查看您是否传递了正确的表名。

希望这有帮助。

Answer 3

const char *sqlStatement = "select * from %@",tableName ;

将上一行更改为

const char *sqlStatement = (const char *) [[NSString stringWithFormat:@"select * from %@", tableName]  UTF8String];

如何在sqlite查询中传递表名

3 个答案: