iPhoneSdk,SQLite数据库搜索查询不起作用

时间:2011-06-24 04:08:58

标签: iphone sqlite

我想知道如何在搜索查询中传递特定的单词或字符来查找记录。 我传递的格式如'%%% @ %%' 但它没有给我输出...
这是我的代码......

searchName = [NSString stringWithFormat:@"'%%%@%%'",[[NSUserDefaults standardUserDefaults] objectForKey:@"key"]];  
-(void)getSearchData  
{  

        NSLog(@"search Name %@", searchName);   

    searchArray = [[NSMutableArray alloc] init];   

    const char *sql = "select * from features where lower(name) like ? limit 500";  
    if(sqlite3_prepare_v2(database, sql, -1, &selectstmt, NULL)!= SQLITE_OK)  
    {  
        NSAssert1(0,@"error : failed to prepare statement with message '%s'.",sql ite3_errmsg(database));  

      }  
    sqlite3_bind_text(selectstmt, 1, [searchName UTF8String], -1, SQLITE_TRANSIENT);  
    //  sqlite3_bind_text(selectstmt, 2, [state UTF8String], -1, SQLITE_TRANSIENT);  

    while (sqlite3_step(selectstmt) == SQLITE_ROW)  
    {   

        NSMutableDictionary *datah = [[NSMutableDictionary alloc] init];  

        [datah setObject:[NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt,0)] forKey:@"number"];  

        [datah setObject:[NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt,1)] forKey:@"name"];  

        [datah setObject:[NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt,2)] forKey:@"type"];  

        [datah setObject:[NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt,4)] forKey:@"country"];  

        [datah setObject:[NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt,5)] forKey:@"lat"];  

        [datah setObject:[NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt,6)] forKey:@"long"];  

        [datah setObject:[NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt,7)] forKey:@"elevation"]; 

        [datah setObject:[NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt,8)] forKey:@"area"];  

        [datah setObject:[NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt,3)] forKey:@"state"];  

        [searchArray addObject:datah];  
        [datah release];  
    }  
    sqlite3_reset(selectstmt);    
}

如果我使用单词“s”传递相同的查询,我在sqlite查询构建器中获得了100条记录但是如果m通过上面的代码则不会搜索任何记录
请帮帮我

1 个答案:

答案 0 :(得分:1)

我处理了上面给出的代码并修复如下

-(void)getSearchData:(NSString*)searchText
{  

    **NSString* searchName = [NSString stringWithFormat:@"%%%@%%",searchText];**  
    NSLog(@"search Name %@", searchName);   
    sqlite3 *database;
    sqlite3_stmt *selectstmt; 
    NSString* searchQuery = @"SELECT * FROM contacts WHERE lower(name) like ?";
    const char *sql =   [searchQuery UTF8String];
      **if(sqlite3_open([databasePath UTF8String], &database) == SQLITE_OK) {**
          if(sqlite3_prepare_v2(database, sql, -1, &selectstmt, NULL) != SQLITE_OK)  
          { 
            //NSAssert1(0,@"error : failed to prepare statement with message '%s'.",sql ite3_errmsg(database));  

          }  
          sqlite3_bind_text(selectstmt, 1, [searchName UTF8String], -1, SQLITE_STATIC);  
          while (sqlite3_step(selectstmt) == SQLITE_ROW)  
          {     
            NSLog(@"%@",[NSString stringWithUTF8String:(char *)sqlite3_column_text(selectstmt,0)]);
          }  
          sqlite3_reset(selectstmt);    
      }
    sqlite3_close(database);
}

可以在Double *之间找到更改的部分。您还想阅读this

我改变的两个步骤是打开数据库连接,你必须在其他地方完成,然后从searchName中删除''。如果这样可以解决问题,请告诉我们。