不能得到选择查询工作

Question

我想在C选用显示一个从MySQL数据库的选择查询：

mysql_query(conn,"SELECT SI AS SUBSCRIBER_ID ,TG2 AS TAG_ID, SUM(CTR) AS NBR FROM (SELECT H.SUBSCRIBER_ID AS SI, TG.TAG_ID AS TG1,T.TAG_ID AS TG2, COUNT(TG.TAG_ID) AS COUNTER,CASE WHEN (TG.TAG_ID = T.TAG_ID) THEN COUNT(TG.TAG_ID) ELSE 0 END AS CTR from content_hits H left join CONTENT_TAG TG ON TG.CONTENT_ID = H.CONTENT_ID LEFT JOIN TAG T ON 1= 1 GROUP BY H.SUBSCRIBER_ID, TG.TAG_ID,T.TAG_ID) AS TAB GROUP BY SI,TG2"); 

在那之后，我想用“NBR”，以填补一维数组。

我尝试这样做：

result = mysql_store_result(conn); 

while ((row = mysql_fetch_row(result))) 
{ 
    t[i]=*row['NBR']; 
    printf("%d",t[i]); 
} 

但没有奏效。

Answer 1

您不能像名称那样通过名称访问行列t[i]=*row['NBR'];。使用例如fields = mysql_fetch_fields(result);来获取列名并遍历字段数组以找到哪个列ID为'NBR'。这个id然后可以在t[i]=row[id];中使用。这全是在mysql连接器中doc http://dev.mysql.com/doc/refman/5.0/en/mysql-fetch-fields.html