从php中的mysql表中选择count（*）

本文介绍了从php中的mysql表中选择count（*）的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我可以得到mysql查询结果的值和行。

但我很难得到查询的单个输出。例如：

$ result = mysql_query（SELECT COUNT（*）FROM Students;）;

我需要显示结果。但我没有得到结果。

我尝试了以下方法：

mysql_fetch_assoc（）

mysql_free_result

mysql_fetch_row（）

您需要使用别名来对聚合进行别名，作为关键字，以便从 mysql_fetch_assoc

中调用

$ result = mysql_query（SELECT count（*）as total from Students）; $ data = mysql_fetch_assoc（$ result）; echo $ data ['total'];

I am able to get both the value and row of the mysql query result.

But I am struggling to get the single output of a query. e.g.:

$result = mysql_query("SELECT COUNT(*) FROM Students;");

I need the result to display. But I am not getting the result.

I have tried with the following methods:

mysql_fetch_assoc()

mysql_free_result()

mysql_fetch_row()

But I didn't succeed to display (get) the actual value.

解决方案

You need to alias the aggregate using the as keyword in order to call it from mysql_fetch_assoc

$result=mysql_query("SELECT count(*) as total from Students"); $data=mysql_fetch_assoc($result); echo $data['total'];