从php中的mysql表中选择count(*)我能够获得mysql查询结果的值和行。但我正在努力获得查询的单个输出。例如:$result = mysql_query("SELECT COUNT(*) FROM Students;");我需要显示结果。但我没有得到结果。我尝试过以下方法:mysql_fetch_assoc()mysql_free_result()mysql_fetch_row()但我没有成功显示(获得)实际值。
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慕哥6287543
TA贡献1831条经验 获得超10个赞
您需要使用as
关键字对聚合进行别名以便从中调用它mysql_fetch_assoc
$result=mysql_query("SELECT count(*) as total from Students");$data=mysql_fetch_assoc($result);echo $data['total'];
![?](http://img1.sycdn.imooc.com/545863f50001df1702200220-100-100.jpg)
尚方宝剑之说
TA贡献1788条经验 获得超4个赞
如果您只需要值:
$result = mysql_query("SELECT count(*) from Students;");echo mysql_result($result, 0);
![?](http://img1.sycdn.imooc.com/54584cfb0001308402200220-100-100.jpg)
慕村9548890
TA贡献1884条经验 获得超4个赞
$result = mysql_query("SELECT COUNT(*) AS `count` FROM `Students`");$row = mysql_fetch_assoc($result);$count = $row['count'];
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