选择并显示表中具有相同值的行数 [英] Selecting and displaying the count of rows with the same value in a table
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问题描述
基本上这是一个调查系统，我想从我的结果表中计算是和否，它有
Basically this is a survey system and I want to count the Yes's and No's from my results table which has
| q1 | q2 | q3 | q4 | q5 | q6 | q7 | q8 | q9 | q10 |
| Yes | Yes | No | No | No | Yes | No | Yes | Yes | No |
| No | Yes | No | Yes | Yes | Yes | No | No | Yes | No |
| Yes | Yes | No | No | No | Yes | No | Yes | Yes | No |
| No | Yes | No | Yes | Yes | Yes | No | No | Yes | No |
例如我想要的结果是q1 = 2否2是和q2 = 4是0否等等
然后，我希望显示在一个简单的表，但我仍然无关的查询
For example I want my result to be q1 = 2 No's 2 Yes and q2 = 4 yes 0 No's and so on Then after that I was hoping to display it in a simple table but I'm still clueless about the query
<body>
<form method="POST">
<table border="1">
<th>Answered Yes</th>
<th>Answered No</th>
</tr>
<?php
include("testdb.php");
$result= mysqli_query($con, "SELECT * FROM results");
if($result){
?>
</table>
</body>
推荐答案
真是 1 和 false 是 0 ，因此对条件求和有多少次是真的：
In MySQL (only) true is 1 and false is 0, so summing a condition counts how many times it's true:
select
sum((q1 = 'Yes') + (q2 = 'Yes') + ... etc) yeses,
sum((q1 = 'No') + (q2 = 'No') + ... etc) nod
from mytable
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