Php 计算总价值的百分比

好的，我想知道行总数中正确的百分比是多少（result=1）。result 2表示它仍然处于活动状态，所以只需要result！=二,

我有以下代码，但没有显示任何内容 我做错了什么（公平地说，我仍有一半是清醒的）

---

您需要为

COUNT（*）

添加别名，然后需要使用该别名来获取值

您需要

echo

您的最终输出

你可以这样做：-

$successq = mysqli_query($con,"SELECT COUNT(*) as success FROM `table` 
WHERE `result` ='1'");//success used as alias

$success = mysqli_fetch_assoc($successq);

$totalq = mysqli_query($con,"SELECT COUNT(*) as total FROM `table` 
WHERE `result` !='2'");// total use as alias

$total = mysqli_fetch_assoc($totalq);


echo $percent = ($success['success'] /$total['total'] ) * 100;// use aliases and echo the output

---

注意：-

mysqli\u fetch\u assoc（）

返回关联数组，因此不能将其用作变量。您必须使用它的关联索引从中获取值（就像我在

echo$percent=（$success['success']/$total['total']）*100；

）。

您需要为

COUNT（*）

添加别名，然后您需要使用该别名来获取值

您需要

echo

您的最终输出

你可以这样做：-

$successq = mysqli_query($con,"SELECT COUNT(*) as success FROM `table` 
WHERE `result` ='1'");//success used as alias

$success = mysqli_fetch_assoc($successq);

$totalq = mysqli_query($con,"SELECT COUNT(*) as total FROM `table` 
WHERE `result` !='2'");// total use as alias

$total = mysqli_fetch_assoc($totalq);


echo $percent = ($success['success'] /$total['total'] ) * 100;// use aliases and echo the output

---

注意：-

mysqli\u fetch\u assoc（）

返回关联数组，因此不能将其用作变量。您必须使用它的关联索引从中获取值（就像我在

echo$percent=（$success['success']/$total['total']）*100；

）。

因为

mysqli\u fetch\u assoc

返回数组不是单个值。因此，请按以下方式更改代码：

$successq = mysqli_query($con,"SELECT COUNT(*) as total FROM `table` 
WHERE `result` ='1'");
$success = mysqli_fetch_assoc($successq);
$success_cnt = $success["total"];
$totalq = mysqli_query($con,"SELECT COUNT(*) as total FROM `table` 
WHERE `result` !='2'");
$total = mysqli_fetch_assoc($totalq);
$total_cnt = $total["total"];
$percent = ($success_cnt /$total_cnt ) * 100;

---

因为

mysqli\u fetch\u assoc

返回数组不是单个值。因此，请按以下方式更改代码：

$successq = mysqli_query($con,"SELECT COUNT(*) as total FROM `table` 
WHERE `result` ='1'");
$success = mysqli_fetch_assoc($successq);
$success_cnt = $success["total"];
$totalq = mysqli_query($con,"SELECT COUNT(*) as total FROM `table` 
WHERE `result` !='2'");
$total = mysqli_fetch_assoc($totalq);
$total_cnt = $total["total"];
$percent = ($success_cnt /$total_cnt ) * 100;

---

再次感谢你，我知道我不是很傻spotting@ChrisYates很乐意帮忙。请仔细阅读php手册。你会明白自己做错了什么。再次感谢你，我知道我不是在做傻事spotting@ChrisYates很乐意帮忙。请仔细阅读php手册。你会知道自己做错了什么。