计算多列中特定变量的出现次数答案

【问题标题】：Counting the occurence of a specific variable in multiple columns计算多列中特定变量的出现次数
【发布时间】：2017-05-21 12:28:11
【问题描述】：

我非常感谢你们中的任何一位出色的程序员可以帮助我解决这个问题。我在 mysql/php 方面的编码专业知识有限，但我很固执。

到目前为止：下面这个成功的查询为名为“zmon”的企业提供了仅在一个“rsmed”列中具有“严重”的员工数量，我现在需要从企业“zmon”的多个列中计算“严重”：

$host="localhost";
$username="user"; 
$password="pass";
$db_name="dbase";
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

$query = "SELECT COUNT(*) FROM forearm WHERE business='zmon' AND rsmed = 'severe' "; 

$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result))
{
echo "There are ". $row['COUNT(*)'] ." employees severe in rsmed.";
}

我被困在这里：我需要为名为 zmon 的企业计算名为“forearm”的表中多个列（rslat、rsmed、rscentral、rselbow）中的“severes”数量。

因此，业务列包含业务名称。同一个企业可以有多行，每行对应于他们的不同员工。其他列（rslat、rsmed、rscentral、rselbow）包含 4 个变量中的任何一个：不显着、低、中、高和严重。

我希望这对你来说已经足够了。

谢谢，保罗

【问题讨论】：

标签： php mysql count multiple-columns

【解决方案1】：

您可以操纵查询以使用SUM(criteria) 或SUM(IF(condition, 1, 0)) 单独计算每一列。

SELECT 
    SUM(rslat = 'severe') as rslat_count,
    SUM(rselbow = 'severe') as rselbow_count,
    SUM(rsmed = 'severe') as rsmed_count,
    SUM(rscentral = 'severe') as rscentral_count
FROM forearm
WHERE business='zmon'

数据：

| id | business |  rslat | rselbow |  rsmed | rscentral |
|----|----------|--------|---------|--------|-----------|
|  1 |     zmon | severe |  severe | severe |      good |
|  2 |     zmon | severe |  severe |   good |      good |
|  3 |     zmon |   good |  severe |   good |      good |
|  4 |     zmon | severe |  severe |   good |      good |

结果：http://sqlfiddle.com/#!9/093bd/2

| rslat_count | rselbow_count | rsmed_count | rscentral_count |
|-------------|---------------|-------------|-----------------|
|           3 |             4 |           1 |               0 |

然后你可以使用 php 在 php 中显示结果

$sentence = 'There are %d employees severe in %s';
while ($row = mysql_fetch_assoc($result)) {
    printf($sentence, $row['rslat_count'], 'rslat');
    printf($sentence, $row['rselbow_count'], 'rselbow');
    printf($sentence, $row['rsmed_count'], 'rsmed');
    printf($sentence, $row['rscentral_count'], 'rscentral');
}

更新

要获得各个列的派生总数，只需将它们相加即可。

SELECT 
   SUM(counts.rslat_count + counts.rselbow_count + counts.rsmed_count + counts.rscentral_count) as severe_total,
   counts.rslat_count,
   counts.rselbow_count,
   counts.rsmed_count,
   counts.rscentral_count
FROM (
   SELECT 
      SUM(rslat = 'severe') as rslat_count,
      SUM(rselbow = 'severe') as rselbow_count,
      SUM(rsmed = 'severe') as rsmed_count,
      SUM(rscentral = 'severe') as rscentral_count
   FROM forearm
   WHERE business='zmon'
) AS counts

结果http://sqlfiddle.com/#!9/093bd/10

| severe_total | rslat_count | rselbow_count | rsmed_count | rscentral_count |
|--------------|-------------|---------------|-------------|-----------------|
|            8 |           3 |             4 |           1 |               0 |

然后显示严重总数

$sentence = 'There are %d employees severe in %s';
while ($row = mysql_fetch_assoc($result)) {
    printf($sentence, $row['rslat_count'], 'rslat');
    printf($sentence, $row['rselbow_count'], 'rselbow');
    printf($sentence, $row['rsmed_count'], 'rsmed');
    printf($sentence, $row['rscentral_count'], 'rscentral');
    echo 'business in ' . $row['severe_total'] . ' severe conditions';
}

【讨论】：

如果您希望获得所有状态的计数，也可以对它们使用不同的查询。
谢谢你。我现在意识到我的问题没有完全表达。我需要汇总表前臂中的严重次数。
@Paul 简单调整 - 已更新，但您也可以通过 PHP 汇总各个列以显示每个列的总数。

【解决方案2】：

如果您计算不同列（rslat、rsmed、rscentral、rselbow）中有多少“严重”，您可以尝试将查询修改为以下内容：

SELECT COUNT(*) AS employee_count, "rsmed" AS rtype 
FROM forearm WHERE business='zmon' AND rsmed = 'severe'
UNION
SELECT COUNT(*) AS employee_count, "rslat" AS rtype 
FROM forearm WHERE business='zmon' AND rslat = 'severe'
UNION
SELECT COUNT(*) AS employee_count, "rscentral" AS rtype 
FROM forearm WHERE business='zmon' AND rscentral = 'severe'
UNION
SELECT COUNT(*) AS employee_count, "rselbow" AS rtype 
FROM forearm WHERE business='zmon' AND rselbow = 'severe'

然后你现在可以像这样编写你的循环：

while($row = mysql_fetch_array($result))
{
    echo "There are {$row['employee_count']} employees severe in {$row['rtype']}.";
}

【讨论】：

【解决方案3】：

试试这个：

<?php

$host="localhost";
$username="user"; 
$password="pass";
$db_name="dbase";

$conn = mysql_connect("$host", "$username", "$password")or die("cannot connect"); 

mysql_select_db("$db_name")or die("cannot select DB");

$query = "SELECT COUNT(*) FROM forearm WHERE business='zmon' AND rsmed = 'severe' "; 

$result = mysqli_query($conn,$query);

if($result){

  // Return the number of rows in result set

  $rowcount=mysqli_num_rows($result);

  printf("Result set has %d rows.\n",$rowcount);
  // Free result set

  mysqli_free_result($result);
  }

mysqli_close($conn);


?>

【讨论】：

谢谢。这引发了一个错误。我认为这是在 msql/msqli 的混乱部门。干杯，保罗