php – 计算多列中特定变量的出现次数

我会非常感谢你们中任何一位出色的编码员可以帮助我.我在mysql / php中的编码专业知识有限,但我很固执.

至今：
下面这个成功的查询给出了名为’zmon’的业务只有一列“rsmed”中“严重”的员工数量,我现在需要从业务’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：
对于名为zmon的企业,我需要在名为“forearm”的表中计算多列(rslat,rsmed,rscentral,rselbow)中的“severes”数量.

因此,列业务包含业务名称.
同一个企业可以有多行,每行对应他们的不同员工.
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.

我希望这对你来说足够了.

谢谢,保罗

解决方法:

您可以操纵查询以使用SUM(条件)或SUM(IF(条件,1,0))来单独计算每列.

SELECT     SUM(rslat = 'severe') as rslat_count,    SUM(rselbow = 'severe') as rselbow_count,    SUM(rsmed = 'severe') as rsmed_count,    SUM(rscentral = 'severe') as rscentral_countFROM forearmWHERE 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

要获得各列的派生总计,您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_countFROM (   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 |

然后显示严重的总数

$sentence = 'There are %d employees severe in %s';while ($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';}

来源：https://www.icode9.com/content-1-419151.html