我有两个数据表，1表示员工数据，另一个表示员工工时。不幸的是，我收到了行格式的员工小时数据，我需要在员工数据旁边的列中表示......我正在努力...请看下面...
employeehours表：
ID -------- ------------- Paycode小时---- workdate
089999 01普通-------- 4.00 ----- 2013-09-16
089999 02加时赛1.5 --- 2.00 ----- 2013-09-16
089999 03加班2.0 --- 0.50 ----- 2013-09-16
083131 01普通-------- 7.60 ----- 2013-09-16
083131 02加时赛1.5 --- 0.43 ----- 2013-09-16
employeedata table：
ID ------姓氏 - 名字--- salarygrade
------------------- baserate --- otherrate089999 SMITH ----- JOHN -------- XXX TWU EBA Perm Gr6 - 23.8508 ---- 0.0000
我想要一个查询来生成如下结果，我已经尝试了一切我的能力，但我不能让它工作......我知道我可能需要一些PHP编程来产生$值但是它也可以通过SQL做到这一点？非常感谢所有帮助...
ID ------姓氏 - 名字--- salarygrade
------------------- 01小时--- 01 $ ---- 02小时---- 02 $ --- 03小时-03 $089999史密斯-----约翰-------- XXX TWU EBA Perm Gr6 --- 4.00 ---- 95.4032 - 2.00 ---- 71.5524--0.50--23.8508
对于某些具有ID的员工，某些数据可能不存在，并且在某些列中将以0.00表示......
这是我尝试过的：
<?php
$result1=mysql_query("SELECT `ID Number`, Surname, `First Name`, `Base Rate`, `Other Rate`, `Salary Code Description`, workdate, employeehours
FROM payroll.employeedatanew INNER JOIN payroll.employeehours ON employeedatanew.`ID Number` = employeehours.employeeid
GROUP BY Surname ORDER BY Surname Asc
");
$result2=mysql_query(" SELECT `ID Number`, paycode, workdate, employeehours
FROM payroll.employeedatanew INNER JOIN payroll.employeehours ON employeedatanew.`ID Number` = employeehours.employeeid
ORDER BY Surname Asc
");
while($show1=mysql_fetch_array($result1)){
echo("<tr><td>".$show1['ID Number']."</td><td>".$show1['Surname']."</td><td>".$show1['First Name']."</td><td>".substr($show1['Salary Code Description'],11)."</td></tr>");
}
?>
</table>
<table>
<tr>
<td>1.0x Hrs</td>
<td>1.0x $</td>
<!--<td>1.5x Hrs</td>
<td>1.5x $</td>
<td>2.0x Hrs</td>
<td>2.0x $</td>
<td>Crib Hrs</td>
<td>Crib $</td>
<td>Meal Hrs</td>
<td>Meal $</td> -->
</tr>
<?php
$resultset = array();
while ($row = mysql_fetch_assoc($result3)) { $resultset[] = $row; }
while($show2=mysql_fetch_array($result2)){
if ($show2['paycode'] == "01 Ordinary" && $show2['ID Number'] == $resultset['ID Number']) {
echo ("<tr><td>".$show2['employeehours']."</td>");
$normhourspay = ($show2['employeehours'] * $show3['Base Rate']);
echo ("<td>".$normhourspay."</td></tr>");
} //else {echo("<tr><td>0</td><td>0</td></tr>");}}
和此：
<table>
<tr class="tabletitles">
<td>Employee ID</td>
<td>Surname</td>
<td>First Name</td>
<td>Pay Grade</td>
<td>1.0x Hrs</td>
<td>1.0x $</td>
<!--<td>1.5x Hrs</td>
<td>1.5x $</td>
<td>2.0x Hrs</td>
<td>2.0x $</td>
<td>Crib Hrs</td>
<td>Crib $</td>
<td>Meal Hrs</td>
<td>Meal $</td> -->
</tr>
<?php
$result1=mysql_query("SELECT `ID Number`, Surname, `First Name`, `Base Rate`, `Other Rate`, `Salary Code Description`, workdate, employeehours
FROM payroll.employeedatanew INNER JOIN payroll.employeehours ON employeedatanew.`ID Number` = employeehours.employeeid
GROUP BY Surname ORDER BY Surname Asc
");
$result2=mysql_query(" SELECT `ID Number`, paycode, workdate, employeehours
FROM payroll.employeedatanew INNER JOIN payroll.employeehours ON employeedatanew.`ID Number` = employeehours.employeeid
ORDER BY Surname Asc
");
while($show1=mysql_fetch_array($result1)){
echo("<tr><td>".$show1['ID Number']."</td><td>".$show1['Surname']."</td><td>".$show1['First Name']."</td><td>".substr($show1['Salary Code Description'],11)."</td></tr>");
}
echo ("</table>");
while($show2=mysql_fetch_array($result2)){
if ($show2['paycode'] == "01 Ordinary" && $show2['ID Number'] == $show1['ID Number']) {
echo ("<td>".$show1['employeehours']."</td>");
$normhourspay = ($show1['employeehours'] * $show1['Base Rate']);
echo ("<td>".$normhourspay."</td></tr>");
} else {echo("<td>0</td><td>0</td>");}
}
请帮忙！
答案 0 :(得分：0)
SELECT `ID Number`, Surname, `First Name`, `Salary Code Description`,
`Base Rate`, `Other Rate`,
MAX((case when employeehours.paycode = '01 Ordinary' then (employeehours.employeehours) end)) AS `1.0 Hours`,
MAX((case when employeehours.paycode = '02 Overtime 1.5' then (employeehours.employeehours) end)) AS `1.5 Hours`,
MAX((case when employeehours.paycode = '03 Overtime 2.0' then (employeehours.employeehours) end)) AS `2.0 Hours`,
MAX((case when employeehours.paycode = '78 Crib' then (employeehours.employeehours) end)) AS `Crib`,
MAX((case when employeehours.paycode = 'CZ Meal Allowance PS' then (employeehours.employeehours) end)) AS `Meal Allowance`,
MAX((case when employeehours.paycode = '86Y Sick with Cert' then (employeehours.employeehours) end)) AS `Sick with Cert`,
MAX((case when employeehours.paycode = '86N Sick without Cert' then (employeehours.employeehours) end)) AS `Sick without Cert`,
MAX((case when employeehours.paycode = '87 Sick without Pay' then (employeehours.employeehours) end)) AS `Sick without Pay`,
MAX((case when employeehours.paycode = '83 Annual Leave' then (employeehours.employeehours) end)) AS `Annual Leave`,
MAX((case when employeehours.paycode = '95 RDO Taken' then (employeehours.employeehours) end)) AS `RDO Taken`,
MAX((case when employeehours.paycode = '85 LSL' then (employeehours.employeehours) end)) AS `LSL`,
MAX((case when employeehours.paycode = '61 Shift 17.5' then (employeehours.employeehours) end)) AS `Shift 17.5`
FROM payroll.employeedatanew INNER JOIN payroll.employeehours ON employeedatanew.`ID Number` = employeehours.employeeid
Group By Surname
ORDER BY Surname asc, `1.0 Hours` desc, `1.5 Hours` desc, `2.0 Hours` desc
这就是诀窍，感谢大家观看：）