我有一个表用于存储本周的飞机检查时间表,它称为aircraft_sched
。还有另外两个相关的,一个称为aircraft_sched_options,我需要将其加入第一个表,最后一个表称为aircraft .
aircraft_sched
:
column 1: AC_Reg (VARCHAR)(10),(PK),(FK -> `aircraft` PK)
column 2: Sched_Day1 (INT)(1),(FK -> `aircraft_sched_options` PK)
column 3: Sched_Day2 (INT)(1),(FK -> `aircraft_sched_options` PK)
column 4: Sched_Day3 (INT)(1),(FK -> `aircraft_sched_options` PK)
column 5: Sched_Day4 (INT)(1),(FK -> `aircraft_sched_options` PK)
column 6: Sched_Day5 (INT)(1),(FK -> `aircraft_sched_options` PK)
column 7: Sched_Day6 (INT)(1),(FK -> `aircraft_sched_options` PK)
column 8: Sched_Day7 (INT)(1),(FK -> `aircraft_sched_options` PK)
aircraft_sched_options
:
column 1: SchedOpt_ID (INT)(1),(PK)
column 2: SchedOpt_Name (VARCHAR)(10)
column 3: SchedOpt_Color (VARCHAR)(7),
飞机
column 1: AC_Reg (VARCHAR)(10),(PK)
column 2: AC_SN (VARCHAR)(6)
column 3: AC_Year (VARCHAR)(4)
当一架新飞机添加到系统中时,我拥有它,因此它也会将其添加到 aircraft_sched
表中。我认为这不是正确的方式,但现在就是这样。因此,aircraft_sched
表始终填充 AC_Reg
,而 Sched_DayX
单元格要么是 0
(对于 ) NULL
或与所选计划类型对应的 SchedOpt_ID
数字。
我面临的问题是当我尝试将 Sched_DayX
列JOIN
到 SchedOpt_ID
列时。当我只 JOIN
一列时,它有点有效,但如果我尝试执行不止一列,那么该行就会从我的结果中消失。
这是我的代码,“有点”有效:
<?php
$sql = ("
SELECT
*
FROM
aircraft_sched
INNER JOIN aircraft_sched_options AS aso1
ON aircraft_sched.Sched_Day1 = aso1.SchedOpt_ID
");
if(!$result_sql = $mysqli->query($sql))
{
echo QueryCheck("getting the aircraft schedule ","from the aircraft sched",$mysqli) . "Error No: " .$mysqli->errno;
}
while($ACSched = $result_sql->fetch_assoc())
{
echo "<tr>";
echo "<td class=\"ACSched_Reg\">" . $ACSched['AC_Reg'] . "</td>";
echo "<td align=\"center\" style=\"background:" . $ACSched['SchedOpt_Color'] . ";\">" . $ACSched['SchedOpt_Name'] . "</td>";
echo "<td align=\"center\" style=\"background:" . $ACSched['SchedOpt_Color'] . ";\">" . $ACSched['SchedOpt_Name'] . "</td>";
echo "<td align=\"center\" style=\"background:" . $ACSched['SchedOpt_Color'] . ";\">" . $ACSched['SchedOpt_Name'] . "</td>";
echo "<td align=\"center\" style=\"background:" . $ACSched['SchedOpt_Color'] . ";\">" . $ACSched['SchedOpt_Name'] . "</td>";
echo "<td align=\"center\" style=\"background:" . $ACSched['SchedOpt_Color'] . ";\">" . $ACSched['SchedOpt_Name'] . "</td>";
echo "<td align=\"center\" style=\"background:" . $ACSched['SchedOpt_Color'] . ";\">" . $ACSched['SchedOpt_Name'] . "</td>";
echo "<td align=\"center\" style=\"background:" . $ACSched['SchedOpt_Color'] . ";\">" . $ACSched['SchedOpt_Name'] . "</td>";
echo "</tr>";
}
?>
当我说它“有点”有效时,我的意思是它实际上显示了一些结果。它不起作用的原因是,即使只有一天分配了计划类型,它也会在每个单元格中显示相同的结果。
当我将第二个 JOIN
添加到查询中时,如下所示:
$sql = ("
SELECT
*
FROM
aircraft_sched
INNER JOIN aircraft_sched_options AS aso1
ON aircraft_sched.Sched_Day1 = aso1.SchedOpt_ID
INNER JOIN aircraft_sched_options AS aso2
ON aircraft_sched.Sched_Day2 = aso2.SchedOpt_ID
");
...那么它不会返回任何有记录的行。
我一直在阅读JOIN
及其工作原理,我能想到解决问题的唯一方法是为每个AC_Reg
建立一个单独的表,所以一周中的每一天都可以设置为 UNIQUE
列,但我不认为这是完成任务的最佳方式。
编辑:
以下是一些屏幕截图,可以提供更好的视觉效果。
aircraft_sched
:
aircraft_sched_options
:
我的代码:
屏幕显示:
最佳答案
如果我正确理解你的问题,你需要与表tbl_two
连接对于每列 Day
:
SELECT
aircraft_sched.AC_SN,
IF(
aso1.SchedOpt_Name IS NULL OR aso1.SchedOpt_ID = 0,
'-',
aso1.SchedOpt_Name
) as option1,
IF(
aso2.SchedOpt_Name IS NULL OR aso2.SchedOpt_ID = 0,
'-',
aso2.SchedOpt_Name
) as option2
FROM
aircraft_sched
LEFT JOIN aircraft_sched_options AS aso1
ON aircraft_sched.Sched_Day1 = aso1.SchedOpt_ID
LEFT JOIN aircraft_sched_options AS aso2
ON aircraft_sched.Sched_Day2 = aso2.SchedOpt_ID
....
编辑:我已更新查询并使用 LEFT JOIN
而不是INNER JOIN
获取选项,因为正如您所说,有些可能是 NULL
更新:删除了 aircraft
的连接并添加检查选项是否为 null 或 id 为 0 -
将显示
关于php - 我如何正确地 `JOIN` 这些表?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16056736/