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我有三个mysql表。

1. inbox - ID(int), FileNo(varchar), FolioNo(smallint), Reference(varchar), Received(date), Sent(date), Description(text), Forwarded(varchar),...etc

2. outbox - ID(int), Reference(varchar), Sent(date), FileNo(varchar), FileName(varchar), FolioNo(smallint), Description(text), Receiver(varchar),...etc

3. status - ID(int), SN(int), Date(date), Minutes(varchar), Instructed(varchar), Responsible(varchar), Action(boolean), FileNo(varchar),...etc

我想使用 FileNo 加入这三个表,并按 FolioNo 显示它们的顺序,如下所示。

File No: 123
File Name: abc
Folio No | Date     | Description | Sender/Receiver | Minutes   | Instructed | Responsible | Action
----------------------------------------------------------------------------------------------------
1        | 13-09-08 | Something   | Someone         | Something | Someone    | Someone     | Yes
2        | 13-09-10 | Something   | Someone         | Something | Someone    | Someone     | Yes
......etc

我在php中的SQL查询如下,

$sql="SELECT * FROM 
(SELECT I.FileNo, I.FolioNo, I.Received AS Date, I.Description, I.Forwarded 
FROM Inbox I WHERE I.FileNo='$ID' 

UNION SELECT O.FileName, O.FolioNo, O.Sent AS Date, O.Description, O.Receiver 
FROM Outbox O WHERE O.FileNo='$ID' 

UNION SELECT S.Date, S.Minutes, S.Instructed, S.Responsible, S.Action 
FROM Status S WHERE S.FileNo='$ID') 

AS A ORDER BY A.FolioNo";

输出,

<table>
<tr><th>Folio No</th><th>Date</th><th>Description</th><th>Sender/ Receiver</th><th>Minutes</th><th>Instructed</th><th>Responsible</th><th>Action</th></tr>
<?php                           
while ($list = mysql_fetch_assoc($result)) {
echo "<tr><td>" . $list['FolioNo'] . "</td><td>" . $list['Date'] . "</td><td>" . $list['Description'] . "</td><td>" . $list['Receiver'] . "</td><td>" . $list['Minutes'] . "</td><td>" . $list['Instructed'] . "</td><td>" . $list['Responsible'] . "</td><td>" . ($list['Action']=1?'Yes':'No') . "</td></tr>";
$x++;
}
echo "</table>";
?>

但这显示了一个错误,

注意:未定义索引:第 145 行 C:\xampp\htdocs\sp\viewMinutes.php 中的接收者 注意:未定义索引:第 145 行 C:\xampp\htdocs\sp\viewMinutes.php 中的分钟 注意:未定义索引:指示在第 145 行的 C:\xampp\htdocs\sp\viewMinutes.php 中注意:未定义的索引:在第 145 行的 C:\xampp\htdocs\sp\viewMinutes.php 中负责注意:未定义的索引:C:\xampp\ 中的接收者htdocs\sp\viewMinutes.php 在第 145 行 注意:未定义索引:分钟在 C:\xampp\htdocs\sp\viewMinutes.php 在第 145 行 注意:未定义索引:在 C:\xampp\htdocs\sp\viewMinutes 中指示。 php on line 145 注意:未定义的索引:在第 145 行的 C:\xampp\htdocs\sp\viewMinutes.php 中负责 注意:未定义的索引:接收者在 C:\xampp\htdocs\sp\viewMinutes.php 的第 145 行 注意:未定义索引:第 145 行 C:\xampp\htdocs\sp\viewMinutes.php 中的分钟未定义索引:在第 145 行的 C:\xampp\htdocs\sp\viewMinutes.php 中指示注意:未定义索引:在第 145 行的 C:\xampp\htdocs\sp\viewMinutes.php 中负责

FOLIO NO |  DATE    | DESCRIPTION   | SENDER/ RECEIVER  | MINUTES   | INSTRUCTED    | RESPONSIBLE   | ACTION
-------------------------------------------------------------------------------------------------------------------------------------
1        | 13-09-08 |   Something   |                   |           |               |               | Yes
2        | 13-09-08 |   Something   |                   |           |               |               | Yes
         | 13-09-08 |   Something   |                   |           |               |               | Yes
Something|  Someone |   Anyone  |               Yes

这里有什么问题?

更新:

$sql = "SELECT 
Inbox.FolioNo, Inbox.Received, Inbox.Description, Inbox.Forwarded, 
Outbox.FileNo, Outbox.FileName, Outbox.FolioNo, Outbox.Sent, Outbox.Description, Outbox.Receiver, 
Status.Date, Status.Minutes, Status.Instructed, Status.Responsible, Status.Action 
FROM Status INNER JOIN Outbox ON Status.FileNo = Outbox.FileNo 
INNER JOIN Inbox ON Status.FileNo = Inbox.FileNo WHERE Status.FileNo = '$ID' ORDER BY Outbox.FolioNo ASC";
4

2 回答 2

1

您的 SELECT 子句没有提及别名。因此,SQL 引擎不知道要生成哪些列名。我假设它采用第一个 SELECT 的列名。因此,您会看到“Receiver”的未定义索引,它是 2nd SELECT 的一部分。

另外,我不明白 UNION 背后的逻辑。您正在尝试 UNION 具有完全不同列的 3 个不同表的结果。

于 2013-09-11T06:25:18.763 回答
0

我不得不对输出视图做些小改动。现在一切正常。我的sql查询如下。

$sql="SELECT I.FileNo, I.FolioNo, I.Received AS Date, I.Sender AS Person, I.Description, I.Forwarded,  
S.Minutes, S.Instructed, S.Responsible, S.Action FROM Inbox I 
LEFT JOIN Status S ON I.ID=S.SN 
WHERE I.FileNo='$ID' 
UNION SELECT O.FileName, O.FolioNo, O.Sent AS Date, O.Description, O.Receiver AS Person, O.Reference, O.FileNo, O.Signed, O.Subject, O.ID 
FROM Outbox O WHERE O.FileNo='$ID'";

多谢你们!

于 2013-09-11T15:51:41.403 回答