PHP/MYSQL查询id"duplicate ids"
我试图从3个表中获得结果,但它重复了PART_ID并反复显示相同的id.我怎样才能解决这个问题？
<?php
$product_list = "";
$sql = mysql_query("SELECT * FROM PART, PART_TYPE, RACK");
$productCount = mysql_num_rows($sql);
if ($productCount >0){
while($row= mysql_fetch_array($sql)){
$id = $row["PART_ID"];
$PART_DESC = $row["PART_DESC"];
$SERIAL_NUM = $row["SERIAL_NUM"];
$RACK_NUM = $row["RACK_NUM"];
$PART_TYPE_ID = $row["PART_TYPE_ID"];
$PART_TYPE_DESC = $row["PART_TYPE_DESC"];
$product_list .= " <strong>PART_ID:</strong> $id -<strong>$PART_DESC</strong> -<strong>Product Type</strong> $SERIAL_NUM - <em><strong>RACK_NUM</strong> $RACK_NUM - <em> <strong>PART_TYPE_ID</strong> $PART_TYPE_ID - <em> <strong>PART_TYPE_DESC </strong> $PART_TYPE_DESC - <em> <a href='inventory_edit.php?pid=$id'>edit</a> • <a href='inventory_list.php?deleteid=$id'>delete</a><br />";
}
}else
$product_list = "You have not items in the inventory yet"
?>
结果
PART_ID: 1001 -Power Mac G4 Desktop -Product Type XBO31 1WAJ3B - RACK_NUM R1S1 - PART_TYPE_ID 101 - PART_TYPE_DESC MAC - edit • delete
PART_ID: 1001 -Power Mac G4 Desktop -Product Type XBO31 1WAJ3B - RACK_NUM R1S2 - PART_TYPE_ID 101 - PART_TYPE_DESC MAC - edit • delete
PART_ID: 1001 -Power Mac G4 Desktop -Product Type XBO31 1WAJ3B - RACK_NUM R1S3 - PART_TYPE_ID 101 - PART_TYPE_DESC MAC - edit • delete
PART_ID: 1001 -Power Mac G4 Desktop -Product Type XBO31 1WAJ3B - RACK_NUM R1S4 - PART_TYPE_ID 101 - PART_TYPE_DESC MAC - edit • delete
PART_ID: 1001 -Power Mac G4 Desktop -Product Type XBO31 1WAJ3B - RACK_NUM R1S5 - PART_TYPE_ID 101 - PART_TYPE_DESC MAC - edit • delete
PART_ID: 1001 -Power Mac G4 Desktop -Product Type XBO31 1WAJ3B - RACK_NUM R1S6 - PART_TYPE_ID 101 - PART_TYPE_DESC MAC - edit • delete
PART_ID: 1001 -Power Mac G4 Desktop -Product Type XBO31 1WAJ3B - RACK_NUM R2S1 - PART_TYPE_ID 101 - PART_TYPE_DESC MAC - edit • delete
PART_ID: 1001 -Power Mac G4 Desktop -Product Type XBO31 1WAJ3B - RACK_NUM R2S2 - PART_TYPE_ID 101 - PART_TYPE_DESC MAC - edit • delete
PART_ID: 1001 -Power Mac G4 Desktop -Product Type XBO31 1WAJ3B - RACK_NUM R2S3 - PART_TYPE_ID 101 - PART_TYPE_DESC MAC - edit • delete
PART_ID: 1001 -Power Mac G4 Desktop -Product Type XBO31 1WAJ3B - RACK_NUM R2S4 - PART_TYPE_ID 101 - PART_TYPE_DESC MAC - edit • delete
PART_ID: 1001 -Power Mac G4 Desktop -Product Type XBO31 1WAJ3B - RACK_NUM R2S5 - PART_TYPE_ID 101 - PART_TYPE_DESC MAC - edit • delete
PART_ID: 1001 -Power Mac G4 Desktop -Product Type XBO31 1WAJ3B - RACK_NUM R2S6 - PART_TYPE_ID 101 - PART_TYPE_DESC MAC - edit • delete
PART_ID: 1001 -Power Mac G4 Desktop -Product Type XBO31 1WAJ3B - RACK_NUM R3S1 - PART_TYPE_ID 101 - PART_TYPE_DESC MAC - edit • delete
PART_ID: 1001 -Power Mac G4 Desktop -Product Type XBO31 1WAJ3B - RACK_NUM R3S2 - PART_TYPE_ID 101 - PART_TYPE_DESC MAC - edit • delete
PART_ID: 1001 -Power Mac G4 Desktop -Product Type XBO31 1WAJ3B - RACK_NUM R3S3 - PART_TYPE_ID 101 - PART_TYPE_DESC MAC - edit • delete
如果您有任何一对多关系(您看起来像这样),那么您的查询格式不正确.根据某种标准链接表格,如下所示:
SELECT * FROM PART
JOIN PART_TYPE on PART.part_type = PART_TYPE.ID
JOIN RACK ON PART.part_rack = RACK.ID
或类似的东西.您想告诉数据库您希望这些表链接在一起的方式.这样做.
您应该使用某些条件将表连接在一起，例如：
SELECT PART_ID, PART_DESC, SERIAL_NUM, RACK_NUM, PART.PART_TYPE_ID, PART_TYPE_DESC
FROM PART
INNER JOIN PART_TYPE ON PART.PART_TYPE_ID = PART_TYPE.PART_TYPE_ID
基本上，这将完成的是从 PART 表中获取所有行，并且对于我们找到的每一行，将该行与 PART_TYPE 表中的行进行匹配（条件是它们具有相同的 PART_TYPE_ID）。如果在 PART 表和 PART_TYPE 表之间找不到 PART 表中给定行的匹配项，则该行将不会包含在结果中。实际上，这意味着您只会获得具有有效的相应零件类型的零件。
注意：使用SELECT *从表中选择所有列通常不受欢迎，因为它使维护变得困难。如果您要添加、删除或重新排列列，您的所有代码都会崩溃。因此，我更改了您的 select 语句，以按照代码中引用的顺序显式检索您引用的列。
编辑：我在代码中省略了与 RACK 表的联接，因为您没有引用 RACK.LOCATION 列。如果这是一个错误，只需添加另一个连接，如下所示：
INNER JOIN RACK ON RACK.RACK_NUM = PART.RACK_NUM
并将 LOCATION 列添加到要在 SELECT 语句中检索的列的列表中。