Question

我正在尝试从3个表中获取结果，但它重复了PART_ID并反复显示相同的id。我该如何解决这个问题？

<?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> $PART_TYPE_DESC   - <em> &nbsp; &nbsp; &nbsp; <a href='inventory_edit.php?pid=$id'>edit</a> &bull; <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

enter image description here

Answer 1

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：

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。

Answer 2

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，这将实现从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表中的行匹配（条件是它们具有相同的PART_TYPE_ID）。如果在PART表中的给定行中找不到PART和PART_TYPE表之间的匹配项，则该行不会包含在结果中。实际上，这意味着您只能获得具有有效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赞成，因为它会使维护变得困难。如果您要添加，删除或重新排列列，则所有代码都会中断。因此，我更改了您的select语句，以显式检索您引用的列，其顺序与代码中引用的顺序相同。

编辑：我在代码中省略了与RACK表的连接，因为您没有引用RACK.LOCATION列。如果这是一个错误，只需抛出另一个连接：

INNER JOIN RACK ON RACK.RACK_NUM = PART.RACK_NUM

将LOCATION列添加到SELECT语句中要检索的列列表中。

Answer 3

根据您的图表

SELECT * FROM PART
INNER JOIN PART_TYPE ON PART.PART_ID=PART_TYPE.PART_TYPE_ID
INNER JOIN RACK ON PART.RACK_NUM=RACK.RACK_NUM

应该做你想做的事。

Answer 4

为了不反复显示相同的结果，你需要使用DISTINCT，所以你可能想尝试：

SELECT DISTINCT * FROM PART, PART_TYPE, RACK

PHP / MYSQL查询id“duplicate ids”

4 个答案: