在一个查询中从两个 mysql 表中选择数据

Posted 2023-05-09

【中文标题】在一个查询中从两个 mysql 表中选择数据

【英文标题】：Selecting data from two mysql tables in one query

【发布时间】：2010-01-22 17:45:37

【问题描述】：

我有两张桌子，房东和财产。我的属性表有； ID，地址，邮政编码，租约和房东ID。我面临的问题是：如果我想搜索以 Mr.Spina 作为房东的所有房产，我需要搜索名为“spina”的房东数据库，以获取保存在房产数据库中的他的 ID，我可以从中获取提取属性详细信息。

我认为这会起作用，但它不正确：

> SELECT property.ID, property.address, property.postcode, property.lease, landlords.firstName, landlords.lastName FROM property INNER JOIN landlords ON landlords.firstName LIKE '%spina%' OR landlords.lastName LIKE '%spina%'

我附上了表格结构的图片。

房东：

只允许一个链接

属性：

http://img5.imageshack.us/img5/7199/propertyn.gif

在字段中插入“spina”的结果应该是： 只允许一个链接

这是我提取的代码...

>  if($field=="landlord")
>     
>     $sql="SELECT property.ID, property.address, property.postcode,
> property.lease, landlords.firstName,
> landlords.lastName FROM ".$do." INNER
> JOIN landlords ON landlords.firstName
> LIKE '%".$q."%' OR landlords.lastName
> LIKE '%".$q."%'";
>             else
>     $sql="SELECT * FROM ".$do." WHERE " . $field . " LIKE '%" . $q . "%'";  
>     //end special case     $result =
> mysql_query($sql);
>       echo "$sql";
>       echo "<table border='1'>
>     <tr>
>      <th>ID</th>
>      <th>Address</th>
>      <th>Post Code</th>
>      <th>Lease</th>
>      <th>Landlord</th>
>     </tr>";
> 
>    while($row =
> mysql_fetch_array($result))
>      
>      echo "<tr>";
>      echo "<td>" . $row['ID'] . "</td>";
>      echo "<td>" . $row['address'] . "</td>";
>      echo "<td>" . $row['postcode'] . "</td>";
>      echo "<td>" . $row['lease'] . "</td>";
>      echo "<td>" . $row['firstName'] ." ". $row['lastName'] ."</td>";
>      echo "</tr>";
>          echo "</table>";
> 
>    mysql_close();

提前非常感谢！

【参考方案1】：

在您的示例查询中，您需要加入房东 ID。

SELECT property.ID, property.address, property.postcode, property.lease,
    landlords.firstName, landlords.lastName 
FROM property INNER JOIN landlords ON landlords.id = property.landlordID
WHERE landlords.firstName LIKE '%spina%' OR landlords.lastName LIKE '%spina%'

【参考方案2】：

您似乎错误地使用了INNER JOIN；你想要的是这样的：

SELECT
    property.ID, property.address, property.postcode, property.lease, landlords.firstName, landlords.lastName
FROM
    property
INNER JOIN
    landlords
ON 
    landlords.ID = property.landlord
WHERE
    landlords.firstName LIKE '%spina%'
    OR
    landlords.lastName LIKE '%spina%'

【讨论】：

非常感谢，您的帖子与上面的帖子相结合！

【参考方案3】：

你说：

> SELECT property.ID, property.address, property.postcode, property.lease, firstName, lastName FROM property INNER JOIN landlords ON firstName LIKE '%spina%' OR lastName LIKE '%spina%'

例如，您的连接表达式需要说明这两个表是如何相关的

... LEFT JOIN landlords ON landlords.ID == property.landlord_ID ...

然后将您的选择移动到WHERE 子句中：

... WHERE firstName LIKE '%spina%' OR lastName LIKE '%spina%'

示例：

> cat > landlords.csv
1,Alex,Spina
2,Spina,Brown
3,Katell,Jentreau

> cat > properties.csv
1,toytown
2,hogwarts
3,mars
4,new york
2,sheffield

> sqlite3

> CREATE TABLE landlords (ID,firstname,lastname);
> .import "landlords.csv" "landlords";
> CREATE TABLE properties (landlord,address);
> .import "properties.csv" "properties";
> SELECT * FROM properties JOIN landlords ON landlord = ID
    WHERE (firstname LIKE "Spina" OR lastname LIKE "Spina");
landlord,address,ID,firstname,lastname
1,toytown,1,Alex,Spina
2,hogwarts,2,Spina,Brown
2,sheffield,2,Spina,Brown