Question

我有两张桌子，房东和房产。我的属性表有;其中的ID，地址，邮政编码，租赁和房东ID。我面临的问题是：如果我想搜索所有以Mr.Spina为房东的房产，我需要搜索房东数据库，名称为“spina”，以获取他的ID，该ID保存在属性数据库中，我可以从中查找提取属性详细信息。

我认为这样可行，但不正确：

> 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%'

我附上了桌子结构的图像。

业主：

只允许一个链接

属性：

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

将“spina”插入字段的结果应为：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();

非常感谢提前！

Answer 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%'

Answer 2

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%'

您的连接表达式需要说明这两个表是如何相关的，例如

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

然后将您的选择移到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

Answer 3

好像你错误地使用了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%'

在一个查询中从两个mysql表中选择数据

3 个答案: