SELECT table1.id , table1.price , table2.price ,table3.price 
FROM table1 
JOIN table2 ON table.id = table2.id 
JOIN table3 ON table.value = table3.value

结果

你需要做的是做一个 JOIN 在表中，使用一个称为主键和外键的一致值。在本例中，将表链接在一起的一致值是名为value的列。所以这就是你在工作中需要用到的 JOIN ```
SELECT
table1.id as TABLE1_ID,
table1.price as TABLE1_PRICE,
table2.price as TABLE2_PRICE,
table3.price as TABLE3_PRICE
FROM
table1
LEFT JOIN
table2 ON table1.value = table2.value
LEFT JOIN
table3 ON table2.value = table3.value

输出：
![](https://i.stack.imgur.com/TgDoM.png)
工作sql fiddle
更多关于 `JOIN` 在这里。
既然您在评论部分提到了php，请尝试以下操作：

$sql="SELECT
table1.id as TABLE1_ID,
table1.price as TABLE1_PRICE,
table2.price as TABLE2_PRICE,
table3.price as TABLE3_PRICE
FROM
table1
LEFT JOIN
table2 ON table1.value = table2.value
LEFT JOIN
table3 ON table2.value = table3.value"
$result_set=mysqli_query($conn, $sql);
$row=mysqli_fetch_array($conn, $result_set);

$rowcount=mysqli_num_rows($result_set);
$count=0;

while($rowcount > $count) {
$count++;

echo 'id: '.$row['id'].'<br />';
echo 'price: '.$row['price'].'<br />';

$row=mysqli_fetch_array($conn, $result_set);

}

记住 `include` 你的数据库连接文件 `$conn` 可访问。
如果最初不起作用，请尝试用定义的别名替换实际列名，如下所示： `$row['id']` 至 `$row['TABLE1_ID']` 等。

尝试内部连接

select id,table1.value, table1.price as price1,table2.price as price2,table3.price as price3 from table1 
inner join table2 on table1.value=table2.value
inner join table3 on table2.value=table3.value

select table1.id,table1.value, table1.price,table2.price,table3.price

来自表1的表1内部联接表2。value=table2.value表2的表3内部联接表3.value=table3.value