如何选择三个列名相同但值不同的表

Question

Hi i'm so confused to select 3 table data value.

Ex: In table1:

Id    value   price  <br> 
1     101     30   <br> 
2     102     40  <br>
3     103     50  <br>

In table 2

value       price   <br>
101       25  <br>
102       35   <br>
103       45  <br>

In table 3

value       Price   <br>
101          5%  <br>
102         6%   <br>
103         7%  <br>

in this how can i get all price value with id

I need a result like

ID    Price   Price   Price <br>
1  --   30  --   25 --     5%<br>
2  --    40  --   35   --   6%<br>
3  --    50  --   45   --   7%<br>

Answer 1

Try with inner join

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

Answer 2

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

Result

Answer 3

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

从 table1 内连接 table2 上 table1.value=table2.value 内连接 table3 上 table2.value=table3.value

Answer 4

What you need to do is perform a JOIN on your tables, using a coherent value known as primary and foreign key. The coherent values that links your tables together in this case is the column named value . So that's what you need to be using in your 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`

Outputs:

Working SQL fiddle

More about JOIN 's here .

Since you mentioned PHP in the comment section, try something like this:

$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);
}

Remember to include your database connection file so that $conn is accessible.

If it doesn't work initially, try to swap out the actual column names with the defined aliases, like so:

$row['id'] to $row['TABLE1_ID'] etc.