如何在PHP中找到最接近的值

Question

How to determine the RESULTS field in table users, base on USER SCORE field with the provisions of the value closest to SCORE BRAND field.

This is table Brand

    <table>
<tr>
<th>BRAND NAME</th>
<th>SCORE BRAND</th>
</tr>";
$sql = mysql_query("SELECT * FROM brand");
while($m=mysql_fetch_array($sql)){ 
echo "<tr> 
<td>$m[brand_name]</td>
<td>$m[score]</td><tr>";
}
</table>

This is table users

    <table>
<tr>
<th>USER NAME</th>
<th>SCORE USER</th>
<th>RESULT</th>
</tr>";
$sql2 = mysql_query("SELECT * FROM users");
while($u=mysql_fetch_array($sql2)){ 
echo "<tr> 
<td>$u[username]</td>
<td>$u[score]</td>
<td> ??? </td>
<tr>";
}
</table>

Answer 1

You can use subquery in selection to find proper brand for every selected user like this:

SELECT u.*, (
    SELECT b.id
    FROM brand AS b
    ORDER BY ABS(b.score - u.score) ASC, b.score DESC -- selects brands ordered by their difference from user's score
    LIMIT 1 -- get just the first brand (with score closest to user's)
)
FROM user AS u

Answer 2

select * 
from table 
order by abs(value - $myvalue)
limit 1

Answer 3

您应该有一些预定义的阈值来匹配（因此得分1与得分1000不匹配）。

SELECT .. WHERE score_brand >= score_user - :epsilon AND score_brand <= score_user + :epsilon

Answer 4

Finally I got some pretty solution:

SELECT u.*, b.nama_lengkap as result 
FROM `users` AS u 
JOIN `brands` AS b 
WHERE (b.score - u.score) > 0 
AND (b.score - u.score) < 10 
ORDER BY u.id;

b.nama_lengkap -> brand name

I have just joined two tables and did some arithmetic operation to get the appropriate row.

EDIT

SELECT v.id, v.username, v.score, b.nama_lengkap AS result
FROM (SELECT users.id, users.score, users.username, min(brands.score) nextSc
      FROM users
      LEFT JOIN brands
        ON users.score <= brands.score
     GROUP BY users.id, users.score) AS v
LEFT JOIN brands b
  ON v.nextSc = b.score
ORDER BY v.id, v.score;

This is the complete dynamic query for any multiplicated values. It will get you closure score of brands for users.

You can find LIVE DEMO HERE