如何在PHP中找到最接近的值
如何根据USER SCORE字段以及最接近SCORE BRAND字段的值的规定来确定表用户中的RESULTS字段。
这是餐桌品牌
<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>
这是桌面用户
<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>
您可以在选择中使用子查询为每个选定的用户找到适当的品牌，例如：
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
select *
from table
order by abs(value - $myvalue)
limit 1
你应该有一些预先定义的阈值来匹配周围（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