Mysql查询选择值小于X%或大于X%的值 - Mysql query to select from where value is X% less or X% more than a value

Mysql query to select from where value is X% less or X% more than a value

我想要完成的是选择所有比$influencer_account['followed_by']多20%或少20%的用户。

我尝试使用下面的代码，但它不工作。

$matchquery = mysql_query("SELECT * FROM `publishers_instagram_accounts` WHERE `pid` != '$publisher_id' AND ((".$influencer_account['followed_by']."+(".$influencer_account['followed_by']."*0.20)) <= followed_by) OR ((".$influencer_account['followed_by']."+(".$influencer_account['followed_by']."*0.20)) > followed_by) ORDER BY `id` DESC");

我认为你在代码中没有正确地进行数学运算。我还建议您使用mysqli函数。看看这段代码，我没有测试它，但我认为它会工作:

<?php
        $mysqli = new mysqli("localhost", "my_user", "my_password", "my_db");
        $query = "SELECT * FROM `publishers_instagram_accounts` WHERE `pid` != :pub_id
            AND (followed_by >= :start ) AND (followed_by <= :end) ORDER BY `id` DESC";
        //You need users with followed_by in the range:
        //      [start,end]
        // where:
        //      start = $influencer_account['followed_by'] - $influencer_account['followed_by'] * 0.2
        //      end = $influencer_account['followed_by'] + $influencer_account['followed_by'] * 0.2
        $percent = 0.2;
        $start = $influencer_account['followed_by'] * (1 - $percent); //20% less
        $end = $influencer_account['followed_by'] * (1 + $percent);//20% more       
        $stmt = $mysqli->prepare($query);
        $stmt->bind_param('idd',$publisher_id,$start,$end);
        $stmt->execute();
        $result = $stmt->get_result();
      while ($row = $result->fetch_array(MYSQLI_ASSOC))
      {
            var_dump($row);
      }
?>

编辑:

如果你想使用mysql_query，试试:

$percent = 0.2;
$start = $influencer_account['followed_by'] * (1 - $percent); //20% less
$end = $influencer_account['followed_by'] * (1 + $percent);//20% more       
$query = "SELECT * FROM `publishers_instagram_accounts` WHERE `pid` !=  $publisher_id AND (followed_by >= $start ) AND (followed_by <= $end ) ORDER BY `id` DESC";
mysql_query($query);

我建议你先计算一下。

$reference_value = $influencer_account['followed_by'] * 1.2;
$matchquery = mysql_query(
    "SELECT * FROM `publishers_instagram_accounts` 
     WHERE `pid` != '$publisher_id' AND
         (($reference_value <= followed_by) OR ($reference_value > followed_by)) 
     ORDER BY `id` DESC");

您使用了!=表示not =，而应该使用<>。还有，我想，这里放错了大括号....

$followedby=$influencer_account['followed_by'];
$matchquery = mysql_query("
    SELECT * FROM `publishers_instagram_accounts` 
    WHERE 
        `pid` <> '$publisher_id' AND (
            ( ".$followedby." + ( ".$followedby." * 0.20 ) ) <= followed_by ) 
            OR
            ( ".$followedby." + (".$followedby." * 0.20 ) ) > followed_by )
        )
    ORDER BY `id` DESC;");

如果可以的话试试

   $matchquery = mysql_query(
        "SELECT * FROM `publishers_instagram_accounts` WHERE `pid` != ".$publisher_id."
    AND (
         ".$influencer_account['followed_by']."+".$influencer_account['followed_by']."*0.20 <= followed_by OR ".$influencer_account['followed_by']."+".$influencer_account['followed_by']."*0.20 > followed_by
        )
    ORDER BY `id` DESC");