php - how can i get two id from 1st select query and then use this to 2nd select query? -

i'm using below code

<?php  $site_id = '7af099d94576f8c4';  // check monitor id site id  $sql = "select * status site_id='$site_id'";  $result = $conn->query($sql);  while ($row = $result->fetch_assoc()) {      $st_id = $row['st_id'];        $mon_id = $row['mon_id'];        $mon_site_id = $row['mon_site_id'];      echo $mon_id;        $result = mysql_query("select * monitors mon_id='$mon_id'")              or die(mysql_error());    // keeps getting next row until there no more      while ($row = mysql_fetch_array($result)) {          // print out contents of each row          echo $row['name'] . "<br />";      }  }  ?>

and i'm getting 2 mon_id 1 , 2 next want use these 2 mon_id in 2nd select query 2nd query give me 1 result mean give result one name. how can 2nd name ?

you overwriting $result variable in second query, losing data first query. change names second query variables , you're done.

<?php $site_id = '7af099d94576f8c4'; // check monitor id site id $sql = "select * status site_id='$site_id'"; $result = $conn->query($sql); while($row = $result->fetch_assoc()) {     $st_id = $row['st_id'];     $mon_id = $row['mon_id'];     $mon_site_id = $row['mon_site_id'];     echo $mon_id;      /// change here     $result2 = mysql_query("select * monitors mon_id='$mon_id'") or die(mysql_error());      /// change here       // keeps getting next row until there no more      /// change here     while($row2 = mysql_fetch_array( $result2 )) {     /// change here         // print out contents of each row     /// change here         echo $row2['name']."<br />";     /// change here     } } ?>