PHP Output data from 3 different tables

Question

I'm trying to display a HTML table with information from 3 different tables in my mysql DB. However I am unsure on how to display the information from the third table.

Currently what I am using is:

        $SQL = "SELECT members.*, exp.*, lvl.* 
          FROM members 
          INNER JOIN exp ON members.id = exp.member_id
         INNER JOIN lvl ON members.id = lvl.member_id
        ORDER BY lvl.level DESC, 
      lvl.total DESC, xp.total DESC";
        $result = mysql_query($SQL) or die(mysql_error());
        $count = 1;
        while ($row = mysql_fetch_assoc($result)) { 
        $level = $row['level'];
        $exp = $row['exp.overall'];
        }

the $level is from the second table which grabs correctly, and the $exp is what I want to grab from the third table which is "exp" but it doesn't return anything

How can I change this because at the moment it just seems to be focusing on the data from the "lvl" table when using $row[]

Edit: Both the lvl and exp tables have a row in called 'overall' which is why using $row['overall'] doesn't return what I want as it returns the data from lvl table rather than exp.

share your 3 tables sample data and expected output, without sample its not possible to guide — Zaynul Abadin Tuhin
– Zaynul Abadin Tuhin, Commented Sep 28, 2018 at 17:41

vhoxha · Accepted Answer · 2018-09-28 17:42:44Z

1

First off I believe you have a typo in your last order column: should be exp.total DESC. Secondly, unless you specify the columns to be named with dot notation explicitly they will retain their column names so try changing the last line to: $exp = $row['overall'];.

Also consider using mysqli or PDO.

answered Sep 28, 2018 at 17:42

vhoxha

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2 Comments

Henry D. Over a year ago

$exp = $row['overall']; results in the overall row from the lvl table being returned EDIT: lvl being the second table which also has an 'overall' row. I'm trying to grab the 'overall' row from the exp table

vhoxha Over a year ago

You can simply do this then: $SQL = "SELECT members.*, exp.*, lvl.*, exp.overall AS exp_overall

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