php mysql commands not returning anything

Question

I am trying to setup an application availability report, I have an array of with app names in it and a number. I am going through the list and finding the duration of issues for each one and saving it a corresponding array and then doing a calculation on it. I try to print it out to a table but no dice. I have my test echos and for loops still in there and I get a the app array printed out but inside the while look doesn't execute and I can't pinpoint why. Thanks for the help and I can answer questions about it. I am going to put a month/year form input at the bottom with a submit button that calls this same page back.

<?php
                $applist = array('default1','default2','default3');
                $numofapps = 3;
                $i=0;
                $downtime;
                if($_POST['submitted']==1)
                {
                    $month = $_POST['startmonth'];
                }   
                else {
                    $month = date("m");
                }
                echo "<br />";

                switch ($month){
                    case 01:
                    case 03:
                    case 05:
                    case 07:
                    case 08:
                    case 10:
                    case 12:
                        $totaltime = 31 * 1440;
                        break;
                    case 04:
                    case 06:
                    case 09:
                    case 11:
                        $totaltime = 30 * 1440.0;
                        break;
                    case 02:
                        $year = $_POST['startyear'];
                        if(($year%4)==0)
                        {
                            $totaltime = 29 * 1440.0;
                        }
                        else {
                            $totaltime = 28 * 1440.0;
                        }
                        break;

                }
                echo "<br />";
                while($numofapps > $i)
                {
                    $query="Select `duration` from `issuetrack` WHERE `app`='" . $applist[$i] . "'";
                    $result=mysql_query($query);

                    while($row = mysql_fetch_array($result))
                    {
                        echo $downtime[$i];
                        echo "TEST downtime <br />";
                        echo $row[$i];
                        echo "ETST DURATION <br />";
                        $downtime[$i]+=$row['duration'];
                    }
                    $appavail[$i]=($downtime[$i] / $totaltime) * 100;

                    $i++;                   

                    echo "<br />";
                }
                $j=0;
                echo "<table border ='1'>";
                        echo "<tr><th>Application Affected</th><th>Application Availability</th>";

                        while ($numofapps > $j)
                        {
                            echo "<tr><td>";
                            echo $applist[$i];
                            echo "</td><td>";
                            echo $appavail[$i];
                            echo "</td></tr>";
                        }
                        echo "</table>";



            ?>

What have you done to bug check? What is the output supposed to look like? Can you remove code that's not related to the problem area? More info, please. — rockerest
– rockerest, Commented Nov 22, 2011 at 5:52
Ok I took at the test code. My while($row = fetch) code isn't working and I know I have data to be returned as I have ran the sql command in the phpmyadmin console — Intelwalk
– Intelwalk, Commented Nov 22, 2011 at 6:01
What does echo mysql_error(); print out after your $result = mysql_query($query)? — Dirk
– Dirk, Commented Nov 22, 2011 at 6:04
Did you try $count = mysql_num_rows($query); and then echo $count; ? — David Bélanger
– David Bélanger, Commented Nov 22, 2011 at 6:45

vkGunasekaran · Accepted Answer · 2011-11-22 12:38:36Z

4

Check the $result variable like,

if($result)
 echo "1";
else
 echo mysql_error();

Even if you don't have any data, if your query is right then it prints 1, otherwise it prints error if any.

Also check your mysql_connection to find out you've a connection to db, like this:

mysql_connect('localhost','username','password') or die (mysql_error());

edited Nov 22, 2011 at 12:38

answered Nov 22, 2011 at 7:40

vkGunasekaran

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Comments

John Watson · Accepted Answer · 2011-11-22 07:48:31Z

1

If your script isn't entering while($row = mysql_fetch_array($result)) then it's because the query didn't return any results. Try printing out your query (echo $query) and run it directly in mysql to see what it gives back.

The variables in the while ($numofapps > $j) loop are indexed by $i but they should be indexed by $j. Also, you need a $j++ there or it will loop until the script execution time limit is reached.

answered Nov 22, 2011 at 7:48

John Watson

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Comments

Manigandan Arjunan · Accepted Answer · 2012-01-19 06:56:29Z

0

as a first step check the mysql connection

mysql_connect('localhost','username','password') or die (mysql_error());

if there is connection then check the query is returning something by using the following query

 $count=mysql_num_rows($result);
       if($count!=0)
       {
           while($row = mysql_fetch_array($result))
          {
               echo $downtime[$i];
               echo "TEST downtime <br />";
               echo $row[$i];
               echo "ETST DURATION <br />";
               $downtime[$i]+=$row['duration'];
          }
      }
     else
     {
         echo "The query returned empty";
     }

answered Jan 19, 2012 at 6:56

Manigandan Arjunan

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