Nested While Loops php

Question

I've been trying to get a series of nested loops working to select the database name from one table then query the selected table in that database, and add up the results and display the number of them and the database name.

I have gotten the code to work but it keeps displaying:

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in

I have tried every way possible I have found online to help, but none work.

$resulta = mysql_query("SELECT dbname AF012 FROM Customer");

while($data = mysql_fetch_array($resulta))
  {
    $db = $data[' dbname '];

$result = null;
$result2 = mysql_query("SELECT changemade FROM $db.orders");
//looping through the results
while($row = mysql_fetch_array($result2))
  {
    //checking if any record is 1,2 or 3
  if( ($row[‘changemade’]== 1) || ($row[‘changemade’]== 2) || ($row[‘changemade’]== 3) )          {
      //if any match the if adding 1 to the counter
      $counter ++;
    }
     }
 unset($result2);
echo $db."  ".$counter;
 echo "<br>";
 $counter = 0;
 $result = null;
 $result2 = null;
}

All the database connections are made and work fine, so it has nothing to do with that. Any help would be great.

You have a bunch of curly apostrophes instead of real single quotes $row[‘changemade’]== 1) should be $row['changemade']== 1) — Michael Berkowski
– Michael Berkowski, Commented Apr 11, 2012 at 13:25

hakre · Accepted Answer · 2012-04-11 13:47:39Z

You need to introduce error handling, also you can streamline your code. The current point of failure is querying the database and fetching from it, so you can encapsulate it into a function of it's own which will also reduce your code:

function mysql_query_array($query)
{
    if (!$result = mysql_query($query)) {
        throw new Exception(sprintf('Invalid query "%s", error: %s.', $query, mysql_error()));
    }
    return function () use ($result) {
        return mysql_fetch_array($result);
    };
}

With that little helper function, your code is now more compact and it has actual error handling:

$queryA = mysql_query_array("SELECT dbname AF012 FROM Customer");
while ($data = $queryA())
{
    $counter = 0;
    $queryB = mysql_query_array("SELECT changemade FROM {$data['dbname']}.orders");
    while ($row = $queryB())
    {
        if (in_array($row['changemade'], range(1, 3))) {
            $counter++;
        }
    }
    printf("%s  %s<br>\n", $db, $counter);
}

Nicola Peluchetti · Accepted Answer · 2012-04-11 13:25:57Z

1

This is caused by the fact that mysql_query return false if the query fails, so one of your query fails and mysql_fetch_array() gets a boolean. Try changing $db = $data[' dbname ']; to $db = $data['dbname'];

answered Apr 11, 2012 at 13:25

Nicola Peluchetti

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