need help with array

[elentz]

I have put together a method to reboot my SIP phones.  I can use an array to feed the foreach and it works fine.  Now I need to take a step or two further.  I want to use the result of a query in the array.  The next step would be to instead of having all the extensions in the array  , to be able to select extensions from a form.  But for now getting the initial array is what I am needing help with.

//*****This doesn't work *******
//$phones = "SELECT extension FROM `extensions`IN(".implode(',',$array).")";

//******This works********
$phones  = array(201,202,204);

foreach ($phones as $phone)
{
$out = shell_exec("asterisk -rx 'sip notify CQ_Phone-Reboot $phone'");
echo str_replace("\n", "<br/>", $out);
}

Thanks

[taquitosensei]

$phones = "SELECT extension FROM `extensions` where extension IN(".implode(',',$array).")"; 

or whatever your field is that matches the values in the array

Edited February 28, 2019 by cyberRobot
Removed quote, which included the entire post above.

[Barand]

To use the results of a query you first have to execute the query (not just define a string with incorrect query syntax)

[elentz]

Thanks for the reply.

Using your suggestion I am getting this:

Undefined variable: array in /var/www/html/cqadmin/Assign/restart.php on line 23 Warning: implode(): Invalid arguments passed in /var/www/html/cqadmin/Assign/restart.php on line 23

Line 23 is the new code.

 

Barand

Point taken

 

 

[cyberRobot]

What does your code look like now? Is $array defined...before the query?

[elentz]

Here's all I have at the moment.  The table does populate with the correct info

$sql = "select extension from extensions";
$result = mysqli_query($link,$sql) or die(mysql_error());
echo "<table border='3'>
    <tr>
        <th>Extension #</th>
    </tr>";

while($row = mysqli_fetch_array($result))
{
    echo "<tr>";
    echo "<td>" . $row['extension'] . "</td>";
             
    }
       
echo "</table>";

//$phones = "SELECT extension FROM extensions where extension IN(".implode(',',$array).")";
//$phones  = array(201,202,204);

foreach ($result as $phone)
{
$out = shell_exec("asterisk -rx 'sip notify CQ_Phone-Reboot $phone'");
echo str_replace("\n", "<br/>", $out);
}
?>

Thanks

[cyberRobot]

3 minutes ago, elentz said:

echo "<tr>";
echo "<td>" . $row['extension'] . "</td>";

In case this is more than just test code, the above code is missing the closing tag for the table row.

As for the error, $array isn't defined in your code. Assuming that you are looking for extensions 201, 201, and 204, try something like this

$extensions = array(201,202,204);
$sql = "SELECT extension FROM extensions where extension IN(".implode(',',$extensions).")";

Note that you'll need to run the query, like you did for the extensions HTML table. The foreach loop that runs through the phone numbers can be replaced with a while loop, like you have for the HTML table.

[cyberRobot]

1 hour ago, elentz said:

...able to select extensions from a form...

I should mention that you'll want to use caution when you get to the step where you are populating the $extensions array using form input. Information supplied by a user should never be trusted. Forms, for example, can be tampered with, even ones using the POST method.

[elentz]

This did the trick!

$sql = "select extension from extensions";
$result = mysqli_query($link,$sql) or die(mysql_error());

while($row = mysqli_fetch_array($result))
{
    $out = shell_exec("asterisk -rx 'sip notify CQ_Phone-Reboot " . $row['extension'] . "'");
    //echo str_replace("\n", "<br/>", $out);         
    }

Thanks so much for the help !