Returning values from functions

[chomedey]

Hi all,

 

I am calling a function from a script, and wanting that function to return a variable back to the script so I can then use it in another function called by the script.  Although I can echo the variable at the end of the function, when I try and return it to the original script, using "return", I get an error message - unrecognized variable.  Any ideas?  I've been trying different things for hours.

 

Below is  the code (I'm using BBEdit but I don't know how to retain the styled text in this text box):

 

This is the original script:

 

if ((isset($_POST['submitted']) )  &&  (!empty($_POST['category']) )) {

 

// Handle the form

 

$category = ($_POST['category']);

$limit_number = 0;

 

$dbc = mysql_connect('localhost', 'root', 'One4Two');

mysql_select_db('mds');

 

$query = "SELECT * FROM deepshares WHERE category = '$category'

ORDER BY date_entered

DESC LIMIT $limit_number,11";

 

display_results($query);

 

display_prev_next('categories', $limit_number, $num_results, $category);

 

}

 

And this is the function:

 

function display_results($query) {

 

if ($r = mysql_query($query))

 

{

while ($row = mysql_fetch_array($r))

{

$entry = $row['entry'];

$id = $row['entryID'];

$entry_length = str_word_count($entry);

$num_results = mysql_num_rows($r);

 

if ($entry_length>=20)

{

echo "<p><h3>Title: {$row['title']}</h3>

Author: {$row['username']}<br />

Date submitted: {$row['date_entered']}<br />

Views: {$row['views']}<br />

Category: {$row['category']}<br />";

echo substr($entry, 0,100);

echo "... <a href='view_entry.php?id=".($id)."'>(more)</a><br />";

 

}

else

{

echo "<p><h3>Title: {$row['title']}</h3>

Author: {$row['username']}<br />

Date submitted: {$row['date_entered']}<br />

Views: {$row['views']}<br />

Category: {$row['category']}<br />";

echo $entry;

}

}

}

else

{

print '<p>Could not retrieve the data because:<br />' . mysql_error(). '.</p>

<p>The query being run was: ' . $query . '</p>';

}

return $num_results;

}

 

And here is the function I want to call once the variable $num_results has been returned:

 

function display_prev_next($link_page, $limit_number, $num_results, $category)

{

if (($limit_number >= 10)  &&  ($num_results > 10))

{

echo "<p><a href='".($link_page).".php?ln=".($limit_number - 10)."&category=".($category)."'>Prev</a>";

echo "  ";

echo "<a href='".($link_page).".php?ln=".($limit_number + 10)."&category=".($category)."'> Next</a></p>";

}

 

else if (($limit_number >= 10)  && ($num_results <=10))

 

{

echo "<p><a href='".($link_page).".php?ln=".($limit_number - 10)."&category=".($category)."'>Prev</a>";

echo "  ";

}

 

else if (($limit_number <= 10) && ($num_results <=10))

 

{

echo "";

}

 

else

{

echo "<p><a href='".($link_page).".php?ln=".($limit_number + 10)."&category=".($category)."'>Next</a></p>";

}

}

 

[Fergal Andrews]

Hi chomedy,

 

You need to assign a variable to the function call in your script. At the moment there is nothing in your script to catch the returned value. Try changing

 

 

display_results($query);

 

to

 

$num_results = display_results($query);

 

Alternatively you can make $num_results global, but unless you intend using that value somewhere else, I'd just do it as above.

 

I also noticed that the conditional statement at the start of display_results is incorrect. At moment it is setting the value of $r and not checking it.

 

if ($r = mysql_query($query))

 

{

 

should be

 

 

if ($r == mysql_query($query))

 

{

 

I haven't had time to check your script, so I hope that works.

 

Fergal

[wildteen88]

return does not return the variable itselft. It only return the variables value. You'll need to capture the return value with a variable. For example

$results = display_results($query);

 

For more information go to http://php.net/manual/en/functions.returning-values.php

[chomedey]

Once again awesome help from php freaks.

 

Thanks so much.

 

Julian