php function not returning an array

Question

I have the following code, sorry for the length but it saves explaining most of what i need... the $result in the pages() function is not being returned. When I call for it all I get is undefined variable. Why? What i need is to pass the $start and $display variables to the mysql query.

<?php
if(isset($_POST['get_records_submit'])) {
    $pages; $start; $display; $result;
    function pages($stmt) {
        //set how many records per page
        $display = 10;
        //determine how many pages there are
        if(isset($_GET['p']) && is_numeric($_GET['p'])) { //already determined
            $pages = $_GET['p'];
        }
        else {
            $records = mysqli_stmt_num_rows($stmt);
        }
        //calculate the number of pages
        if($records > $display) { //if there are more records than will fit on one page
            $pages = ceil($records/$display); //ceil() rounds up to the nearest integer
        }
        else { //records will fit on one page
            $pages = 1;
        }
        //determine which row to start returning results from
        if(isset($_GET['s']) && is_numeric($_GET['S'])) { //already determined
            $start = $_GET['s'];
        }
        else {
            $start = 0;
        }
        $result = array(0=>$display, 1=>$pages , 2=>$start);
        return $result;
    }
    $searchby = $_POST['search_by'];
    $searchfor = $_POST['search_for'];
    $contact = 1;
    $i = 0;
    //set the initial query
    $query = "SELECT  client_id, title_desc, firstname, surname, house, street, town, city, county, postcode as count FROM address LEFT JOIN client USING (client_id) LEFT JOIN client_title USING (title_id) LEFT JOIN address_street USING (address_id) LEFT JOIN address_town USING (address_id) LEFT JOIN address_city USING (address_id) LEFT JOIN address_county USING (address_id) WHERE is_contact = ?";
    //depending on search terms, amend the query
    if($searchby == 'all') {
        $query .= " ORDER BY surname ASC, firstname ASC";
        $stmt = mysqli_prepare($dbc, $query);
        mysqli_stmt_bind_param($stmt, 'i', $contact);
        mysqli_stmt_execute($stmt);
        mysqli_stmt_store_result($stmt);
        pages($stmt);
        var_dump ($result);
        foreach ($result as $var) { echo $var.' ';}
        mysqli_stmt_close($stmt);
        $query .= " ORDER BY surname ASC, firstname ASC LIMIT ?, ?";
        $stmt = mysqli_prepare($dbc, $query);
        mysqli_stmt_bind_param($stmt, 'iii', $contact, $start, $display);
        mysqli_stmt_execute($stmt);
        mysqli_stmt_bind_result($stmt, $client_id, $stitle, $sfname, $ssname, $shouse,$sstreet, $stown, $scity, $scounty, $spostcode);      

        if($searchfor != '') {
            echo "<p>You searched under <span class=\"bold\">\"All\"</span>, therefore your search term <span class=\"bold\">\"$searchfor\"</span> has not been included.</p>";
        }
    }
}

$result = [$display, $pages, $start]; No need for array() or numbers for keys. — m59
– m59, Commented Sep 19, 2013 at 18:05
@RocketHazmat I always forget that :) But that's only the [], right? The key things is still odd? — m59
– m59, Commented Sep 19, 2013 at 18:08
@m59: Yeah. array($display, $pages, $start) is what you need for < PHP 5.4. Arrays start at 0 by default. — gen_Eric
– gen_Eric, Commented Sep 19, 2013 at 18:08

theftprevention · Accepted Answer · 2013-09-19 18:14:42Z

5

In this line:

pages($stmt);

The function pages() returns the array, but this result is not being set to a variable. Try using:

$result = pages($stmt);

And then accessing the array via the variable $result.

That $result variable that you declared up top? The reason it's not being set is because the function pages() doesn't have access to it because it's not within the function's scope.

Edit

As Lyth pointed out, if you don't want the function to return anything, you can modify your function as follows:

function pages($stmt) {
    global $result;
    // ...
    // The rest of your function
    // ...
    $result = array(0=>$display, 1=>$pages , 2=>$start);
    // Remove the 'return $result' line
}

edited Sep 19, 2013 at 18:14

answered Sep 19, 2013 at 18:06

theftprevention

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4 Comments

Adrien Gorrell Over a year ago

Or maybe he didn't want to make a return at first.

theftprevention Over a year ago

If that's the case, he can add global $result; to the beginning of the pages() function, and then remove the return $result; line.

user2417303 Over a year ago

@theftprevention I'm avoiding using 'thanks' but I'm really glad there are all the stackoverflowers like you out there to stop beginners like me going mentile.

theftprevention Over a year ago

I'll avoid using "you're welcome", but if I can keep PHP from making someone else go mental, I'll gladly do so. :)

Janitra Panji · Accepted Answer · 2013-09-19 18:17:00Z

0

I do not know why you write $pages; $start; $display; $result;? If you want to get the value of array try to use this...

$result = array($display, $pages , $start);
return $result;

You can check the array with

print_r ($result);

answered Sep 19, 2013 at 18:17

Janitra Panji

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