How to append count to variables in php

Question

I am just starting to learn PHP and to compound the issue I am trying to modify existing code to match what I am attempting to do. I have multiple variables I am trying to define all while partially using a counter. My problem is appending the count to my individually defined variables. Also, I'm sure there is a cleaner way to do this (like a loop or nested array), I just can't see how.

<?php                   
$pageId0 = opt('first_tab_page');
    $post0 = get_post($pageId0);  
    $content0 = apply_filters('the_content', $post0->post_content); 
    $icon0 = wp_get_attachment_image_src( get_post_thumbnail_id( $post0->ID ), 'full' ); 

$pageId1 = opt('second_tab_page');
    $post1 = get_post($pageId1);            
    $content1 = apply_filters('the_content', $post1->post_content); 
    $icon1 = wp_get_attachment_image_src( get_post_thumbnail_id( $post1->ID ), 'full' ); 

$pageId2 = opt('third_tab_page');
    $post2 = get_post($pageId2);        
    $content2 = apply_filters('the_content', $post2->post_content); 
    $icon2 = wp_get_attachment_image_src( get_post_thumbnail_id( $post2->ID ), 'full' );        
?>

<div>
<?php echo $icon0[0]; ?>
</div>

<div>
<?php echo $icon1[0]; ?>
</div>

<div>
<?php echo $icon2[0]; ?>
</div>

<?php 
    $tab_position = opt('tab_position');
    if ($tab_position == '' || count($tab_position) != 3) {
        $tab_position = array(0, 1, 2);
    }
    for($i=0; $i < count($tab_position); $i++)
        {
        $selected = '';
        if (opt('default_selected_tab') == $tab_position[$i]){
                $selected = 'class="selected"';
        }
?>  
<a <?php echo $selected; ?> href="#tab<?php echo $tab_position[$i];?>">
<?php echo $content//should be 0 to start; ?>
</a>
<?php } ?>

Is there anything I can do to clear it up a little? I'm confused on how to use an array with the opt('first_tab_page') pieces — Heather
– Heather, Commented Mar 9, 2016 at 7:23

bansi · Accepted Answer · 2016-03-09 08:02:26Z

1

Here is a short example on using arrays. comment in code

// Storing tab names to an array, just add more tabs if required 
// and details for all those will be loaded in the following arrays
$tabs = array('first_tab_page', 'second_tab_page', 'third_tab_page');

//declare arrays to store (not required, but better practice)
$pageids = array();
$posts = array();
$content = array();
$icons = array();
//Iterate through the tabs array
foreach ($tabs as $tab){
    //Store page id and post in a variable, we require it
    $pageid = opt($tab);
    $post = get_post($pageid);
    // Store pageid in pageids. note the [] this means it will store pageid at next index
    //also store other items
    $pageids[] = $pageid;
    $posts[] = $post;
    $contents[] = apply_filters('the_content', $post->post_content);
    $icons[] = wp_get_attachment_image_src( get_post_thumbnail_id( $post->ID ), 'full' );
}

now $icons[0] will contain first icon $icons[1] the second one and so on. Same thing applies to other variables.

Note: this code is just typed in here and not checked. Please fix if there is any syntax errors. Also note this is one way and there are more ways.

But I would suggest keeping data of each page together.

Edit: Here is how you can keep things together (only relevant parts shown)

$tabs = array('first_tab_page', 'second_tab_page', 'third_tab_page');
$pages = array();
foreach ($tabs as $tab){
    $pageid = opt($tab);
    $post = get_post($pageid);
    $content = apply_filters('the_content', $post->post_content);
    $icon = wp_get_attachment_image_src( get_post_thumbnail_id( $post->ID ), 'full' );
    //Store everything to $pages array
    $pages[] = array('pageid' => $pageid, 'post' => $post, 'content' => $content, 'icon', $icon);
}

foreach ($pages as $page){
?>

<div>
<?php echo $page['icon'][0]; ?>
</div>

<?php } ?>


foreach ($pages as $page){
?>

<a <?php echo $selected; ?> href="#tab<?php echo $tab_position[$i];?>">
<?php echo $page['content']; //echo the content of page ?>
</a>

<?php } ?>

Instead of foreach to display you can use index also like

<a <?php echo $selected; ?> href="#tab<?php echo $tab_position[$i];?>">
<?php echo $pages[0]['content']; //echo the content of the first page ?>
</a>

<a <?php echo $selected; ?> href="#tab<?php echo $tab_position[$i];?>">
<?php echo $pages[1]['content']; //echo the content of the second page ?>
</a>

edited Mar 9, 2016 at 8:02

answered Mar 9, 2016 at 7:43

bansi

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2 Comments

Heather Over a year ago

This seems to be the correct answer but now I have two loops so everything is shown twice. How do I modify the tab_position code to still count through the number of posts and append the proper number in the <a href="tab<?php echo $tab_position[$i];?>"> and applies the selected field but doesn't reprint everything another 2 times. Here's what I have so far: pastebin.com/q20xPGCA

bansi Over a year ago

looked into your code. I didn't understand the tab_position stuff. what does the opt('default_selected_tab') return?

zypro · Accepted Answer · 2016-03-09 07:38:52Z

0

I guess there is no other way then (nested) arrays. Also I'm thinking that arrays are a acceptable solution.

foreach ($tabs as $tab) {
    $tab['content'] = 'somecontent';
    $tab['counter'] = 'counter';
}

or

foreach ($tabs as $tab) {
    $tab = [
      "content" => "content",
      "counter" => "counter"];
}

answered Mar 9, 2016 at 7:38

zypro

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Comments

Halayem Anis · Accepted Answer · 2016-03-09 13:36:42Z

0

I'm not sure what you are trying to do as your question is not very clear to me.
But i think you are trying to echo $content in your last line of code. If yes, just put your all your variables $content1, $content2 and $content3 in an array like below:

$contents = array($content1, $content2, $content3);

Then use foreach loop and traverse this array like this.

$count = 0;
foreach($contents as $content) {
    echo $content.$count;
    $count++;   
}

Try and let me know if it works for you.

edited Mar 9, 2016 at 13:36

Halayem Anis

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answered Mar 9, 2016 at 7:30

Krishna

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2 Comments

Heather Over a year ago

I'm confused how to plug this in, it's already inside some sort of loop (I think) so it just repeats all the $content variables inside each <a>

Krishna Over a year ago

No, the first few lines of your code defines 3 variables $content0, $content1 and $content2 and they are not inside a loop. just put them in $contents array and then where ever you want them to display just lop them using foreach loop above in my code.

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