How to display an array in PHP?

Question

This is the code I currently have:

<?php
 echo file_get_contents("http://example.com/bin/serp.php?engine=google&phrase=stackoverflow&format=ARRAY");
?>

It is displaying this on my page:

Array
(
[0] => Array
    (
        [idx] => 0
        [title] => Stack Overflow
        [description] => A language-independent collaboratively edited question and answer site for programmers.
        [url] => http://stackoverflow.com/
    )

[1] => Array
    (
        [idx] => 1
        [title] => Stack Overflow - Wikipedia, the free encyclopedia
        [description] => Stack Overflow website logo.png &middot; Stack Overflow.png. Screenshot of Stack Overflow as of December 2011. Web address &middot; stackoverflow.com. Commercial? Yes.
        [url] => http://en.wikipedia.org/wiki/Stack_Overflow
    )
)

What do I need to change to have it displayed like this instead?

1.  <a href="http://stackoverflow.com/">Stack Overflow</a>
A language-independent collaboratively edited question and answer site for programmers.

2.  <a href="http://en.wikipedia.org/wiki/Stack_Overflow">Stack Overflow - Wikipedia, the free encyclopedia</a>
Stack Overflow website logo.png &middot; Stack Overflow.png. Screenshot of Stack Overflow as of December 2011. Web address &middot; stackoverflow.com. Commercial? Yes.

Any help with this would be highly appreciated.

That sounds like a third-party API. What does their documentation say? — Álvaro González
– Álvaro González, Commented Aug 7, 2014 at 6:55
It seems output of print_r. Is it right or it's your print_r ? — Kevin Labécot
– Kevin Labécot, Commented Aug 7, 2014 at 6:55
$result = file_get_contents(.....); foreach($result as $value) { do something with $value } — Mark Baker
– Mark Baker, Commented Aug 7, 2014 at 6:55
@hd1 I've probably tried 100 different things and nothing seems to work. I'm guessing that I simply don't understand how to pull arrays from pages and actually use their data. — User
– User, Commented Aug 7, 2014 at 7:04
@ÁlvaroG.Vicario This isn't a third party API, it's an older PHP script that I got awhile back and it doesn't look like the original author is active anymore. — User
– User, Commented Aug 7, 2014 at 7:05

Álvaro González · Accepted Answer · 2014-08-07 07:13:26Z

1

The print_r() function is a debug helper. It isn't intended as serialisation format:

There aren't built-in parsers (and it isn't trivial to write a robust one)

It can cause data loss, e.g.:

$data = array(true, 1, false, 0, '');
print_r($data);

Array
(
    [0] => 1
    [1] => 1
    [2] => 
    [3] => 0
    [4] => 
)

You have this in the URL:

format=ARRAY

Just look at whatever place you use $_GET['format']. You'll possibly have more useful formats to choose from. If you don't, it'll be trivial to implement a sensible format, e.g. JSON.

answered Aug 7, 2014 at 7:13

Álvaro González

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2 Comments

User Over a year ago

It looks like there is an option for JSON. Would posting that output make it easier? It doesn't matter to me which I use...

Álvaro González Over a year ago

If you have to ask whether calling json_decode() is easier than writing a custom parser for a bogus pseudo-format, it means that I've failed to properly explain the problem :)

Amit Horakeri · Accepted Answer · 2014-08-07 07:11:26Z

0

You can use the foreach control statement of PHP in order to achieve this, and here it goes:

Say that the $array is the resultant of file_get_contents and supposing that it holds the result to be displayed.

    <?php foreach($array as $arr){?>
     <a href="<?php echo $arr['url'];?>"><?php echo $arr['title'];?></a>
     <?php echo $arr['description']; ?>
    <?php } ?>

Hope this helps you....

edited Aug 7, 2014 at 7:11

answered Aug 7, 2014 at 6:56

Amit Horakeri

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1 Comment

User Over a year ago

I got this error with that code:

[07-Aug-2014 02:01:31 America/Chicago] PHP Parse error:  syntax error, unexpected $end in /home/site/public_html/folder/folder/index.php

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