Convert 1d array of strings into anchor hyperlinks by applying the strings as the href and text of <a> tags

Question

I would like to wrap all elements of an array with something like but I don't want a lot of lines or foreach loop

$links = array('london','new york','paris');

The outcome should be

<a href="#london">london</a>
<a href="#new york">new york</a>
<a href="#paris">paris</a>

So, having ruled out the canonical and sensible approaches, what else did you try? — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Dec 8, 2011 at 13:57
well i am confortable with foreach loop I just want to see if it is possible. — Val
– Val, Commented Dec 8, 2011 at 13:58

Lightness Races in Orbit · Accepted Answer · 2011-12-08 14:07:26Z

34

How about array_map?

$links   = array('london', 'new york', 'paris');
$wrapped = array_map(
   function ($el) {
      return "<a href=\"#{$el}\">{$el}</a>";
   },
   $links
);

Demo (Click source)

Without PHP > 5.3, you can't use a lambda function, so you'll need something like this:

function wrap_those_links($el)  { 
      return "<a href=\"#{$el}\">{$el}</a>"; 
}

$links   = array('london', 'new york', 'paris');
$wrapped = array_map('wrap_those_links', $links);

Demo for PHP 5.2 (Again, click Source)

edited Dec 8, 2011 at 14:07

Lightness Races in Orbit

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answered Dec 8, 2011 at 14:00

nickb

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4 Comments

Val Over a year ago

Parse error: syntax error, unexpected T_FUNCTION, expecting ')' in ... on line 183, gives an error but that's something cool :)

nickb Over a year ago

You must not be running PHP 5.3.0 :) - It's an easy fix, one sec I'll update my answer

Mike Causer Over a year ago

Do you mean "Without PHP >= 5.3"?

Zakaria Acharki Over a year ago

The links are down.

Ed Heal · Accepted Answer · 2014-11-06 04:08:48Z

7

Try join('\n', array_map(function($a) { return "<a href=\"#$a\",>$a<\\a>";}, $links));

edited Nov 6, 2014 at 4:08

answered Dec 8, 2011 at 14:10

Ed Heal

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2 Comments

SEoF Over a year ago

...or because instead of using the named function "implode", he's used the alias "join", and hasn't formatted the answer in an easily readable way, or provided an explanation as to the limitations of the suggestion (PHP 5 >= 5.3). Implode is far more implicit when it comes to self-documentation.

superUntitled Over a year ago

I think that this answer is not quite right, using single instead of double quotes, and a backslash instead of forward slash, it should be: join("\n", array_map(function($a) { return "<a href=\"#$a\",>$a</a>";}, $links));

ya.teck · Accepted Answer · 2023-05-17 15:16:49Z

1

On modern PHP versions it could be shorten as follows.

$wrapper = static fn (string $link): string => \sprintf('<a href="#%1$s">%1$s</a>', $link);
$result = \array_map($wrapper, $links);

answered May 17, 2023 at 15:16

ya.teck

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Comments

Moin Khan · Accepted Answer · 2019-07-09 08:06:54Z

0

Reusable function.

function array_wrap_template($array, $template, $rep="{{el}}")
{
    return array_map(function($el)use($template, $rep){
        return str_replace($rep, $el, $template);
    }, array_values($array));
}

$links = array('london','new york','paris');
$wrapped = array_wrap_template($links, "<a href=\"#{{el}}\">{{el}}</a>");

answered Jul 9, 2019 at 8:06

Moin Khan

12 bronze badges

1 Comment

mickmackusa Over a year ago

This answer is missing its educational explanation.

mickmackusa · Accepted Answer · 2023-05-18 04:45:07Z

0

Another technique not yet mentioned is to use preg_replace() -- which will happily iterate an array of strings -- although, obviously this is not a task that actually demands a regex pattern. (Demo)

$links = array('london','new york','paris');
var_export(
    preg_replace('/.+/', '<a href="#$0">$0</a>', $links)
);

Otherwise, you can go that extra step toward generating encoded urls by calling urlencode() on the city string which is used as the href value.

var_export(
    array_map(
        fn($city) => sprintf('<a href="#%s">%s</a>', urlencode($city), $city),
        $links
    )
);

answered May 18, 2023 at 4:45

mickmackusa♦

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