Filter all other values except a specific URL from array of URLs PHP

Question

I want filter array in PHP.

suppose, i have many URLs in a single array.

eg; $someArray ie;

Array ( 
[0] => javascript:signin('https://login.alibaba.com') 
[2] =>http://www.alibaba.com 
[3] => http://www.alibaba.com/Products?cd=buyhome 
[4] => http://www.alibaba.com/today_new/catalogs/0.html 
[5] => http://www.alibaba.com/help/search-for-products.html 
[6] => http://resources.alibaba.com/trade_safe/home.htm 
[7] => http://us.my.alibaba.com/product/buyoffer/myalibaba/post_buying_lead_no_member.htm [8] => http://www.alibaba.com/sell/sell.htm 
[9] => http://sourcing.alibaba.com/rfq_search_list.htm?availability=y&tracelog=sell_br_20111229 
[10] => http://www.alibaba.com/help/how_to_sell/join_alibaba.html 
[11] => http://us.my.alibaba.com/product/post_product.htm 
[12] => http://resources.alibaba.com/ 
[13] => http://ask.alibaba.com 
[14] => http://resources.alibaba.com/forum/trade_related.htm 
[15] => http://tradeshow.alibaba.com/ 
[16] => http://us.my.alibaba.com/ 
[17] => http://us.my.alibaba.com/mcadmin/inbox/inboxList.htm 
[18] => http://us.my.alibaba.com/product/post_product_interface.htm 
[20] => http://trademanager.alibaba.com/ 
[21] => http://www.alibaba.com/trade/servlet/page/static/paid_memberships/index 
[22] => http://us.favorite.alibaba.com 
[23] => http://www.alibaba.com/trade/help/helpcenter 
[24] => http://www.alibaba.com/help 
[25] => http://resources.alibaba.com/trade_safe/complaint.html 
[26] => http://legal.alibaba.com/legal/site/login/login.htm?site_type=international&language_id=english 
[27] => http://www.alibaba.com/help/contact-us.html#askquestion 
[28] => javascript:showFeedBackWindow() 
[29] => javascript:void(0) 
[30] => http://www.example.com/help
[30] => http://www.example.com/about
);

See the bold values.I only want these values in array... means I want to delete all other values except those values have example.com in value.

please help.

Oleg · Accepted Answer · 2014-11-29 13:25:52Z

1

I hope this will helps

$someArray = array('http://www.alibaba.com', ' http://us.my.alibaba.com/', 'http://www.example.com/help',  'http://www.example.com/di' );

$goodLink = 'example.com';

foreach ($someArray as $key => $link) {
    if(strpos($link, $goodLink) === false) unset($someArray[$key]);
}

Output

Array ( [2] => http://www.example.com/help [3] => http://www.example.com/di )

answered Nov 29, 2014 at 13:25

Oleg

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ekad · Accepted Answer · 2014-11-29 13:38:35Z

0

Use array_filter with a callback function as below

$someArray = array("javascript:signin('https://login.alibaba.com')", "http://www.example.com/help", "http://www.example.com/about", "http://www.alibaba.com");

$new_arr = array_filter($someArray, "filter");

function filter($element) {
    if (strpos($element,'example.com') !== false) {
        return $element;
    }
    else {
        return;
    }

}

print_r($new_arr);

Output

Array
(
    [1] => http://www.example.com/help
    [2] => http://www.example.com/about
)

Working demo: http://codepad.org/UPVhHvlJ

edited Nov 29, 2014 at 13:38

answered Nov 29, 2014 at 13:32

ekad

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MH2K9 · Accepted Answer · 2014-11-29 13:44:12Z

0

Use array_map() & strpos(). Example:

$url_array = [
    'http://www.google.com/',
    'http://www.example.com/help',
    'http://www.example.com/about',
    'http://www.facebook.com/',
];
$result = array_filter(array_map(function($v){
    if(strpos($v, 'example.com') !== false){
        return $v;
    }
}, $url_array));

print '<pre>';
print_r($result);
print '</pre>';

Reference:

edited Nov 29, 2014 at 13:44

answered Nov 29, 2014 at 13:25

MH2K9

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