Asked  7 Months ago    Answers:  5   Viewed   33 times

I am trying to pass percent (%) sign in url like


but when I echo, it returns


I want exact same value as I pass it in URL.

I have tried to use urlencode() function, but it give me output like this...


please help me regarding this




To send a % sign in a url, instead send %25.

In your case, in order for php to see a percent sign, you must pass the character string %25B6011000995504101^SB to the server.


In URLs, the percent sign has special meaning. Is used to encode special characters. For example, & is the separator between parameters, so if you want your parameter to actually contain an &, you instead write %26. Because the percent sign is used to encode special characters, it is also a special character, and so if you want to actually send a percent sign, it must also be encoded. The encoding for a percent sign is %25.

Wednesday, March 31, 2021
answered 7 Months ago

Try this:

$str = ' file [that] needs "to" be (encoded).zip';
$pos = strrpos($str, '/') + 1;
$result = substr($str, 0, $pos) . urlencode(substr($str, $pos));

You're looking for the last occurrence of the slash sign. The part before it is ok so just copy that. And urlencode the rest.

Wednesday, March 31, 2021
answered 7 Months ago

You've double encoded the URL. Running urldecode() on your output string is giving me the following:[tt_news]=13144&cHash=9ef2e9ee006fb16188ebf764232a0ba9

EDIT: try the following

Wednesday, March 31, 2021
answered 7 Months ago

Check this:

getProvisionalTotalRoomsCount($currentRoom, $arrayOfRooms){
  foreach($arrayOfRooms as $key=>$value){
     if($value['provisionalBookingRoomID'] == $currentRoom){
            return $value['totalSpecificRoomCount'];
Saturday, May 29, 2021
answered 5 Months ago
new File(path).toURI().toURL();
Wednesday, June 9, 2021
answered 5 Months ago
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