CODEWITHSUNDEEP
phpapicurl
joseph
joseph

Reputation: 80

How can i have Variable in middle of url in php which is pointing to a api?

i want to add a $var[5]which is containing string value to the {subscription_reference} part of the following url.

$url5 = $api_entry_point ."/subscription/{subscription_reference}/credentials";

it needs to be like this..

$url5 = $api_entry_point ."/subscription/0997f4ea6b6ca709/credentials";

I have tried different ways but all gone wrong.here are those

$url5 = $api_entry_point ."/subscription/'.$var[5].'/credentials";

$url5 = $api_entry_point ."/subscription/$var[5]/credentials";

Upvotes: 0

Views: 349

Answers (3)

RichardAtHome
RichardAtHome

Reputation: 4313

You were pretty close, but got tangled up with your quotes ;-)

$url5 = $api_entry_point ."/subscription/'.$var[5].'/credentials";

Should be...

$url5 = $api_entry_point ."/subscription/" . $var[5] . "/credentials";

or you could have used:

$url5 = $api_entry_point ."/subscription/{$var[5]}/credentials";

Upvotes: 2

DocRattie
DocRattie

Reputation: 1423

The long way:

$url5 = $api_entry_point ."/subscription/".$var[5]."/credentials";

or

$url5 = $api_entry_point .'/subscription/'.$var[5].'/credentials';

They both work the same. But you need to use either "or '. Don't mix them.

The fanzy way: 

$url5 = $api_entry_point ."/subscription/{$var[5]}/credentials";

This onyl works witht he ". They check if anything in the string is a variable. So a normal "asd $var asd"would understand $var as a variable. But this won't work with special signs like [] or () and so on. Thus you have to put the variable in the {} for the string to know which parts a pieces of variables.

Upvotes: 2

Thlbaut
Thlbaut

Reputation: 649

Try this :

$url5 = $api_entry_point."/subscription/".$var[5]."/credentials";

Upvotes: 5

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