I use json_decode to vardump a return which yields

array(3) { ["jsonrpc"]=> string(3) "2.0" ["result"]=> array(13) { ["status"]=> string(7) "Pending" ["amount (BTC)"]=> string(10) "0.05130899" ["index_url"]=> . . . . [more fields here] . . . ["address"]=> string(34) "1GHHojoC4Ai1SPMN3YoNzemLNaaj2XFHKc"

I am trying to fill a variable with the text from the "address" field and I tried

$result = json_decode($response, true); $address = $result->address;

but a varbump of $addrwss is NULL

any other variant like $result->"address" or $result->["address"] is improper syntax

Can someone tell me how to get at the text of the address field of a JSON PRC 2.0 response to put it in a variable in PHP

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    Mar 19, 2018 at 12:31

2 Answers 2


To get the address you would do:

$result = json_decode($response, true);
$address = $result["result"]["address"];
  • Thanks. The [0] was throwing a "undefined offset" error - so I removed it and it returns the address field now with => $address = $result["result"]["address"]; <= the var dump shows a => string(34) <= in front of the address but I can hack the rest of the script fixed now. I would not have figured that one out without your assistance. Thanks - and all that come behind me to find this page thank you too. Aug 19, 2017 at 17:06

As you have convert response to array using json_decode so you can access by array element inside result. You can access by echo $result["result"]["address"];

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