PHP Notice: Array to string conversion when echoing json_decode()'ed data

Question

I have this code in a.json:

{
  "body" : {
    "outboundSMSMessageRequest": {
      "address": [
        "9456654978" /* 1 */
      ],
      "senderAddress": " 64735 ", /* 3 */
      "outboundSMSTextMessage": {
        "message": "Welcome to fgf  Your Confirmation Code - " /* 2 */
      },
      "clientCorrelator": "1", /* 4 */
      "receiptRequest": {
        "notifyURL": "2", /* 5 */
        "callbackData": "3" /* 6 */
      },
      "senderName": "4" /* 7 */
    }
  }
}

I want to print the values as marked above (1, 3, 2, 4, 5, 6, 7, the values only). I have tried this:

$jsonData = file_get_contents("a.json");

$json = json_decode($jsonData,true);

echo $json;

but I end up with the following notice:

PHP Notice: Array to string conversion

How can I print the values as desired?

`echo` expects a string, `$json` is an array at that point. Try `print_r($json);` instead. — rickdenhaan, Aug 11 '17 at 12:20
Possible duplicate of [How do I extract data from JSON with PHP?](https://stackoverflow.com/questions/29308898/how-do-i-extract-data-from-json-with-php) — localheinz, Aug 11 '17 at 13:10

score 0 · Accepted Answer · answered Aug 11 '17 at 13:04

Presumably by:

I want to in a PHP page to print (1, 3, 2, 4, 5, 6, 7, the values only)

... you mean those values you've marked with the numbered comments, e.g. /* 1 */ - if so, this should do the trick (to just echo it out anyway).

$jsonData = file_get_contents("a.json");
$o = json_decode($jsonData);

$outboundSMSRequest = $o->body->outboundSMSMessageRequest;
echo $outboundSMSRequest->address[0] . "\n"                         // outbound address
    . $outboundSMSRequest->outboundSMSTextMessage->message . "\n"   // outbound message
    . $outboundSMSRequest->senderAddress . "\n"                     // sender address
    . $outboundSMSRequest->clientCorrelator . "\n"                  // client correlator
    . $outboundSMSRequest->receiptRequest->notifyURL . "\n"         // notification URL
    . $outboundSMSRequest->receiptRequest->callbackData . "\n"      // callback data
    . $outboundSMSRequest->senderName;                              // sender name

When you use json_decode() you turn the JavaScript object into a PHP object - so you just access it accordingly.

score -1 · Answer 2 · answered Aug 11 '17 at 12:21

-1

Json do not support comments so you need to remove them from your file, see Can comments be used in JSON?

answered Aug 11 '17 at 12:21

rypskar

2,012
13
13

1

Not related to the problem. If the JSON could not be decoded, PHP would not have thrown an "Array to string conversion" notice. – rickdenhaan Aug 11 '17 at 12:23
i want in php page to print or display (1,3,2,4,5,6,7) (values only) i tried many ways to fix it . – Amer Abo Tbk Aug 11 '17 at 12:25
@rickdenhaan that is true, but you had already added a comment that should solve that part :) – rypskar Aug 11 '17 at 12:25
@AmeerAboTbk then you need to pick the values you want from the object you get back from json_decode and echo them. Using print_r as other have suggested to see the data you get should point you in the right direction – rypskar Aug 11 '17 at 12:29

PHP Notice: Array to string conversion when echoing json_decode()'ed data

2 Answers2