Get JSON to print empty array.

Question

I'm new to php and not sure why the go-daddy server treats JSON data differently. When I host it locally on my computer in local host, when I want to echo empty JSON array, I simply put [] shown as below. But when I uploaded the code to go-daddy server and try it out it echoed an error, Parse error: syntax error, unexpected '[', expecting ')' in...I'm wondering how can I put the JSON so it can echo [] when needed. Otherwise it will give "null" and when it parse into AS3, it turn into a JSON parse error.

if (!empty($output)){
        echo json_encode( $output );}
        else{
            echo json_encode( [] );
        }

Shankar Narayana Damodaran · Accepted Answer · 2014-04-14 05:36:16Z

7

That is because your webserver version must be less than 5.4. You are trying to use a new feature of PHP 5.4 called the short array syntax

Use echo json_encode(array()); instead of echo json_encode( [] );

Working Demo on PHP v 5.3

edited Apr 14, 2014 at 5:36

answered Apr 14, 2014 at 5:19

Shankar Narayana Damodaran

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1 Comment

Benyaman Over a year ago

Thanks for the info, i just went to the server and updated the version, now it works, i didnt know it was set to 5.3 before.

Guns · Accepted Answer · 2014-04-14 05:17:02Z

4

You have to use this:

if (!empty($output)){
        echo json_encode( $output );}
        else{
            echo json_encode(array());
        }

answered Apr 14, 2014 at 5:17

Guns

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Comments

Rick Lancee · Accepted Answer · 2014-04-14 05:18:14Z

3

[] for arrays is only supported in php 5.4+.

So that means php is not a high enough version on your host, try asking them to upgrade to 5.4 or try array(); instead of []

answered Apr 14, 2014 at 5:18

Rick Lancee

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Comments

Ferrakkem Bhuiyan · Accepted Answer · 2014-04-14 05:31:00Z

2

if(count(json_encode($jArrary,1))==0) {
echo "empty";
}
//or
if(empty(json_encode($jArrary,1))) {
echo "empty";
}

you can use this

edited Apr 14, 2014 at 5:31

answered Apr 14, 2014 at 5:23

Ferrakkem Bhuiyan

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