php variable converting to JSON object

Question

I want to make an array of json objects having the structure as follows:

{
    "client-mac": "0C:D2:B5:68:73:24",
    "client-dbm": "-82",
    "clientManuf": "unknown"
}

I am using following php code to get this result:

$clientMac = $clients->{'client-mac'};
$clientStrength = $clients->{'snr-info'}->{'last_signal_dbm'};
$clientManuf = $clients->{'client-manuf'};
$jsonObject = array('client-mac' => $clientMac,
                    'client-dbm' => $clientStrength,
                    'clientManuf' => $clientManuf);
$jsonString = json_encode($jsonObject);

The problem is I am getting following json string:

{"client-mac":{
                 "0":"0C:D2:B5:68:73:24"
              },
 "client-dbm":{
                 "0":"-82"
              },
 "clientManuf":{"0":"Unknown"}
}

Why I am getting those extra keys as "0"? And how can I get my desired output? Thank you in advance :)

trincot · Accepted Answer · 2017-07-22 14:32:18Z

2

Apparently your source data has one more nested level with one key/value pair.

You could use reset on them to just pick the first value from that:

array('client-mac' => reset($clientMac),
      'client-dbm' => reset($clientStrength),
      'clientManuf' => reset($clientManuf));

answered Jul 22, 2017 at 14:32

trincot

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Comments

Rajdeep Paul · Accepted Answer · 2017-07-22 14:31:49Z

0

That's because $clientMac, $clientStrength and $clientManuf are objects, not literal strings. You have to change the first three lines in the following way,

$clientMac = $clients->{'client-mac'}->{'0'};
$clientStrength = $clients->{'snr-info'}->{'last_signal_dbm'}->{'0'};
$clientManuf = $clients->{'client-manuf'}->{'0'};
...

answered Jul 22, 2017 at 14:31

Rajdeep Paul

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