PHP - generating JSON output

Question

    $showcaseObject = new stdClass();
    $generalObject = new stdClass();
    $generalObject->roundCorner = 0;
    $generalObject->borderStroke = 2;
    $generalObject->backgroundColor = '#fff';

    $showcaseObject->general = $generalObject;

    echo json_encode($showcaseObject);

and i get something like this

{"general":{
 "roundCorner":"0",
 "borderStroke":"2",
 "backgroundColor":"#ffffff"
 }
}

Now i want to get something like this

{"general":{
 "round-corner":"0",
 "border-stroke":"2",
 "background-color":"#ffffff"
 }
}

i try to change above code like under code, and get syntax error, unexpected '='

 $generalObject->round-corner = $row->general_round_corner_radius;
 $generalObject->border-stroke = $row->general_border_stroke;
 $generalObject->background-color = $row->background_color;

Help me !

Dash / minus is not legal for use in PHP variable (or property, class, function, etc.) names. us2.php.net/manual/en/language.variables.basics.php — cbednarski
– cbednarski, Commented Aug 14, 2010 at 2:01
thank ! i try with $generalObject->{'round-corner'} and it work — Chameron
– Chameron, Commented Aug 14, 2010 at 3:56

Gumbo · Accepted Answer · 2010-08-14 16:50:53Z

1

You could use the brace syntax when defining the object members:

$showcaseObject = new stdClass();
$generalObject = new stdClass();
$generalObject->{'round-corner'} = 0;
$generalObject->{'border-stroke'} = 2;
$generalObject->{'background-color'} = '#fff';

$showcaseObject->general = $generalObject;

echo json_encode($showcaseObject);

This brace syntax allows you to use expressions rather than just identifiers.

answered Aug 14, 2010 at 16:50

Gumbo

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Comments

Delan Azabani · Accepted Answer · 2010-08-14 01:27:19Z

0

To convert first form to second, traverse over each key and convert the keyname.

foreach ($generalObject as $keyName => $keyValue) {
    $newKey = strtolower(preg_replace('/([^A-Z])([A-Z])/', "$1_$2", $keyName));
    $generalObject[$newKey] = $keyValue;
    unset($generalObject[$keyName]);
}

To access them (this is what you're having a problem with), use bracket notation:

$generalObject['hyphenated-name']

answered Aug 14, 2010 at 1:27

Delan Azabani

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1 Comment

Chameron Over a year ago

thank for your anwser. But how i can to use $generalObject[$newKey] = $keyValue.The variable $generalObject is instance of stdClass

Jeffrey Blake · Accepted Answer · 2010-08-14 16:47:31Z

0

I think the easiest way may be to simply do some string replacement on the JSON strings that you have, to get the hyphen characters in where they couldn't be automatically inserted due to php's allow variable charset:

$showcaseObject = new stdClass();
$generalObject = new stdClass();
$generalObject->roundCorner = 0;
$generalObject->borderStroke = 2;
$generalObject->backgroundColor = '#fff';

$showcaseObject->general = $generalObject;

$jsonStr = json_encode($showcaseObject);
$jsonStr = str_replace('"roundCorner":', '"round-corner":', $jsonStr);
$jsonStr = str_replace('"borderStroke":', '"border-stroke":', $jsonStr);
$jsonStr = str_replace('"backgroundColor":', '"background-color":', $jsonStr);

echo $jsonStr;

answered Aug 14, 2010 at 16:47

Jeffrey Blake

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