Question

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

    $showcaseObject->general = $generalObject;

    echo json_encode($showcaseObject);

我得到类似的东西

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

现在我想得到这样的东西

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

我尝试在代码下更改上面的代码，并获得语法错误，意外的'='

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

帮助我！

Answer 1

定义对象成员时可以使用大括号语法：

$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);

此大括号语法允许您使用表达式而不仅仅是标识符。

Answer 2

要将第一个表单转换为第二个表单，请遍历每个键并转换键名。

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

要访问它们（这是您遇到的问题），请使用括号表示法：

$generalObject['hyphenated-name']

Answer 3

我认为最简单的方法可能是简单地在你拥有的JSON字符串上进行一些字符串替换，以便在由于php允许变量字符集而无法自动插入的情况下获取连字符：

$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;

PHP - 生成JSON输出

3 个答案: