在php中调用json_decode后，如何连接和打印json对象中的值？

Question

This is a schema generator that displays the key:value pair

 "default": [
                {
                    "one": "u0001u0000u0000u0000",
                    "two": "u0002u0000u0000u0000",
                    "three": "u0003u0000u0000u0000"
                }
            ]

What I would like to print out is "default": [{"u0001u0000u0000u0000u0002u0000u0000u0000u0003u0000u0000u0000"}]

Similarly if it is an object for instance:

"default": [ "a":
                {
                    "one": "u0001u0000u0000u0000",
                    "two": "u0002u0000u0000u0000",
                    "three": "u0003u0000u0000u0000"
                }
            ]

concatenate only values and print like this:

["a": {"u0001u0000u0000u0000u0002u0000u0000u0000u0003u0000u0000u0000"}]

Sample Code Test: // this method gets the value entered by user in json format. The user can put in a nested json format as well. The above examples that I mentioned works. it then calls for scanForNestedType method which scans whether the format contains any array, array>, map etc... Once it scans for nested type, it internally calls $this->encodeValues($unit)which converts the values entered by user from integer to bytes.

Here is an example:

User enters array [{"one": 1, "two": 2, "three": 3}]. After conversion, the result would be as follows: [ { "one": u0001u0000u0000u0000, "two": u00002u0000u0000u0000, "three: u0003u0000u0000u0000 } ]

Now I am getting the values correctly for each key. All I need is to print in this format: [ { u0001u0000u0000u0000u0002u0000u0000u0000u0003u0000u0000u0000 } ]

private function jsonDecode(array $value)
{
    $strValue = $value['value'];
    $jsonValue = json_decode($strValue);

    $this->scanForNestedType($jsonValue);
    return $jsonValue;
}

private function scanForNestedType(&$value)
{
    foreach ($value as $key => &$unit) {
        if (is_array($unit) || is_object($unit)) {
            $this->scanForNestedType($unit);
        } else {
            $value->$key = $this->encodeValues($unit);
        }
    }
}

private function encodeValues(int $value)
{
    $encodedValue = '';

    $bytesArray = unpack("C*", pack("V", $value));
    foreach ($bytesArray as $byte) {
        $encodedValue .= sprintf('u%04x', dechex($byte));
    }
    return $encodedValue;
}

If I get a working example then it would be great!

Answer 1

You can use json_decode to convert json to an array. Then implode sub-array ['one'....] using foreach loop.

snippet

$json1 = '[{"one": "u0001u0000u0000u0000", "two": "u0002u0000u0000u0000", "three": "u0003u0000u0000u0000"}]';
$json2 = '[{ "a": { "one": "u0001u0000u0000u0000", "two": "u0002u0000u0000u0000", "three": "u0003u0000u0000u0000" } }]';
$arr1 = json_decode($json1,true);
$arr2 = json_decode($json2,true);


foreach($arr1 as $data){
    $default1[] = implode($data);
}
$default1 = json_encode($default1, JSON_FORCE_OBJECT);

foreach($arr2 as $key => $child){
    foreach($child as $childKey => $data){
        $default2[$childKey] = implode($data);
    }
}
$default2 = json_encode($default2, JSON_FORCE_OBJECT);

print_r($default1);
print_r($default2);

Output

{"0":"u0001u0000u0000u0000u0002u0000u0000u0000u0003u0000u0000u0000"}
{"a":"u0001u0000u0000u0000u0002u0000u0000u0000u0003u0000u0000u0000"}

Live Demo

You can now json_decode $default1, $default2 to get php object/array.