php - 在 php 中不使用 json_decode 的 JSON pretty-print

Question

例如，我有这样的 json 字符串(从一开始)。而且它没有格式化。

{"data":[{"id":"14","memo_kondisi":"Kekurangan pekerjaan","total_row":"5","nilai_temuan":"1.000.000","data_sebab":[{"id":"15","id_sebab":"","id_sub_sebab":"","memo_sebab":"coba","data_rekomendasi":[{"id":"25","id_rekomendasi":"10","id_sub_rekomendasi":"","id_s_sub_rekomendasi":"","nilai_rekomendasi":"0"},{"id":"26","id_rekomendasi":"10","id_sub_rekomendasi":"","id_s_sub_rekomendasi":"","nilai_rekomendasi":"0"},{"id":"31","id_rekomendasi":"10","id_sub_rekomendasi":"","id_s_sub_rekomendasi":"","nilai_rekomendasi":"0"}]},{"id":"16","id_sebab":"","id_sub_sebab":"","memo_sebab":"coba","data_rekomendasi":[{"id":"34","id_rekomendasi":"10","id_sub_rekomendasi":"","id_s_sub_rekomendasi":"","nilai_rekomendasi":"0"},{"id":"35","id_rekomendasi":"10","id_sub_rekomendasi":"","id_s_sub_rekomendasi":"","nilai_rekomendasi":"0"}]}]},{"id":"15","memo_kondisi":"Kekurangan pekerjaan","total_row":"2","nilai_temuan":"1.000.000","data_sebab":[{"id":"5","id_sebab":"","id_sub_sebab":"","memo_sebab":"coba","data_rekomendasi":[]},{"id":"10","id_sebab":"","id_sub_sebab":"","memo_sebab":"coba","data_rekomendasi":[]}]},{"id":"16","memo_kondisi":"","total_row":"2","nilai_temuan":"0","data_sebab":[{"id":"9","id_sebab":"","id_sub_sebab":"","memo_sebab":"coba","data_rekomendasi":[]},{"id":"12","id_sebab":"","id_sub_sebab":"","memo_sebab":"coba","data_rekomendasi":[]}]}]}

我看到一些类似的问题，您必须使用 json_decode 并且我必须再次编码并使用 json_encode($json,JSON_PRETTY_PRINT)

有没有一种方法可以使 json 可读，而无需先解码 JSON，然后在 PHP 中再次编码？

注意:我希望结果仍然是 JSON

Answer 1

不是真的。使用其他人的解析器库不会有任何区别，因为他们也会调用 json_decode() 。

您可以创建一个可以调用的小函数:

function prettify($json)
{
    $array = json_decode($json, true);
    $json = json_encode($array, JSON_PRETTY_PRINT);
    return $json;
}

然后 echo prettify($jsonString); 会比不断解码和重新编码更容易。看到这里https://3v4l.org/CcJlf