Question

我想检查用户输入的文本是否是有效的JSON。我知道我可以使用这样的东西轻松地做到这一点：

function IsJsonString(str) {
    try {
        JSON.parse(str);
    } catch (e) {
        return false;
    }
    return true;
}

我的问题是来自Mongo的JSON，它包含在ObjectId，ISODate中，即：

{
    "_id" : ObjectId("5733b42c66beadec3cbcb9a4"),
    "date" : ISODate("2016-05-11T22:37:32.341Z"),
    "name" : "KJ"
}

这不是有效的JSON。如何在允许类似上述内容的情况下验证JSON？

Answer 1

你可以用字符串替换裸函数调用，类似这样

function IsJsonLikeString(str) {
  str = str.replace(/(\w+)\("([^"]+)"\)/g, '"$1(\"$2\")"');
  try {
    JSON.parse(str);
  } ...

来自https://regex101.com/r/fW7iH4/#javascript的

解释：

/(\w+)\("([^"]+)"\)/g
    1st Capturing group (\w+)
        \w+ match any word character [a-zA-Z0-9_]
            Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
    \( matches the character ( literally
    " matches the characters " literally
    2nd Capturing group ([^"]+)
        [^"]+ match a single character not present in the list below
            Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
            " a single character in the list " literally (case sensitive)
    " matches the characters " literally
    \) matches the character ) literally
    g modifier: global. All matches (don't return on first match)

Answer 2

您所拥有的问题不是JSON验证之一，它与数据库是否实际上是ACCEPTS输入数据有关。您已经有了正确的想法来检查语法是否正确，但您必须通过mongo集合运行数据并检查是否存在任何错误。

查看MongoDB db.collection.explain()以验证它是否是有效的Mongo查询

从Mongo验证JSON？

2 个答案: