java使用正则表达式替换json值

Question

String inputJson = "{\r\n" + 
            "  \"fullName\" : \"Hamo\",\r\n" + 
            "  \"staff\" : false,\r\n" + 
            "  \"supr\" : true,\r\n" + 
            "  \"permissions\" : [ \"Perm1\", \"Perm2\" ],\r\n" + 
            "  \"services\" : [ \"Serv1\", \"Serv2\" ],\r\n" + 
            "  \"authToken\" : \"1234567890abcdefghaijklmnopqrstuvwxyz\",\r\n" + 
            "  \"customerId\" : 12345,\r\n" + 
            "  \"clients\" : [ 1, 3, 8 ],\r\n" + 
            "  \"wts\" : false,\r\n" + 
            //yyyy-MM-dd
            "  \"testDate\" : \"1982-09-21\"\r\n" + 
            "}";

我需要用另一个使用正则表达式的字符串替换authToken值。 此问题已被编辑，因为未正确提问

Answer 1

没有JSON类，仍然存在：

inputJson = inputJson.replaceFirst("(\"authToken\"\\s*:\\s*\")[^\"]*\"",
    "$1" + authToken + "\"");

这假设authToken不包含美元$

Answer 2

除了使用方便的json编辑库，还可以使用regexp：

String inputJson = "{\r\n" +
        "  \"fullName\" : \"Hamo\",\r\n" +
        "  \"staff\" : false,\r\n" +
        "  \"supr\" : true,\r\n" +
        "  \"permissions\" : [ \"Perm1\", \"Perm2\" ],\r\n" +
        "  \"services\" : [ \"Serv1\", \"Serv2\" ],\r\n" +
        "  \"authToken\" : \"1234567890abcdefghaijklmnopqrstuvwxyz\",\r\n" +
        "  \"customerId\" : 12345,\r\n" +
        "  \"clients\" : [ 1, 3, 8 ],\r\n" +
        "  \"wts\" : false,\r\n" +
        //yyyy-MM-dd
        "  \"testDate\" : \"1982-09-21\"\r\n" +
        "}";

String desiredAuthToken = "anyAuthTokenYouWant";
inputJson.replaceAll("(\"authToken\"\\s*:\\s*\")\\w*", "$1" + desiredAuthToken);

解释如下：

"(\"authToken\"\\s*:\\s*\")\\w*";
 (                              – start of a first capturing group
  \"authToken\"                 – matches "authToken" string
               \\s*             – 0 to N whitespaces (including normal spaces, tabulations, etc.)
                   :            – matches a single colon
                    \\s*        – 0 to N whitespaces again
                        )       – end of the capturing group
                         \\w*   – matches 0 to N alphanumeric characters. You may want to replace it with a different regexp, that suits you needs in a better way.

最后，将整个匹配表达式替换为"$1" + desiredAuthToken，其中$1基本上是第一个捕获组的内容。（在这种情况下是\"authToken\" : \"）