根据Crazy Dino的评论,这是解决我的问题的正确方法:
String json = str1; // sample json string
Pattern codePattern = Pattern.compile("\"code\"\\s*:\\s*\"([^,]*)\",");
Pattern messagePattern = Pattern.compile("\"message\"\\s*:\\s*\"([^,]*)\",");
Pattern statusPattern = Pattern.compile("\"status\"\\s*:\\s*\"(FAILURE)\"");
Matcher code_matcher = codePattern.matcher(json);
Matcher message_matcher = messagePattern.matcher(json);
Matcher status_matcher = statusPattern.matcher(json);
if (code_matcher.find() && message_matcher.find() && status_matcher.find()) {
System.out.println("\nException found!");
System.out.println("\n" + code_matcher.group(1));
System.out.println("\n" + message_matcher.group(1));
System.out.println("\n" + status_matcher.group(1));
}
并输出:
Exception found!
INVALID_PARAMETER_VALUE
Invalid value for parameter: language_code
FAILURE
本文收集自互联网,转载请注明来源。
如有侵权,请联系[email protected] 删除。
我来说两句