Question

我需要以最有效的方式将其解析为字符串。我目前有这条线

 PowerMockito.mockStatic(Color.class);
        PowerMockito.when(Color.rgb(Mockito.anyInt(), Mockito.anyInt(), Mockito.anyInt())).thenAnswer(new Answer<Object>() {
            @Override
            public Object answer(InvocationOnMock invocation) throws Throwable {
                int red = (int) invocation.getArguments()[0];
                int green = (int) invocation.getArguments()[1];
                int blue = (int) invocation.getArguments()[2];
                return (0xFF << 24) | (red << 16) | (green << 8) | blue;
            }
        });
        PowerMockito.when(Color.alpha(Mockito.anyInt())).thenAnswer(new Answer<Object>() {
            @Override
            public Object answer(InvocationOnMock invocation) throws Throwable {
                return ((int)invocation.getArguments()[0])>>>24;
            }
        });
        PowerMockito.when(Color.red(Mockito.anyInt())).thenAnswer(new Answer<Object>() {
            @Override
            public Object answer(InvocationOnMock invocation) throws Throwable {
                return (((int)invocation.getArguments()[0])>>16) & 0xFF;
            }
        });
        PowerMockito.when(Color.green(Mockito.anyInt())).thenAnswer(new Answer<Object>() {
            @Override
            public Object answer(InvocationOnMock invocation) throws Throwable {
                return (((int)invocation.getArguments()[0])>>8) & 0XFF;
            }
        });
        PowerMockito.when(Color.blue(Mockito.anyInt())).thenAnswer(new Answer<Object>() {
            @Override
            public Object answer(InvocationOnMock invocation) throws Throwable {
                return (int)invocation.getArguments()[0] & 0xFF;
            }
        });

需要分别解析为四个字符串，

D[date-string] T[time-string] N[name-string] M[message-string]

有一种简单的方法吗？

Answer 1

您可以使用正则表达式和组来匹配和提取这类数据：例如，像：

    Pattern p = Pattern.compile("D(\\w+) T(\\w+) N(\\w+) M(\\w+)");
    Matcher m = p.matcher(yourString);
    if (m.matches()){
        String d = m.group(1);
        String t = m.group(2);
        String n = m.group(3);
        String w = m.group(4);
    }

括号内的模式被保存到组中，您可以在匹配后提取它（从1开始，因为0是整个匹配）。然后，您必须将其更改为您要接受的字符和模式。

Answer 2

我不确定我是否理解你，但看起来使用正则表达式是进行这种解析的最简单方法。

Answer 3

以下逻辑可行。如何regix将不确定。但是下面的代码也可以，因为它不是循环的，所以我认为它可能是最佳的解决方案。

String s = "D[date-string] T[time-string] N[name-string] M[message-string]";   
String[] str = s.split(" ");    
 String date = str[0].substring(str[0].indexOf("[")+1, str[0].lastIndexOf("]")); //would equal time-string
 String time=str[1].substring(str[1].indexOf("[")+1, str[1].lastIndexOf("]")); //would equal date-string
 String name=str[2].substring(str[2].indexOf("[")+1, str[2].lastIndexOf("]")); //would equal name-string
 String message=str[3].substring(str[3].indexOf("[")+1, str[3].lastIndexOf("]"));

Answer 4

看看here和here，你必须相应地调整它。

Answer 5

哇，轻松轻松！

D [（。*？）]为日期！

https://regex101.com/r/vM0yW8/1

Java解析输出的消息

5 个答案: