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I have JSON response from server. JSON has exception info inside of it.
I need to parse it to detect that exception occurred and identify which one.
JSON example:
I found the solution to what I need
Pattern errorCodePattern = Pattern.compile("\"code\"\\s*:\\s*\"([^,]*)\",");
Pattern messagePattern = Pattern.compile("\"message\"\\s*:\\s*\"([^,]*)\",");
Pattern statusPattern = Pattern.compile("\"status\"\\s*:\\s*\"(FAILURE)\"");
Matcher errorCodeMatcher = errorCodePattern.matcher(response);
Matcher messageMatcher = messagePattern.matcher(response);
Matcher statusMatcher = statusPattern.matcher(response);
Java JSON Parser Example with Regex
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How do I get the after last two semicolon's occurrences of a string ?
Example:
Mobiles;Students;Test;Yes;1234
The output should be
Yes;1234
Using a regex replacement, we can try:
String input = "Mobiles;Students;Test;Yes;1234";
String output = input.replaceAll("^.*;([^;]+;[^;]+)$", "$1");
System.out.println(output); // Yes;1234
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Suppose i have below Json String:
{
"name":"noor",
"pass":"12345"
}
I want to mask pass value using Regex, like below
{
"name":"noor",
"pass":"*****"
}
How i can do it, using Java regex?
Try this:
"pass":"(.*?)"
As seen on: https://regex101.com/r/cK4bD0/1
try this
String jsonString = "{ \"name\":\"noor\", \"pass\":\"12345\" }";
String result = jsonString.replaceAll("(?<=pass\":\")(.*?)(?=\")", "*****");
System.out.println(result);
{[^}]*"pass"\s*:\s*"(.*?)"[^}]*}
Here is the DEMO: https://regex101.com/r/nL3tP2/2
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I want to write a regex for this below pattern.
of John at /roger/adam/sam
here John is a value(can be alphanumeric) and /roger/adam/sam is an XPath with can change.
I want to replace all such instances with
of $value at $XPath
You can use the following to match
(of\s*)[a-zA-Z\d:-]+(\s*at\s*).*?(?=\s)
and replace with $1$value$2$XPath
See DEMO
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Below are the String values I want to separate the inside brackets value
US Records (100)
Foreign Records (243)
In the string above I want to separate the counts and store into another string
100, 243 using regular expression.
This code schold give you the result:
String s = new String("US Records (100)");
Pattern p = Pattern.compile("\\((\\d+)\\)");
Matcher m = p.matcher(s);
m.find();
System.out.println(m.group(1));
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I have this text:
1 A Maths
And i would like to get only Maths.
I don´t know about regular expressions.
Can any one help me, please?
Thanks
Why don't you split the string ?
String chain="A Maths";
String[] array=chain.split(" ");
String number=array[0];
String course=array[1];
...
How to split a string in Java