how to split string using comma but ignore on certain condition? - java

I want to split below string
{"address":"XXXXXX","amount":"0.25"},{"address":"AAAAAA","amount":"0.25"}
into two values
{"address":"XXXXXX","amount":"0.25"}
{"address":"AAAAAA","amount":"0.25"}

It is not a recommended thing to parse json the way you are doing but in case you really wanna just split, you can do so using lookarounds using regex. Here is the java code for same.
public static void main(String args[]) throws Exception {
String s = "{\"address\":\"XXXXXX\",\"amount\":\"0.25\"},{\"address\":\"AAAAAA\",\"amount\":\"0.25\"}";
String[] data = s.split("(?<=\\}),(?=\\{)");
Arrays.asList(data).forEach(System.out::println);
}
Prints,
{"address":"XXXXXX","amount":"0.25"}
{"address":"AAAAAA","amount":"0.25"}

Related

Easiest way to do this : Put a String into the middle of a String for various lengths

Trying to split a String in certain ways without changing the
String(String embed, String payload)
structure.
System.out.println(embedCenter("<<>>", "Yay")); // => <<Yay>>
This is how it should look, so putting "<<>>" for embed and "Yay" for payload should return <<Yay>>, however for "()" embed and "Yay" Payload it should return "(Yay)" and for ":-)" embed and "Yay" payload it should return ":Yay-)"
So I'm just starting to learn, and kinda stuck at this question - I've tried doing substrings but while I could get one of those results, I cant find a way to get all of them with the same method.
public static String embedCenter(String embed, String payload) {
return ""; //
}
public static void main(String[] args) {
System.out.println(embedCenter("<<>>", "Yay")); // => <<Yay>>
System.out.println(embedCenter("()", "Yay")); // => (Yay)
System.out.println(embedCenter(":-)", "Example")); // :Example-)
Ok, I did it, I thought way too complicated, did it really easy simply by dividing the String with length()/2, perfectly worked. Thanks for the input!
int length = embed.length();
String subembed = embed.substring(0,embed.length()/2);
String finembed = embed.substring(embed.length()/2);
return subembed + payload + finembed;

Better way to replace a set of signs with strings [duplicate]

This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 5 years ago.
I need to make a simple code, which get a string, and an array of String, and replace the ? signs with each element of array.
Here is the test case in junit:
#Test
public void QueryFitterTest() {
ArrayList<String> args=new ArrayList<String>();
args.add("Bad code");
args.add("The code is buggy");
String res = QueryMaker.queryFitter("insert into vulnerability (name,descirption) values(?,?)",args);
String correctQuery="insert into vulnerability (name,descirption) values(Bad code,The code is buggy)";
assertEquals(correctQuery, res);
}
and here is the code:
public static String queryFitter(String query, ArrayList<String> args){
String[] colapsedQuery = query.split("");
int parmNum=0;
for(int i=0;i<colapsedQuery.length;i++){
if(colapsedQuery[i]=="?"){
colapsedQuery[i]=args.get(parmNum);
parmNum++;
}
}
query=concatenator(colapsedQuery);
return query;
}
public static String concatenator(String[] colapsedQuery){
String delimiter = "";
String result = String.join(delimiter, colapsedQuery);
System.out.println("query is: "+result);
return result;
}
The code is working fine but
I don't like my approach, is there an easier way to do it?
There are 2 issues:
1- My code cannot pass the test, it returns the query without any
change.
query is: insert into vulnerability (name,descirption) values(?,?)
2-
I don't like my approach, is there an easier way to do it?
Well, the good news is that your JUnit test discovered a bug in your program. The other good news is that the answer to both of your questions is the same. Just fix your code in the method queryFitter.
Try the following code:
public static String queryFitter(String query, ArrayList<String> args){
for(int i=0;i<args.size();i++){
query = query.replaceFirst("\\?",args.get(i));
}
return query;
}
Almost forgot to tell you. You don't need concatenator method either.

Read text from a string & edit it

Maybe there is already a similar question, but I didn't find it.
So, how can I "read" a string and edit it? For example:
String exampleString = "Monday: Do stuff & things!";
Now, I want to "read and edit" the string, so I get this:
String exampleString = "Do stuff & things!";
Is there a way to edit a string?
Kind regards!
Try this
Examplestring.split(":")[1]
You will get you desired o/p
Better go through java string basics
Use substring() or split() method :
public static void main(String[] args) {
String exampleString = "Monday: Do stuff & things!";
String editedString=exampleString.substring(exampleString.indexOf(":")+1);
System.out.println("Edited String :"+editedString);
//OR
String []edited=exampleString.split(":", 2);
System.out.println("Edited String :"+edited[1]);
}
}

How to use regular expression for fetching specific data?

I have input stream with the following data:
---------------------------------------------
manil#manil-ubvm:~$ db2level
DB21085I Instance "manil" uses "64" bits and DB2 code release "SQL10010" with
level identifier "0201010E".
Informational tokens are "DB2 v10.1.0.0", "s120403", "LINUXAMD64101", and Fix
Pack "0".
Product is installed at "/home/manil/sqllib".
---------------------------------------------
From above i need v10.1.0.0 to be stored in a string variable.
How to do that using java regular expression?
Use something like this to capture the version pattern :
import java.util.regex.*;
public class RTest {
public static void main(String [] args) {
String raw_data = "asdkgjasdbf984 sdkjfashfiu 4qwsadkfjnv w98sa-asdf08gywbfsd v1231.123.12.11.1 fkjsdfn9823isd";
Pattern version_find = Pattern.compile("v[\\d+\\.?]+");
Pattern directory_find = Pattern.compile("[\\/[^\\/]+]+");
Matcher version_finder = version_find.matcher(raw_data);
while(version_finder.find()) {
System.out.println(version_finder.group());
}
}
}
Output is :
v1231.123.12.11.1
/isd/asdasd2903 ajshdaq09r34/adsj 38/
You really need to understand regexes deeply if you are a programmer. They are one of the essentials. They are hard at first, but once you 'crack them' you don't forget it. Like riding a bike.
This will suit your needs:
String version = yourLine.replaceAll(".*(v\\d+([.]\\d+){3}).*", "$1")
You dont need regularExpression here
just use
String .contain() method and String substring()

How to replace tokens in a string without StringTokenizer

Given a string like so:
Hello {FIRST_NAME}, this is a personalized message for you.
Where FIRST_NAME is an arbitrary token (a key in a map passed to the method), to write a routine which would turn that string into:
Hello Jim, this is a personalized message for you.
given a map with an entry FIRST_NAME -> Jim.
It would seem that StringTokenizer is the most straight forward approach, but the Javadocs really say you should prefer to use the regex aproach. How would you do that in a regex based solution?
Thanks everyone for the answers!
Gizmo's answer was definitely out of the box, and a great solution, but unfortunately not appropriate as the format can't be limited to what the Formatter class does in this case.
Adam Paynter really got to the heart of the matter, with the right pattern.
Peter Nix and Sean Bright had a great workaround to avoid all of the complexities of the regex, but I needed to raise some errors if there were bad tokens, which that didn't do.
But in terms of both doing a regex and a reasonable replace loop, this is the answer I came up with (with a little help from Google and the existing answer, including Sean Bright's comment about how to use group(1) vs group()):
private static Pattern tokenPattern = Pattern.compile("\\{([^}]*)\\}");
public static String process(String template, Map<String, Object> params) {
StringBuffer sb = new StringBuffer();
Matcher myMatcher = tokenPattern.matcher(template);
while (myMatcher.find()) {
String field = myMatcher.group(1);
myMatcher.appendReplacement(sb, "");
sb.append(doParameter(field, params));
}
myMatcher.appendTail(sb);
return sb.toString();
}
Where doParameter gets the value out of the map and converts it to a string and throws an exception if it isn't there.
Note also I changed the pattern to find empty braces (i.e. {}), as that is an error condition explicitly checked for.
EDIT: Note that appendReplacement is not agnostic about the content of the string. Per the javadocs, it recognizes $ and backslash as a special character, so I added some escaping to handle that to the sample above. Not done in the most performance conscious way, but in my case it isn't a big enough deal to be worth attempting to micro-optimize the string creations.
Thanks to the comment from Alan M, this can be made even simpler to avoid the special character issues of appendReplacement.
Well, I would rather use String.format(), or better MessageFormat.
String.replaceAll("{FIRST_NAME}", actualName);
Check out the javadocs for it here.
Try this:
Note: The author's final solution builds upon this sample and is much more concise.
public class TokenReplacer {
private Pattern tokenPattern;
public TokenReplacer() {
tokenPattern = Pattern.compile("\\{([^}]+)\\}");
}
public String replaceTokens(String text, Map<String, String> valuesByKey) {
StringBuilder output = new StringBuilder();
Matcher tokenMatcher = tokenPattern.matcher(text);
int cursor = 0;
while (tokenMatcher.find()) {
// A token is defined as a sequence of the format "{...}".
// A key is defined as the content between the brackets.
int tokenStart = tokenMatcher.start();
int tokenEnd = tokenMatcher.end();
int keyStart = tokenMatcher.start(1);
int keyEnd = tokenMatcher.end(1);
output.append(text.substring(cursor, tokenStart));
String token = text.substring(tokenStart, tokenEnd);
String key = text.substring(keyStart, keyEnd);
if (valuesByKey.containsKey(key)) {
String value = valuesByKey.get(key);
output.append(value);
} else {
output.append(token);
}
cursor = tokenEnd;
}
output.append(text.substring(cursor));
return output.toString();
}
}
With import java.util.regex.*:
Pattern p = Pattern.compile("{([^{}]*)}");
Matcher m = p.matcher(line); // line being "Hello, {FIRST_NAME}..."
while (m.find) {
String key = m.group(1);
if (map.containsKey(key)) {
String value= map.get(key);
m.replaceFirst(value);
}
}
So, the regex is recommended because it can easily identify the places that require substitution in the string, as well as extracting the name of the key for substitution. It's much more efficient than breaking the whole string.
You'll probably want to loop with the Matcher line inside and the Pattern line outside, so you can replace all lines. The pattern never needs to be recompiled, and it's more efficient to avoid doing so unnecessarily.
The most straight forward would seem to be something along the lines of this:
public static void main(String[] args) {
String tokenString = "Hello {FIRST_NAME}, this is a personalized message for you.";
Map<String, String> tokenMap = new HashMap<String, String>();
tokenMap.put("{FIRST_NAME}", "Jim");
String transformedString = tokenString;
for (String token : tokenMap.keySet()) {
transformedString = transformedString.replace(token, tokenMap.get(token));
}
System.out.println("New String: " + transformedString);
}
It loops through all your tokens and replaces every token with what you need, and uses the standard String method for replacement, thus skipping the whole RegEx frustrations.
Depending on how ridiculously complex your string is, you could try using a more serious string templating language, like Velocity. In Velocity's case, you'd do something like this:
Velocity.init();
VelocityContext context = new VelocityContext();
context.put( "name", "Bob" );
StringWriter output = new StringWriter();
Velocity.evaluate( context, output, "",
"Hello, #name, this is a personalized message for you.");
System.out.println(output.toString());
But that is likely overkill if you only want to replace one or two values.
import java.util.HashMap;
public class ReplaceTest {
public static void main(String[] args) {
HashMap<String, String> map = new HashMap<String, String>();
map.put("FIRST_NAME", "Jim");
map.put("LAST_NAME", "Johnson");
map.put("PHONE", "410-555-1212");
String s = "Hello {FIRST_NAME} {LAST_NAME}, this is a personalized message for you.";
for (String key : map.keySet()) {
s = s.replaceAll("\\{" + key + "\\}", map.get(key));
}
System.out.println(s);
}
}
The docs mean that you should prefer writing a regex-based tokenizer, IIRC. What might work better for you is a standard regex search-replace.
Generally we'd use MessageFormat in a case like this, coupled with loading the actual message text from a ResourceBundle. This gives you the added benefit of being G10N friendly.

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