Convert String="one,two,three" to String='one','two','three' - java

Need to convert my String values "one,two,three" to 'one','two','three'
I have below code
String input = "One,two,three";
Need to send this input values for an query in hibernate.
So need to send as 'one','two','three' as single string, please provide me a solution easy way to do it

You can do the following:
String result = input.replace(",", "','").replaceAll("(.*)", "'$1'");
input.replace(",", "','") replaces each , with ',' so at this step your string will look like:
One','two','three
next we use a regex to surround the string with ', now it'll look
'One','two','three'
which is what you want.
Regex explanation: We catch the whole string, then we replace it with itself, but surrounded with single quote.
References:
String#replaceAll
String#replace
Regex tutorial

Using regex :
input.replaceAll("(\\w+)", "\'$1\'")
In regex, 'w' is the word character.
http://docs.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html#sum

String[] split = input.split(",");
StringBuilder sb = new StringBuilder();
for (int i = 0 ; i < split.length ; i++ ) {
sb.append("'" + str + "'");
if ( i != split.length-1 ) {
sb.append(",");
}
}
sb.toString();
not smart, but easiest..

Related

How to put Java Regex matches to Resultant String?

How to tokenize an String like in lexer in java?
Please refer to the above question. I never used java regex . How to put the all substring into new string with matched characters (symbols like '(' ')' '.' '<' '>' ") separated by single space . for e.g. before regex
String c= "List<String> uncleanList = Arrays.asList(input1.split("x"));" ;
I want resultant string like this .
String r= " List < String > uncleanList = Arrays . asList ( input1 . split ( " x " ) ) ; "
Referring to the code that you linked to, matcher.group() will give you a single token. Simple use a StringBuilder to append this token and a space to get a new string where the tokens are space-separated.
String c = "List<String> uncleanList = Arrays.asList(input1.split(\"x\"));" ;
Pattern pattern = Pattern.compile("\\w+|[+-]?[0-9\\._Ee]+|\\S");
Matcher matcher = pattern.matcher(c);
StringBuilder sb = new StringBuilder();
while (matcher.find()) {
String token = matcher.group();
sb.append(token).append(" ");
}
String r = sb.toString();
System.out.println(r);
String c = "List<String> uncleanList = Arrays.asList(input1.split('x'));";
Matcher matcher = Pattern.compile("\\<|\\>|\\\"|\\.|\\(|\\)").matcher(c);
while(matcher.find()){
String symbol = matcher.group();
c = c.replace(symbol," " + symbol + " ");
}
Actually if you look deeply You can figure out that you have to separate only not alphabet symbols and space ((?![a-zA-Z]|\ ).)

Extract last number after decimal

I am getting a piece of JSON text from a url connection and saving it to a string currently as such:
...//setting up url and connection
BufferedReader in = new BufferedReader(new InputStreamReader(connection.getInputStream()));
String str = in.readLine();
When I print str, I correctly find the data {"build":{"version_component":"1.0.111"}}
Now I want to extract the 111 from str, but I am having some trouble.
I tried
String afterLastDot = inputLine.substring(inputLine.lastIndexOf(".") + 1);
but I end up with 111"}}
I need a solution that is generic so that if I have String str = {"build":{"version_component":"1.0.111111111"}}; the solution still works and extracts 111111111 (ie, I don't want to hard code extract the last three digits after the decimal point)
If you cannot use a JSON parser then you can this regex based extraction:
String lastNum = str.replaceAll("^.*\\.(\\d+).*", "$1");
RegEx Demo
^.* is greedy match that matches everything until last DOT and 1 or more digits that we put in group #1 to be used in replacement.
Find the start and the end indexes of the String you need and substring(start, end) :
// String str = "{"build":{"version_component":"1.0.111"}};" cannot compile without escaping
String str = "{\"build\":{\"version_component\":\"1.0.111\"}}";
int start = str.lastIndexOf(".")+1;
int end = str.lastIndexOf("\"");
String substring = str.substring(start,end);
just use JSON api
JSONObject obj = new JSONObject(str);
String versionComponent= obj.getJSONObject("build").getString("version_component");
Then just split and take the last element
versionComponent.split("\\.")[2];
Please, your can try the following code :
...
int index = inputLine.lastIndexOf(".")+1 ;
String afterLastDot = inputLine.substring(index, index+3);
With Regular Expressions (Rexp),
You can solve your problem like this ;
Pattern pattern = Pattern.compile("111") ;
Matcher matcher = pattern.matcher(str) ;
while(matcher.find()){
System.out.println(matcher.start()+" "+matcher.end());
System.out.println(str.substring(matcher.start(), matcher.end()));
}

Java String tokens

I have a string line
String user_name = "id=123 user=aron name=aron app=application";
and I have a list that contains: {user,cuser,suser}
And i have to get the user part from string. So i have code like this
List<String> userName = Config.getConfig().getList(Configuration.ATT_CEF_USER_NAME);
String result = null;
for (String param: user_name .split("\\s", 0)){
for(String user: userName ){
String userParam = user.concat("=.*");
if (param.matches(userParam )) {
result = param.split("=")[1];
}
}
}
But the problem is that if the String contains spaces in the user_name, It do not work.
For ex:
String user_name = "id=123 user=aron nicols name=aron app=application";
Here user has a value aron nicols which contain spaces. How can I write a code that can get me exact user value i.e. aron nicols
If you want to split only on spaces that are right before tokens which have = righ after it such as user=... then maybe add look ahead condition like
split("\\s(?=\\S*=)")
This regex will split on
\\s space
(?=\\S*=) which has zero or more * non-space \\S characters which ends with = after it. Also look-ahead (?=...) is zero-length match which means part matched by it will not be included in in result so split will not split on it.
Demo:
String user_name = "id=123 user=aron nicols name=aron app=application";
for (String s : user_name.split("\\s(?=\\S*=)"))
System.out.println(s);
output:
id=123
user=aron nicols
name=aron
app=application
From your comment in other answer it seems that = which are escaped with \ shouldn't be treated as separator between key=value but as part of value. In that case you can just add negative-look-behind mechanism to see if before = is no \, so (?<!\\\\) right before will require = to not have \ before it.
BTW to create regex which will match \ we need to write it as \\ but in Java we also need to escape each of \ to create \ literal in String that is why we ended up with \\\\.
So you can use
split("\\s(?=\\S*(?<!\\\\)=)")
Demo:
String user_name = "user=Dist\\=Name1, xyz src=activedirectorydomain ip=10.1.77.24";
for (String s : user_name.split("\\s(?=\\S*(?<!\\\\)=)"))
System.out.println(s);
output:
user=Dist\=Name1, xyz
src=activedirectorydomain
ip=10.1.77.24
Do it like this:
First split input string using this regex:
" +(?=\\w+(?<!\\\\)=)"
This will give you 4 name=value tokens like this:
id=123
user=aron nicols
name=aron
app=application
Now you can just split on = to get your name and value parts.
Regex Demo
Regex Demo with escaped =
CODE FISH, this simple regex captures the user in Group 1: user=\\s*(.*?)\s+name=
It will capture "Aron", "Aron Nichols", "Aron Nichols The Benevolent", and so on.
It relies on the knowledge that name= always follows user=
However, if you're not sure that the token following user is name, you can use this:
user=\s*(.*?)(?=$|\s+\w+=)
Here is how to use the second expression (for the first, just change the string in Pattern.compile:
String ResultString = null;
try {
Pattern regex = Pattern.compile("user=\\s*(.*?)(?=$|\\s+\\w+=)", Pattern.CASE_INSENSITIVE | Pattern.UNICODE_CASE);
Matcher regexMatcher = regex.matcher(subjectString);
if (regexMatcher.find()) {
ResultString = regexMatcher.group(1);
}
} catch (PatternSyntaxException ex) {
// Syntax error in the regular expression
}

using tokenizer to read a line

public void GrabData() throws IOException
{
try {
BufferedReader br = new BufferedReader(new FileReader("data/500.txt"));
String line = "";
int lineCounter = 0;
int TokenCounter = 1;
arrayList = new ArrayList < String > ();
while ((line = br.readLine()) != null) {
//lineCounter++;
StringTokenizer tk = new StringTokenizer(line, ",");
System.out.println(line);
while (tk.hasMoreTokens()) {
arrayList.add(tk.nextToken());
System.out.println("check");
TokenCounter++;
if (TokenCounter > 12) {
er = new DataRecord(arrayList);
DR.add(er);
arrayList.clear();
System.out.println("check2");
TokenCounter = 1;
}
}
}
} catch (FileNotFoundException ex) {
Logger.getLogger(Driver.class.getName()).log(Level.SEVERE, null, ex);
}
}
Hello , I am using a tokenizer to read the contents of a line and store it into an araylist. Here the GrabData class does that job.
The only problem is that the company name ( which is the third column in every line ) is in quotes and has a comma in it. I have included one line for your example. The tokenizer depends on the comma to separate the line into different tokens. But the company name throws it off i guess. If it weren't for the comma in the company column , everything goes as normal.
Example:-
Essie,Vaill,"Litronic , Industries",14225 Hancock Dr,Anchorage,Anchorage,AK,99515,907-345-0962,907-345-1215,essie#vaill.com,http://www.essievaill.com
Any ideas?
First of all StringTokenizer is considered to be legacy code. From Java doc:
StringTokenizer is a legacy class that is retained for compatibility reasons although its use is discouraged in new code. It is recommended that anyone seeking this functionality use the split method of String or the java.util.regex package instead.
Using the split() method you get an array of strings. While iterating through the array you can check if the current string starts with a quote and if that's the case check if the next one ends with a quote. If you meet these 2 conditions then you know you didn't split where you wanted and you can merge these 2 together, process it like you want and continue iterating through the array normally after that. In that pass you will probably do i+=2 instead of your regular i++ and it should go unnoticed.
You can accomplish this using Regular Expressions. The following code:
String s = "asd,asdasd,asd\"asdasdasd,asdasdasd\", asdasd, asd";
System.out.println(s);
s = s.replaceAll("(?<=\")([^\"]+?),([^\"]+?)(?=\")", "$1 $2");
s = s.replaceAll("\"", "");
System.out.println(s);
yields
asd,asdasd,asd, "asdasdasd,asdasdasd", asdasd, asd
asd,asdasd,asd, asdasdasd asdasdasd, asdasd, asd
which, from my understanding, is the preprocessing you require for your tokenizer-code to work. Hope this helps.
While StringTokenizer might not natively handle this for you, a couple lines of code will do it... probably not the most efficient, but should get the idea across...
while(tk.hasMoreTokens()) {
String token = tk.nextToken();
/* If the item is encapsulated in quotes, loop through all tokens to
* find closing quote
*/
if( token.startsWIth("\"") ){
while( tk.hasMoreTokens() && ! tk.endsWith("\"") ) {
// append our token with the next one. Don't forget to retain commas!
token += "," + tk.nextToken();
}
if( !token.endsWith("\"") ) {
// open quote found but no close quote. Error out.
throw new BadFormatException("Incomplete string:" + token);
}
// remove leading and trailing quotes
token = token.subString(1, token.length()-1);
}
}
As you can see, in the class description, the use of StringTokenizer is discouraged by Oracle.
Instead of using tokenizer I would use the String split() method
which you can use a regular expression as argument and significantly reduce your code.
String str = "Essie,Vaill,\"Litronic , Industries\",14225 Hancock Dr,Anchorage,Anchorage,AK,99515,907-345-0962,907-345-1215,essie#vaill.com,http://www.essievaill.com";
String[] strs = str.split("(?<! ),(?! )");
List<String> list = new ArrayList<String>(strs.length);
for(int i = 0; i < strs.length; i++) list.add(strs[i]);
Just pay attention to your regex, using this one you're assuming that the comma will be always between spaces.

Reformatting a Java String

I have a string that looks like this:
CALDARI_STARSHIP_ENGINEERING
and I need to edit it to look like
Caldari Starship Engineering
Unfortunately it's three in the morning and I cannot for the life of me figure this out. I've always had trouble with replacing stuff in strings so any help would be awesome and would help me understand how to do this in the future.
Something like this is simple enough:
String text = "CALDARI_STARSHIP_ENGINEERING";
text = text.replace("_", " ");
StringBuilder out = new StringBuilder();
for (String s : text.split("\\b")) {
if (!s.isEmpty()) {
out.append(s.substring(0, 1) + s.substring(1).toLowerCase());
}
}
System.out.println("[" + out.toString() + "]");
// prints "[Caldari Starship Engineering]"
This split on the word boundary anchor.
See also
regular-expressions.info/Word boundary
Matcher loop solution
If you don't mind using StringBuffer, you can also use Matcher.appendReplacement/Tail loop like this:
String text = "CALDARI_STARSHIP_ENGINEERING";
text = text.replace("_", " ");
Matcher m = Pattern.compile("(?<=\\b\\w)\\w+").matcher(text);
StringBuffer sb = new StringBuffer();
while (m.find()) {
m.appendReplacement(sb, m.group().toLowerCase());
}
m.appendTail(sb);
System.out.println("[" + sb.toString() + "]");
// prints "[Caldari Starship Engineering]"
The regex uses assertion to match the "tail" part of a word, the portion that needs to be lowercased. It looks behind (?<=...) to see that there's a word boundary \b followed by a word character \w. Any remaining \w+ would then need to be matched so it can be lowercased.
Related questions
Use Java and RegEx to convert casing in a string
Java regex does not support Perl preprocessing operations \l \u, \L, and \U.
Java split is eating my characters.
More examples of using assertions
StringBuilder and StringBuffer in Java
Unfortunately, appendReplacement/Tail only takes StringBuffer
You can try this:
String originalString = "CALDARI_STARSHIP_ENGINEERING";
String newString =
WordUtils.capitalize(originalString.replace('_', ' ').toLowerCase());
WordUtils are part of the Commons Lang libraries (http://commons.apache.org/lang/)
Using reg-exps:
String s = "CALDARI_STARSHIP_ENGINEERING";
StringBuilder camel = new StringBuilder();
Matcher m = Pattern.compile("([^_])([^_]*)").matcher(s);
while (m.find())
camel.append(m.group(1)).append(m.group(2).toLowerCase());
Untested, but thats how I implemented the same some time ago:
s = "CALDARI_STARSHIP_ENGINEERING";
StringBuilder b = new StringBuilder();
boolean upper = true;
for(char c : s.toCharArray()) {
if( upper ) {
b.append(c);
upper = false;
} else if( c = '_' ) {
b.append(" ");
upper = true;
} else {
b.append(Character.toLowerCase(c));
}
}
s = b.toString();
Please note that the EVE license agreements might forbit writing external tools that help you in your careers. And it might be the trigger for you to learn Python, because most of EVE is written in Python :).
Quick and dirty way:
Lower case all
line.toLowerCase();
Split into words:
String[] words = line.split("_");
Then loop through words capitalising first letter:
words[i].substring(0, 1).toUpperCase()

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