Display each words of the string in separate line.

Dear Javaties,
pls help me creating this program.
Display each words of the string in separate line.
Eg.
Input value : Testing this string.
Output : Testing
this
string
Thanks in adavance,
M. Suresh

M Suresh,
We would be delighted to help if only you asked a specific question. Do my work for me is not a specific question.
The Javaties

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Hi,
I wanted to know, how to calculate the length of each word in a string.Suppose for example if the input is like this:
i/p:'Tech on the Net'
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ie Tech-4 on-2 the-3 Net-3
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from(select rownum-1 n
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ya if we are passing three separate string values means we can use
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Blog: baron72.wordpress.com
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CodePlex Corner Series
Please remember to mark your question as "answered" if this solves (or helps) your problem.

Pls help me to count the no of occurence of each word in the data file

I want to count the no of occurence of each word the data file..
anybody help me out to solve this ,,
I have attached my code hereby...
import java.io.*;
import java.util.*;
import java.util.Collections;
import javax.print.attribute.standard.Media;
class WordCounter1 {
String s;
String word;
public static void main (String[] args){
WordCounter1 simpleCounter= new WordCounter1();
try
File firstFile=new File("C:/Documents and Settings/janmala/file4.txt");
simpleCounter.countWords(firstFile);
catch(Exception e)
System.err.println(e);
void countWords(File inputFile)
int count=0;
int key=0;
LinkedHashMap lhm=new LinkedHashMap();
try{
FileInputStream fin=new FileInputStream(inputFile);
DataInputStream din=new DataInputStream(fin);
while(din.available()!=0){
s=din.readLine();
StringTokenizer st=new StringTokenizer(s);
while(st.hasMoreTokens())
word=st.nextToken();
lhm.put(word,count);
if(lhm.containsKey(word))
lhm.get(word);
count++;
System.out.println(word);
System.out.println(count);
int a=lhm.size();
System.out.println("The size of Hash map"+a);
Set set = lhm.entrySet();
Iterator it = set.iterator();
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Map.Entry me = (Map.Entry)it.next();
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catch(Exception e)
System.err.println(e);
}}

jan07 wrote:
anybody help me out to solve this ,,Solve what?
Don't just dump a pile of unformatted code and expect someone to magically fix it for you.
Post code, hightlight it and click the CODE button to retain formatting
Include error messages if you get them
Ask an actual question

Display only part of the STRING field, but search ALL field

Hi all,
I have a field VARCHAR2(4000) that holds a lot of text.
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Thanks

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xtender wrote:
You can use my package xt_regexp - http://github.com/xtender/XT_REGEXP
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select 'ID_A,ID_B,ID_C,' str from dual
union all
select 'ID_A,ID_B,ID_C,ID_D' str from dual
select
t.*,
row_number() over (partition by t.str order by t.str) substr_number,
substring.column_value substring
from t,
table(xt_regexp.split(t.str,',')) substring
That's nice, but you may want to benchmark that against an approach that doesn't make a bunch of external calls (for performance)...
with t as
select 1 as id,'ID_A,ID_B,ID_C' str from dual
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select 2 as id, 'ID_A,ID_B,ID_C,ID_D' str from dual
select
t.*,
regexp_substr (t.str, '[^,]+', 1, s.column_value) as split
from
t,
table( cast (multiset (select level from dual connect by level <= length(regexp_replace(t.str, '[^,]+')) + 1) as sys.odcinumberlist)) s
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      1 ID_A,ID_B,ID_C      ID_B
      1 ID_A,ID_B,ID_C      ID_C
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      2 ID_A,ID_B,ID_C,ID_D ID_B
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TUBBY_TUBBZ?For example.

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Dear friends,
I am looking out for the code of textfield autocompletion in swing which will be applied for each word.

Dear friends,
I am looking out for the code of textfield autocompletion in swing which will be applied for each word.

Is it possible to add a word at the end of a line?

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is it something possible doing by using a script?
i don't script myself but if someone could find me a way to do this it would be really appreciated.
Here's an example to what i need to do:
i have paragraph style X in a current book document, and i need to edit the book and add the copyright source at the end of every line that has the style X to it
so i would need something that asks me
the targeted paragraph style
which word i want to append to it, and then apply it torough the document.
Thanks in advance

Yes, the screen shot shows where you got it wrong:
function(){return A.apply(null,[this].concat($A(arguments)))}
Find: (.)$
and set your paragraph style in the find format area
Change: $1XYZ
where XYZ is the text you want to insert.
.. you forgot the '1' right after the '$'. It's an important omission: "$1" is a GREP code, meaning "insert the found text, parenthesized group #1" -- and there is also a "$2", "$3", etc. up to "$9". It's great if you have to switch two, or even more, items around! But without the digit, the dollar sign indicates ... just a dollar sign.
Change the Change to text to
$1~S(CEMEQ)
and you'll be fine.

Selecting words in the beginning of a LINE (not a paragraph)

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I'm trying to avoid certain combinations to fall at the beginning of any line in a body text.
How do I select a particular character only if it occurs in the beginning of a line (but in the middle of a paragraph.
For instance I want to select the char "a" only if it occurs at the left margin of a column
I only find the location marc "beginning of a paragraph" (^), or beginning of a word, but not the beginning of a line.
It could be as grep, or as a script.
Anybody?
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It would need a script. GREP is not aware of the individual lines in a paragraph.
The good news is, InDesign's scripting language is fully aware of lines. There are, on first sight, two possible ways of locating all a's at the start of a line. Either search for all a's (doesn't need a GREP, but you must remember to set caseSensitive in the FindChangePreferences) and loop over them, checking for the ones that are at the start of a line, or check all lines one by one, checking for the ones that start with an 'a'.
Both options seem equally slow I guess what to use depends on your text, and what you'd want to do when one is found. On that note: you must process them from last found to first, because if you change something, it may have an effect on all text that follows.

How to read each and every word from a string.

Hi all,
I have a string which is having many label numbers. if the string is lv_str, its value is like, 11111111111111##22222222222222##3333333333333.
I need to move the values alone into internal table. each value should be updated as a single row into one internal table. How to read each and every word of the string and move to an internal table.
the internal table should be like this.
11111111111111
22222222222222
3333333333333
Can any one give me a suggestion in this regard.
POINTS PROMISED.
Regards,
Buvana

Hi,
If you know the format and length of the data
Use split at '#' so that you will get the individual values.
Thean append it to internal table.
Reward iof helpful.

Here's how to find the right word in a string

I needed to find the rightmost word in a string. I didn't find a simple formula in these forums, but I now have one, so I wanted to share it. Hope you find it useful.
Assuming that the string is in cell A1, the following will return the rightmost word in the string:
RIGHT(A1,LEN(A1)-FIND("*",SUBSTITUTE(A1," ","*",LEN(A1)-LEN(SUBSTITUTE(A1," ","")))))

I found the problem. Whatever character was being used in the substitution was parsed out by the forum parser. I replaced it with "œ" (option q on my keyboard).
=RIGHT(A1,LEN(A1)-FIND("œ",SUBSTITUTE(A1," ","œ",LEN(A1)-LEN(SUBSTITUTE(A1," ","")))))
Still needs an error check for a single-word "sentence" and to remove the trailing period but it does seem to work. Pretty slick.
Message was edited by: Badunit
I see below that the problem was fixed by the OP.
Message was edited by: Badunit

Match Certain word in a String

Hi all,
I'm actually doing a project regrading e-learning. And i'm suppose to create a set of questions for each topics using labview. So Is there any way that i can match certain word in the string to make sure that answer is correct? Cause i'm sure that every user that input answer will be different. Thus, I want to pick out main point as an answer. Is there anyway i can do it??
Really appreciate your help!!!
Thank you!!
Solved!
Go to Solution.
Attachments:
Match Strings.vi ‏8 KB
Match Strings.vi ‏8 KB

Here's another option (building on Jeff's code). Turn on the Conditional Terminal on the FOR loop and change it to "Continue if TRUE". This way, the loop will exit as soon as a failure is found. Just pass the result straight out of the loop. If none fail, then the FOR loop will exit on its own (from the auto-indexing) and a pass is passed out.
There are only two ways to tell somebody thanks: Kudos and Marked Solutions
Unofficial Forum Rules and Guidelines
Attachments:
Match String.png ‏19 KB

Replace complete words by another string

Hi,
I'm searching for a possibility to replace complete words inside a string.
For example I want to replace "quick" by "slow":
"The <b>quick</b> brown fox is running quicker then the slow brown fox".
The result shall be:
"The <b>slow</b> brown fox is running quicker then the slow brown fox".
The problem is, that the built in "Replace"-Method would also replace "quicker" by "browner".
Furthermore, words beetween different signs (e.g. , . ; : - ! " ? ') shall also be found as complete word.
I already searched for a FM but didn't find anything. Regular expressesions are also not possible because we are running on 6.40.
Maybe anybody can help me with a simple FM or something else.
Best regards,
Stefan

Hi Stefan,
There is a crude way of doing this....with lots of loops..
You can use the SPLIT statement
SPLIT dobj AT sep INTO
{ {result1 result2 ...} | {TABLE result_tab} }
[IN {CHARACTER|BYTE} MODE].
and break the sentence into words using space as delimiter. Once you have all the words in a internal table, you can loop through it and compare each word with the word you want.
This will solve the problem, but will not be performant enough.
Regards,
Vinodh

Display each words of the string in separate line.

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