Problem with index of numbers in BSP Front end
Hello All,
I am facing the below problem. Request your help.
Scenario: When I am trying to add the projects in the Front end , I am facing the problem with the index of numbering of the projects.
Example: If I am adding 6 projects ( at the max only 4 on each page i-e there are 2 pages ), then the index is wrongly numbered and the arrow (for direction of the pages) are wrongly displayed. The direction of the arrow is pointed to the next page rather pointing to the previous page and the index of numbering of the pages is wrong.
Request you to please suggest if this is any code fault in the table maintainance activity in the backend (BSP).
Hi Irina,
If you mean the number range is skipping 10 numbers in between runs then its because of the buffering in the number range.
To avoid the buffer influencing the number range, pass
'X' to the function module parameter: IGNORE_BUFFER.
This should resolve the problem.
Cheers,
Aditya
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Attachments:
zeroiffp.JPG 99 KB
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8
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18
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1
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3
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5
5
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>
>
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We faced some problems in Indexing and Taxonomies for which I would like to know the solution/reason.
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Hi,
i am trying to develop a small search engine... currently i am trying to develop STOPLIST..... which iterates through piece of text and extracts different words and their respective frequencies... here is the piece of code which does this :
======================================
for( int c=0; c < lines.size(); c++) {
temp = lines.elementAt(c).toString();
temp = temp.toLowerCase();
st = new java.util.StringTokenizer(temp);
while( st.hasMoreTokens())
words.addElement(st.nextToken());
for( int b=0; b < words.size(); b++) {
extractedWord = words.elementAt(b).toString();
for( int a=0; a < words.size(); a++) {
if( words.elementAt(a).toString().equalsIgnoreCase(extractedWord))
extractedWordFreq++;
extractedWords.addElement(extractedWord);
extractedWordsFreq.addElement(Integer.toString(extractedWordFreq));
extractedWordFreq=0;
for( int z=0; z < extractedWords.size(); z++)
System.out.println("Extracted Terms : " + extractedWords.elementAt(z).toString() + " Freq : " + extractedWordsFreq.elementAt(z).toString());
========================================
the only problem with above code is that .... it takes a long time to
execute... around 30 seconds....
Here is the spec of my pc :
P4 : 512 MB RAM
The piece of text has around 7500 words
when i execute this code... and look at my Windows Task Manager....
CPU usage goes to 100 %.....
All I wanna know is that ... is there any way to change this code so that it executes more swiftly.....
thanx in advance
have funHi
I read your previous reply: Vector, Array, StringTokenizer, etc. They are "heavy weight" not concerning to peformance so your code uses brute force.
I found an example dealing with this issue and I believe to be faster than your but you must to modify details:
import java.io.*;
import java.util.TreeMap;
import java.util.Iterator;
public class WC {
public static void main(String args[]) throws IOException {
if (args.length != 1) {
System.err.println("Usage: java WC <textfile>");
System.exit(10);
File file = new File(args[0]);
StreamTokenizer stream = new StreamTokenizer(
new FileReader(file));
// treat any punctuation as word delimiters
stream.whitespaceChars('!', '/');
stream.whitespaceChars(':', '@');
WordFrequencyCounter wfr = new WordFrequencyCounter();
String word;
int countNumbers = 0;
while (stream.nextToken() != StreamTokenizer.TT_EOF)
if (stream.ttype == StreamTokenizer.TT_WORD) {
word = stream.sval;
wfr.count(word);
else if (stream.ttype == StreamTokenizer.TT_NUMBER) {
countNumbers++;
System.out.println("File: " + args[0] + " (" + file.length() + " bytes)"
+ "\n#lines = " + stream.lineno()
+ "\n\n#words = " + wfr.size()
+ "\n#numbers = " + countNumbers + "\n");
for (Iterator it = wfr.keySet().iterator(); it.hasNext();) {
word = (String) it.next();
System.out.println(wfr.frequency(word) + " " + word);
class WordFrequencyCounter extends TreeMap {
// first see if this word has already been encountered;
// if not, then create a fresh counter for it;
// otherwise, increment its counter
public synchronized int count(String word) {
Integer counter = (Integer) get(word);
counter = (counter == null) ? new Integer(1)
: new Integer(counter.intValue() + 1);
put(word, counter);
return counter.intValue();
// find out how many times this word has been encountered
public synchronized int frequency(String word) {
Integer counter = (Integer) get(word);
return (counter == null) ? 0 : counter.intValue();
}Regards. -
Problem with total page numbers in Scripts
Hi,
I have a problem with total number of pages in the Script.
I have a script of 15 pages.
On every page, I am using &PAGE& of &SAPSCRIPT-FORMPAGES(C)&.
Problem is : I am getting only single digit for SAPSCRIPT-FORMPAGES as long as the page number remains in single digit - like -
1 of 1
2 of 1
3 of 1... till 9 of 1.
But from 10th page, it is printing :
10 of 15
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So, i think the symbol SAPSCRIPT-FORMPAGES(C) is filled with value 15 , but it is prinitng only the first digit '1' for pages 1 to 9 and then printing both the digits '15' for pages from 10 onwards.
So, please help me out in this problem.
Thank you.Hi Ven,
The CONDENSE option cannot be used on the program icon SAPSCRIPT-FORMPAGES
(C) without an explicit length specification: This symbol is
only replaced with a value after the form has been completely edited, as the
total number of pages of a form is first known in the program function
END_FORM or CLOSE_FORM. However, the size of the symbol (number of
characters) is reserved correctly when the symbol first occurs, with the
current page number. Therefore, only one character is reserved for option C
(CONDENSE) on pages 1-9, two characters on pages 10-99, and so on.
If you are sure that the total number of pages doesn't cross 99(i.e 2-digit number), then use the FORM pages as SAPSCRIPT-FORMPAGES(2CZ).
Here Z will suppress the leading zeros, for the total number of pages.
Hope this information would be useful for you.
Reward points if this is useful.
Pavan.
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