Count of characters in a string

Hi all,
Is there any way to find the repeating characters in the string like 'L' in the example Hello ..
Thanks in advance,
Nalla !
Message was edited by: Nalla
Nalla (Nallasivam)

test@ORA10G>
test@ORA10G> with t as (
2    select 'hello' as str from dual union all
3    select 'abracadabra' from dual union all
4    select 'the quick brown fox jumps over the lazy dog' from dual union all
5    select 'abcdefghijklmnopqrstuvwxyz' from dual union all
6    select 'facetious' from dual)
7 --
8 select
9    str
10 from t
11 where (
12    length(str) - length(replace(lower(str),'a')) > 1 or
13    length(str) - length(replace(lower(str),'b')) > 1 or
14    length(str) - length(replace(lower(str),'c')) > 1 or
15    length(str) - length(replace(lower(str),'d')) > 1 or
16    length(str) - length(replace(lower(str),'e')) > 1 or
17    length(str) - length(replace(lower(str),'f')) > 1 or
18    length(str) - length(replace(lower(str),'g')) > 1 or
19    length(str) - length(replace(lower(str),'h')) > 1 or
20    length(str) - length(replace(lower(str),'i')) > 1 or
21    length(str) - length(replace(lower(str),'j')) > 1 or
22    length(str) - length(replace(lower(str),'k')) > 1 or
23    length(str) - length(replace(lower(str),'l')) > 1 or
24    length(str) - length(replace(lower(str),'m')) > 1 or
25    length(str) - length(replace(lower(str),'n')) > 1 or
26    length(str) - length(replace(lower(str),'o')) > 1 or
27    length(str) - length(replace(lower(str),'p')) > 1 or
28    length(str) - length(replace(lower(str),'q')) > 1 or
29    length(str) - length(replace(lower(str),'r')) > 1 or
30    length(str) - length(replace(lower(str),'s')) > 1 or
31    length(str) - length(replace(lower(str),'t')) > 1 or
32    length(str) - length(replace(lower(str),'u')) > 1 or
33    length(str) - length(replace(lower(str),'v')) > 1 or
34    length(str) - length(replace(lower(str),'w')) > 1 or
35    length(str) - length(replace(lower(str),'x')) > 1 or
36    length(str) - length(replace(lower(str),'y')) > 1 or
37    length(str) - length(replace(lower(str),'z')) > 1
38 );
STR
hello
abracadabra
the quick brown fox jumps over the lazy dog
3 rows selected.
test@ORA10G>
test@ORA10G>pratz

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Counting Characters in a string

HI all!
I am new working with java and i have an application where I have to count specific characters in a string e.g. 'a','.','H','4' and so on! I found an entry in the forum with this script:
String text = "Hello World!";
int count = 0;
for (int x = 0; x < text.length(); x++)
if (text.charAt(x) == 'l')
count++;
This Script count all the 'I', but I could not get CharAt to count different characters at once!
Is there a way?
Thx for your help!
LeBite

String text = "Hello World!";
int countl = 0;
int counto = 0;
for (int x = 0; x < text.length(); x++)
    if (text.charAt(x) == 'l') countl++;
    if (text.charAt(x) == 'o') counto++;
[/code[

How to count number of repeated characters in a String

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it is workingYes, but... attention to surprises...
SQL> var v1 varchar2(1000);
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PL/SQL procedure successfully completed.
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Method to count number of characters in a string?

hi.
i'm trying to write some script that will count the number of characters in a string passed into a constructor and then compare it against another value to check whether the string passed in is valid.
something like this
public class Stringcheck
public static int maxStringLength;
public static double minSize;
public static boolean testStringLength(String x)
// Insert method for counting number of chars in String x and name as 'int numberOfChars
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ummm, you dont need anything fancy. try:
String test = "this is the test string";
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How to count number of characters in a string?

Hi
I need to display values in a script. but the alignment getting distrubed if it is less than 20 charaters, for exmple material descirption is only displaying 20 characters in script. if it is less than 20 charcters the alignment getting changed.
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thanks in advance...

solution to this problem in script is not to count the characters but to align them based on positions .
1.
Use the tab positions for that paragraph format to display the values in a finite format .
In this u need to give the position start and position end length and use the same in the script as
p ,,&itab-f1&
say my value has to print from position 125mm to 155 mm
i take one tab position for this paragraph format P and maintain the same in the output.
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Hi,
I need to replace all occurrences of control characters except space,newline,tabs in a string . I can give a replace statement for each of these characters but I want to avoid this by making use of regular expressions. Can anyone help me in this regard.
I tried using the following replace statements with regular expression, but i am not getting the required results:
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---> this replaces even the spaces
replace all occurrences of REGEX '[[:cntrl:]][^[:space:]]' in lv_char with space replacement count lv_count_r.
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Shankar

is there anyway to do this without using regular
expressions.. regular expressions are the last
solution for me..Remember that you can never really replace the characters of a String. Strings are immutable. Once created they cannot change.

Counting the character in a string

How to find the number of repeated characters in a string
example- 'RAJA' In this string how many 'A' are there

A beautiful query posted by a guru in this forum long time ago.
SQL> set line 1000
SQL> l
1     SELECT ename
2          , Letter_repetition
3          , COUNT(*)repetition_count
4     FROM ( SELECT ename
5                   , Letter_repetition
6                FROM emp
7               MODEL
8                     RETURN UPDATED ROWS
9                     PARTITION BY (ename)
10                     DIMENSION BY (0 i)
11                     MEASURES (ename Letter_repetition)
12                     ( Letter_repetition[FOR i FROM 1 TO
13                                         length(Letter_repetition[0])
14                                         INCREMENT 1] = SUBSTR(Letter_repetition[0],CV(i),1)
15                     )
16             )--End of FROM clause
17      GROUP BY ename
18          , Letter_repetition
19*    HAVING COUNT(*) > 1
SQL> /
ENAME                               LETTER_REPETITION                   REPETITION_COUNT
SCOTT                               T                                                  2
TURNER                              R                                                  2
ALLEN                               L                                                  2
MILLER                              L                                                  2
ADAMS                               A                                                  2
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Count of characters clipped or corrupted in conventional composition

In conventional Modern English composition there is more drawing from character codes whereas in classic Early Modern English composition there is more drawing from default glyph codes.
If Apple Advanced Typography or Microsoft OpenType is used in Apple Pages then between 10% and 40% of the source character string is clipped.
If Apple Advanced Typography or Microsoft OpenType is used in Apple TextEdit 1.4 for Mac OS X 10.4 and Apple TextEdit 1.5 for Mac OS X 10.5, then some 4% of the source character string is clipped.
These counts do not include substitution of correct public characters for incorrect private charactoids and incorrect public characters.
Because more of the drawing must be done in glyph space for Arabic and Indic than for Latin, Greek, Cyrillic and Chinese-Japanese-Korean, one would assume that the scale of the clipping is lowest for the simple scripts and highest for the complex scripts. If someone wants to spend a couple of days counting clipping for Arabic and Indic, it might be worthwhile to know the figures.
/hh

Could you explain what you mean by "clipping" and give an example of how you have gone about measuring it?
Modern imaging models distinguish between content information and imageable appearance, or in technical shorthand terms between code space and data space where the transform between code space and data space is evaluatable (: there are alternate stylistic appearances available as rendering intents in the transform) and where data space is almost any data model (: e.g. prescaled bitmap, TrueType splines, Type 1 splines, indeed any spline programming language for which the system has a scaler / similar data space format flexibility in ICC imaging).
If content information is mapped to imageable appearance, and if imageable appearance is mapped back to content information, clipping of content information can occur either in the first or in the second step of the transform. In the current context, clipping occurs because PDF does not preserve the source character string complete. The amount of clipping depends on the genre of the copy (annual report or history thesis with heavy use of tabular lower case numbers, proportional lower case numbers, superiors numbers, inferior numbers, small upper case, and lower case ligatures) and on how much of the GLYF Glyph table is drawn off the CMAP Character Map versus how much is drawn off the MORX Metamorphosis Extended or GSUB Glyph Substitution tables which is where clipping occurs since there may be no cross-font glyph identifiers for a PDF consumer to synthesise the source character string from.
Currently, the system supposedly uses cross-font glyph identifiers for CFF-housed SFNT and the CMAP for TTF-housed SFNT. Driver-level Unicode imaging works like driver-level ICC imaging, which has been much discussed. Basically, in driver-level Unicode imaging the tags of the tagged SFNT file format are stripped and the CFF font program dictionary (a font within a font) is decompressed into Type 1 and streamed into the PostScript language page description program which leaves only the PostScript name space (PostScript is deliberaty designed not to use a coded character set). If the data space is not CFF but TTF, then either the splines are converted to Type 1 (up to and including OS X 10.3) or a Type42 font is generated where it is legal among other things to strip the POST PostScript glyph name table.
With regard to the calculation method, it is basic. Download and install the Word Counter widget from SuperMagnus Software, available through the Download section of Apple's web site (for whatever reason the Word Counter applet does not appear as reliable). Set the Word Counter widget to exclude word spaces since PDF is winging word spaces anyway (they are synthesised). Input a source character string and count the characters it contains. Compose the source character string with Use Ligatures enabled (whether in Pages or Word 2008 or in TextEdit), use the Typography Palette (not for Word, though) for additional local and manual line layout that cannot be accomplished globally and automatically, save to PDF 1.3, attempt to recover the source character string using e.g. Apple Preview, copy into TextEdit to strip the formatting, copy the character string stripped of formatting into the Word Counter widget, calculate the difference, divide the source number with the difference, multiply by 100, and you have the percentage of clipped characters not counting incorrect public characters and incorrect private charactoids added by the PDF decomposition.
For example, for a passage from Shlomo Perets' discussion of defect decomposition in the Adobe OpenType Guide published in 2002 see http://www.planetpdf.com/enterprise/article.asp?ContentID=6521 and the following count :
3. HTML Composition in CMAP Character Map Table of Microsoft Georgia
copy 12pt on 14pt
To see potential problems with products that support advanced typography features, such as ligatures, small caps and old-style numerals, see the Adobe OpenType User Guide, authored with Adobe InDesign and exported directly to PDF. "2002" is present in the first page below the title—but cannot be located as since old-style figures are used. The SFNT acronym present in the first paragraph in page 2 cannot be located, as it uses small-caps. "Microsoft" is present 5 times in this document—but none of the instances can be located due to the use of ligatures (ft in this case). Even the word "This" in the opening paragraph in page 2 (line before last) cannot be located due to the use of ligatures. The more common fi, fl, ffi ligatures are searchable in the case of this document, but this is not the case in other documents using these ligatures (this depends on the applications used to author/create the PDF). While these OpenType features result in a superior typography, they should be avoided in online documents, until Acrobat Find and Search functions are enhanced to support the additional characters.
Shlomo Perets, PDF Best Practices 5: Acrobat Find & Search, PlanetPDF: 22 April 2002
type product
Adobe Minion Pro, neither ligatures, nor small capitals, nor lower case numbers ('Old Style Figures') are searchable.
typesetting logic
Microsoft OpenType logic (12pt Common Ligatures, Small Capitals, Small Capitals From Capitals, Old Style Figures).
type version
Type version 2.012; PS 002.000; core 1.0.38; makeotf.lib1.6.6565. System version Apple Mac OS X 10.4.11. Application version Apple Pages 4.0.
[Count of 1374 correct public characters.]
3.1. Composition in CMAP Character Map Table and GSUB Glyph Substitution Table of Adobe Minion Pro 2, Apple Pages 4.0, and Apple Mac OS X 10.4.11
Please see parallel PDF.
3.2. Copied PDF Decomposition Composed in CMAP Character Map Table of Microsoft Georgia : Clipped 9.8% off Source Character String
To see potential problems with products that support advanced typography features, such
as ligatures, small caps and old-style numerals, see the Adobe OpenType User Guide,
authored with Adobe InDesign and exported directly to . "" is present in the rst
page below the title—but cannot be located as since old-style gures are used. e
acronym present in the rst paragraph in page cannot be located, as it uses small-caps.
"Microso" is present times in this document—but none of the instances can be located
due to the use of ligatures ( in this case). Even the word "is" in the opening paragraph in
page (line before last) cannot be located due to the use of ligatures. e more common ,
, ligatures are searchable in the case of this document, but this is not the case in other
documents using these ligatures (this depends on the applications used to author/create the
). While these OpenType features result in a superior typography, they should be avoided
in online documents, until Acrobat Find and Search functions are enhanced to support the
additional characters.
Shlomo Perets, Best Practices : Acrobat Find & Search, PlanetPDF: April
Adobe Minion Pro, neither ligatures, nor small capitals, nor lower case numbers ('Old Style
Figures') are searchable.
Microso OpenType logic (pt Common Ligatures, Small Capitals, Small Capitals From
Capitals, Old Style Figures).
Type version .; .; core ..; makeotf.lib... System version Apple Mac
OS X ... Application version Apple Pages ..
[Count of 1239 characters, including incorrect public characters and incorrect private charactoids.]

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Sumanth Nag Kristam wrote:
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