Counting particular character in a string
Hi all
how can we count the number of characters in a perticular string in simple SQL statement
example
how many commas are there in this string 'a,b,s,d,e,f,g,h'
select length('a,b,c,d,e')-length(replace('a,b,c,d,e',','))
from dual
Similar Messages
-
Count a character within a string
Give a string, what function can I use to count how many a particular character in the string. For example:
"abc;def;ghi;xyz"
I want to find out how many ";" in this string.
Thank youAnd of course, if the researched character is not in the given string, then you'll return null, to return 0, you'll need a NVL.even if it is in the string you might need a nvl in Vamsi Kasina's solution:
SQL> select length (';') - length (replace (';', ';')) cnts
from dual
CNTS
1 row selected.Better:
SQL> select nvl(length (';'),0) - nvl(length (replace (';', ';')),0) cnts
from dual
CNTS
1 -
Count of character in a string
Hi,
Here is a string.
Str = 'test,123,1-Jan-2008',sql,oracle,test,date
Can we count the number of commas(,) in the above string without passing it to any loop? I want to use replace/translate and length for this purpose.
Regards,
RiteshSQL> var a varchar2(500)
SQL> exec :a := '''test,123,1-Jan-2008'',sql,oracle,test,date'
PL/SQL procedure successfully completed.
SQL> print a
A
'test,123,1-Jan-2008',sql,oracle,test,date
SQL> select length(:a)-length(replace(:a,',')) comma_count
2 from dual
3 /
COMMA_COUNT
6 -
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 thereA 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
SQL> -
How to remove /delete a particular character in a String
Hii
i have this problem.In order to specify the number of decimal palces for a double type variable,i have used the
NumberFormat class.
My code ;
Double x=234566.4 ;
NumberFormat n1= NumberFormat.getInstance();
n1.setMaximumFractionDigits(2);
n1.setMinimumFractionDigits(2);
String st=n1.format(x);
out.print(st) ;
is giving me 2,34,566.40 which I dont want I want to have it lik 234566.40
since i will again put it in function Integer.parseInt/Double.parseDouble...so those commas are giving me errors.So can anybody suggest how to remove those commas from that string...It would be of gr8 help.thanx
Arnabhi
You can use StringTokenizer
Your code
Double x=234566.4 ;
NumberFormat n1= NumberFormat.getInstance();
n1.setMaximumFractionDigits(2);
n1.setMinimumFractionDigits(2);
String st=n1.format(x);
out.print(st) ;
StringTokenizer sT = new StringTokenizer(st,",");
String newstring="";
while(sT.hasMoreTokens())
newstring += sT.nextToken();
out.println(newstring);
this may help out
cheers
rambee -
To find out the POSITION OF Nth OCCURANCE OF A PARTICULAR CHARACTER
Hi,
Is it possible to find out the POSITION OF Nth OCCURANCE OF A PARTICULAR CHARACTER IN A STRING in crystal reports?
Eg:
I have a string abcdabcdabcd
Now I want to know the position of the 2nd occurance of u2018au2019 in the string u2018abcdabcdabcdu2019hi,
dim MainStr as String=u2018abcdabcdabcdu2019
dim SubStr as String='a' 'Give which alphabet do you want.
dim Num as integer=2 'Give which 'a' do you want. i gave for second a.
dim ResultStr as string
dim ResultNum as integer=1
dim Position as integer=0
for i as integer=0 to MainStr.length-1
ResultStr=MainStr.SubString(i,1)
if ResultStr=SubStr then
ResultNum+=1
if ResultNum=Num then
position=i+1
exitfor
endif
endif
next
hope this solves your problem
regards,
varma -
Replace a character from a string
Hi,
how can I replace a particular character from a string on a particular occurrence? Say for eg, if i wan to replace 'a' of second occurrence from 'abcabcabc', then what should i do? I guess the normal REPLACE function replaces every occurrence in a given string. So is there any other way to solve this?
Thanks!If you are in 9i:
Then you got to split the string into based on the nth position.
SQL> select replace('abddefabc','a','x') from dual;
REPLACE('
xbddefxbc
SQL> select instr('abddefabc','a',2) from dual;
INSTR('ABDDEFABC','A',2)
7
SQL> select substr('abddefabc',1,instr('abddefabc','a',2)-1),substr('abddefabc',instr('abddefabc','a',2)) from dual;
SUBSTR SUB
abddef abc
SQL> select substr('abddefabc',1,instr('abddefabc','a',2)-1),replace(substr('abddefabc',instr('abddefabc','a',2)),'a','X') as replaced from dual;
SUBSTR REP
abddef XbcOfcourse, later on you can join the string..
Jithendra -
How to count number of occurences of a character in a string
I want to get the count of occurences of a character within a string. Is there any function for this or should i go for a PLSQL.
For example is the sting is "occurences" & if i search for character "c", the answer should be 3.
Regards,
Sunil.SQL> select length('occurence') - length(replace('occurence','c')) from dual;
LENGTH('OCCURENCE')-LENGTH(REPLACE('OCCURENCE','C'))
3 -
Count the repeated character in a string
Hi Great Gurus,
Please tell me how to know how many times a given character repeats in a string.
Please mail me .
Thanks Gurus,
Rahulhi,
same thread----
Re: find number of occurances of a particular character
data: var1(30) type c value 'ghghj#ghjgjgh#ghjghjg#ghjg#'.
data : totcnt type i,
cnt type i,
v type c,
n type i.
cnt = strlen( var1 ).
do cnt times.
move var1+n(1) to v.
if v eq '#'.
totcnt = totcnt + 1.
endif.
n = n + 1.
if n = cnt .
exit.
endif.
enddo.
write:/ 'No of #s', totcnt .
Regards
Reshma -
Count the occurance of a character in a string
What is the function to count the occurance of a character in a string.
like 'test test1 test2 test3' with in this string there are 3 white spaces.
instr will give the first one, replace will change the all... and what's for the occurance (number of white spaces, which is 3 in this case) count
Message was edited by:
gladnnSQL> var a varchar2(25)
SQL> exec :a := 'gghhhh999jjjj'
 
PL/SQL procedure successfully completed.
 
SQL> select length(:a) - nvl(length(replace(:a,'h','')),0) from dual;
 
LENGTH(:A)-NVL(LENGTH(REPLACE(:A,'H','')),0)
4Rgds. -
Count the occurence of a character in a string
Post Author: halfpat
CA Forum: Formula
Hello to everyone,
I use CR 8,0. Is anyone can tell me if this is possible:
I have a string "12345-C-1 23456-C-2 34567-C-4", how can I count the occurence of the "C" character in this string ?Is the InStr function can help me to do this ?
I appreciate any suggestion,
Thanks.Post Author: bettername
CA Forum: Formula
How about measuring the length of the string, and subtracting the length of the same string where C is replaced with <nothing>?
len("12345-C-1 23456-C-2 34567-C-4") - len(replace("12345-C-1 23456-C-2 34567-C-4" , "C", "")) -
Looking for the last item in a string set in a particular character style
I should find each space that is the last character in a string set in a particular character style. For example, in the passage "123 456 789" (the underlined part here indicating a passage set in a particular character style), my ideal GREP search would yield a match after the digit 6.
How do this?Excellent! This works perfectly. I have used your string with the bold character formatting, and replaced the matches with [nothing], set in None style.
On second thought, I was surprised that it indeed had worked: why was the space replaced, and not deleted? Shouldn't I have used
(?<!.)(\s)(?=\w)
and replaced it with
$2 -
Which method to count the number of character in a string?
Hi all,
I want to know which class and method to count the number of character in a string?
Garyhttp://java.sun.com/j2se/1.4.2/docs/api/java/lang/String.html
-
Counting character in a string
I'm trying to count how many times a characters occurs in a string, and I'm having a lot of trouble. This is what I have so far:
note: the method call has to be public static int countOccs(char f, String s) and I know that it can be done using only loops, for, charAt, and length () and I want to learn it like that.
[I just chose f, s randomly.]
public class Counting {
public static void main(String [ ] args) {
String s = "Testing";
int countOccs = s.length( );
char [ ] charArray = new new char[counting];
for (int i = 0; i < counting; i++) {
charArray[i]
public static int counting(char f, String s) {
I'm really lost. When I try to define char f = "f" or String s = "Testing" in the method call (public static int counting...) I get messages saying that it has already been declared in the call, but there has been nothing assigned to it. Anyways, I feel really lost. I did find a pattern by looking at my notes from awhile ago, but am having trouble filling out parts of it. (Thanks, memory...) I think the problem comes from having to have the method call be public static int counting (char f, String s). That part confuses me a lot because I want it to print out how many characters of f are in a string, but it doesn't have "void" in it so it doesn't return a value.
for (int i=0; i < s.length(); i++)
char f = s.charAt(i);
if (something that looks at the characters)
do something like count
Edited by: MAresJonson on Feb 6, 2009 6:50 PMh1. The Ubiquitous Newbie Tips
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Counting a particular word in a string
Hi All,
Have a query to find the occurence of a particular word in a string using a query.
ex: str := 'a,b,a,c,d'
search_str := 'a'
=> need to get the number of times 'a' is getting repeated in the string str.
output: 2
Hoping for the best support as always.
Edited by: Aparna16 on Jul 17, 2012 5:55 AM
Edited by: Aparna16 on Jul 17, 2012 5:56 AMSQL> ed
Wrote file afiedt.buf
1 with sample_data as
2 (
3 select 'aaa,abcdd,abc,abc,asdasd' cola, 'abc' matchCol from dual union all
4 select 'a,b,a,c,d', 'a' from dual union all
5 select 'a,b,a,c,d', 'e' from dual union all
6 select 'aaa,abcdd,abc,abc,asdasd,abc', 'ab' from dual union all
7 select 'aaa,abcdd,abc,abc,asdasd,abc', 'abc' from dual
8 )
9 select cola, matchcol, length(cola||',') - length(replace(cola||',',matchcol||',',substr(matchcol,2)||',')) cnt
10* from sample_data
SQL> /
COLA MAT CNT
aaa,abcdd,abc,abc,asdasd abc 2
a,b,a,c,d a 2
a,b,a,c,d e 0
aaa,abcdd,abc,abc,asdasd,abc ab 0
aaa,abcdd,abc,abc,asdasd,abc abc 3
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