Count the number of instances a record appears in a string

I have a move that imports Name, Department and Site from an
external CSV file, this part I managed to get working after many
weeks. I would like to be able to count the number of instances of
a specific Site “Glasgow” appears in the String. This
way I know how many lines to set.
Also can you add Dynamic text to a rollover button? I can get
the the text to display on the root but not on a button.
I am new to programming and Flash so apologies in
advance.

Hi,
length(REGEXP_REPLACE(':XXX:CCC:BBB:','[[:alnum:]]'))counts all kinds of punctuation, spaces, etc., not just colons. Change any (or all) of the colons to periods and it will still return 4. Use '[^:]' instead of '[[:alnum:]]' if you really want to count just colons.
Also, "SELECT ... FROM dual" is usually needed only in SQL*Plus or similar front end tools. In PL/SQL, you can call functions without a query, like this:
x := NVL (LENGTH (REGEXP_REPLACE (txt, '[^:]')), 0);

Similar Messages

How to count the number of instances a record appears in a string

I have a move that imports Name, Department and Site from an
external CSV file, this part I managed to get working after many
weeks. I would like to be able to count the number of instances of
a specific Site “Glasgow” appears in the String. This
way I know how many lines to set.
Also can you add Dynamic text to a rollover button? I can get
the the text to display on the root but not on a button.
I am new to programming and Flash so apologies in
advance.

Those which took time to read carefully *_iWork Formulas and Functions User Guide_* are aware of the availability of wildcard characters.
=COUNTIFI(range;"=text")
will do the trick.
Yvan KOENIG (VALLAURIS, France) lundi 26 avril 2010 17:36:34

I need to WAP to count the number of occurences of an alphabet in a string.

I need to WAP to count the number of occurences of an alphabet in a string.I tried a lot and have surfed a lot regarding this problem.
I m not the most proficient with java.but this is all i could come up with,and would appreciate some help here.I hope you guys would help me find a solution to this.
e.g String : abcabrty
Result should be
a:2
b:2
c:1
r:1
t:1
y:1
public class chkoccurences
     public static void main(String args[ ])
          String user_Data=args[0];
          int counter=0;
          try
               for(int i=0;i<user_Data.length( );i++)
                    for(int j=0;j<user_Data.length( );j++)
                         if(user_Data.charAt(i) == user_Data.charAt(j))
                         counter++;
                    System.out.println(user_Data.charAt(i)+" exists "+counter+" time(s) in the word.");
               System.out.println(" ");
          catch(ArrayIndexOutOfBoundsException e)
               System.out.println("Check the array size.");
}This is the output i get out of the program:
a exists 2 time(s) in the word.
b exists 4 time(s) in the word.
c exists 5 time(s) in the word.
a exists 7 time(s) in the word.
b exists 9 time(s) in the word.
r exists 10 time(s) in the word.
t exists 11 time(s) in the word.
y exists 12 time(s) in the word.What i think is i need an array to store the repeated characters because the repeated characters are getting counted again in the loop,if you know what i mean.
Please, i would appreciate some help here.

Criticism is welcomed
public class tests {
     final int min = 10;
     final int max = 35;
     final char[] chars = new char[] {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h',
               'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u',
               'v', 'w', 'x', 'y', 'z'};
     public static void main(String[] args){
          tests t = new tests();
          String[] strings = new String[] {"aabcze", "att3%a", ""};
          for(String s : strings){
               System.out.println(t.getAlphaCount(s));
     public String getAlphaCount(String s){
          int[] alphaCount = new int[26];
          int val;
          for(char c : s.toCharArray()){
               val = Character.getNumericValue(c);
               if( (val>=min) && (val<=max)){
                    alphaCount[val-min]++;
          StringBuilder result = new StringBuilder();
          for(int i=0; i<alphaCount.length; i++){
               if(alphaCount[i] > 0){
                    result.append(chars[i] + ":" + alphaCount[i] + ", ");
          if(result.length() != 0){
               result.delete(result.length()-2, result.length());
          return result.toString();
}

Find the Number of times a '¯'character appears in a string.

Hi,
In my staging table I am having data like below
ABL¯ABL¯0¯0¯ABL¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯
ABL¯ABQ¯480¯825¯DLS¯AMA¯ABQ¯ ¯ ¯ ¯ ¯ ¯ ¯
ABL¯ACD¯808¯1255¯DLS¯ELP¯TCS¯PHX¯ACD¯ ¯ ¯ ¯ ¯
ABL¯ADE¯1256¯471¯DLS¯AMA¯ABQ¯LSV¯ADE¯ ¯ ¯ ¯ ¯
ABL¯AFT¯1140¯1744¯DLS¯LAX¯FON¯AFT¯ ¯ ¯ ¯ ¯ ¯
ABL¯AHM¯1178¯1637¯DLS¯LAX¯AHM¯ ¯ ¯ ¯ ¯ ¯ ¯
ABL¯ALB¯1769¯1825¯DLS¯WIL¯ALB¯ ¯ ¯ ¯ ¯ ¯ ¯
ABL¯ALE¯1041¯1150¯DLS¯ALE¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯
Now I want to find the Number of times a '¯'character appears in a string.
I should get output 14
thanks & regards,
Vipin Jha
Thankx & regards, Vipin jha MCP

Hi Vipin,
You can try below sql code also
code:- len(data)-len(replace(data,'¯','')) -(len(data)-len(replace(data,' ','')))
select data,len(data)-len(replace(data,'¯','')) -(len(data)-len(replace(data,' ',''))) CountofCharacter from
select 'ABL¯ABL¯0¯0¯ABL¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ' data
UNION
SELECT 'ABL¯ABQ¯480¯825¯DLS¯AMA¯ABQ¯ ¯ ¯ ¯ ¯ ¯ ¯ '
UNION
SELECT 'ABL¯ACD¯808¯1255¯DLS¯ELP¯TCS¯PHX¯ACD¯ ¯ ¯ ¯ ¯ '
UNION
SELECT 'ABL¯ADE¯1256¯471¯DLS¯AMA¯ABQ¯LSV¯ADE¯ ¯ ¯ ¯ ¯ '
UNION
SELECT 'ABL¯AFT¯1140¯1744¯DLS¯LAX¯FON¯AFT¯ ¯ ¯ ¯ ¯ ¯ '
UNION
SELECT 'ABL¯AHM¯1178¯1637¯DLS¯LAX¯AHM¯ ¯ ¯ ¯ ¯ ¯ ¯ '
UNION
SELECT 'ABL¯ALB¯1769¯1825¯DLS¯WIL¯ALB¯ ¯ ¯ ¯ ¯ ¯ ¯ '
UNION
SELECT 'ABL¯ALE¯1041¯1150¯DLS¯ALE¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯'
) t
Thanks
Prasad

Count the number of times a character is in a string using pl/sql

I need to count the number of times ":" appers in the string ":XXX:CCC:BBB:".
I have sound some solution using SQL but I do not want the context switch.
Also I am on 10g so I can not use REGEXP_COUNT.
Any help would be great.

Hi,
length(REGEXP_REPLACE(':XXX:CCC:BBB:','[[:alnum:]]'))counts all kinds of punctuation, spaces, etc., not just colons. Change any (or all) of the colons to periods and it will still return 4. Use '[^:]' instead of '[[:alnum:]]' if you really want to count just colons.
Also, "SELECT ... FROM dual" is usually needed only in SQL*Plus or similar front end tools. In PL/SQL, you can call functions without a query, like this:
x := NVL (LENGTH (REGEXP_REPLACE (txt, '[^:]')), 0);

How do I count the number of times a word appears in a column?

What I have is a spreadsheet logging work history. Let's say the work location is "office", "home", or "travel" and I want to have a separate cell in another table for totals, count up the number of time each appears in the data for the year. So it would look like this:
Date
Location
sept 1
office
sept 2
home
sept 3
home
sept 4
travel
sept 5
office
sept 6
office
sept 7
office
sept 8
travel
sept 9
home
sept 10
office
Totals
Days
10
Office
5
Home
3
Travel
2
I just can't figure out how to develop the formula to put in the Totals column to make it work this way. If you can help me achieve this I'd really appreciate it!

Hi SpartanAntarctican,
Table 1 to record your locations
Date
Location
sept 1
office
sept 2
home
sept 3
home
sept 4
travel
sept 5
office
sept 6
office
sept 7
office
sept 8
travel
sept 9
home
sept 10
office
Table 2 to sum your locations
Totals
Days
10
Office
5
Home
3
Travel
2
In Table 2 the formula in B2
=ROWS(Table 1::A)−1
Minus 1 because there is a Header Row in Table 1.
In Table 2 the formula in B3 (and Fill Down)
=COUNTIF(Table 1::B,A3)
You can check this by adding 5 + 3 + 2 to arrive at 10
Or maybe delete Row 2 put the Days calculation into a Footer Row in Table 2
=SUM(B) .
Regards,
Ian.

Dynamic Pie Chart: Counting the number of instances for each value

Hello my name is Tim and I am at a roadblock concerning a chart for a dashboard design.
I have a list with two fields:
Platform and Status. The list is customized in InfoPath. The
Platform field is located on the form as a drop-down menu with 6 entries.
Status is a choice field with 3 selection that is updated through the lifespan of the project (once item created, it is updated to "In process", once the project is closed, the field is updated to "closed."
My hurdle is viewing these dynamically in a Pie Chart. I am trying to display the number of project's "In process" for each Platform. So, if Platform A has 6 projects "In process" and Platform B has 3 projects "In process"
how can this be displayed in a pie chart?
At the moment, I figured there are two options: Chart View WebPart or importing an Excel worksheet.
Any ideas or solution on how to accomplish this?
-Tim Dempsey

Three options:
1. Create a list that holds your statuses. Then in your original list, create a lookup to the status list. In your status list, make a lookup to the lookup. This creates a circular reference which, for whatever reason, returns a count. You can then connect
a chart web part to this. Repeat for the other column.
2. Go to your sharepoint list and export to spreadsheet. Save the file. This is a query file that you can use as a data connection. From that file, set it to auto-update and then create a chart from a pivot table. Then use Excel Services Web Part to display
the file.
3. The other option is to use Google Charts API. Connect to the list, run some jQuery and JavaScript and then export out a chart. Pretty cool tool.
Andy Wessendorf SharePoint Developer II | Rackspace [email protected]

Calculate the number of times a character appears in a string

Please help.
Is there a way to count a number of occurences of a particular character in a string? e.g. If you have 'DATABASE' as a string, can you calculate the number of times 'A' occurs in it? The answer is supposed to be 3, but can you calculate it in Oracle?
Thanks in advance.

Re: Any built-ins for occurance

Counting the number of occurences in a table column

Hi All
I have a table with a column that contains approx. 5000 6-digit codes. A number of these codes are duplicted in the column, and I want to count the number of occurences of each code. The column looks a bit like -
WCID
940042
920012
940652
940199
188949
155146
155196
174196
152148
151281
196209
174015
182163
195465
195318
182008
189589
150675
There can be mulitple instances of each WCID and I need to count the number of instances of each. I also have access to another table that also has a column of each WCID, but only once - ie no multiple instances. The second table is identical except that there are only single instances of each WCID.
I thought I could either loop through on the table to be counted, from 100000 to 999999 and count each occurence that way, but it would be very inefficient. The other way I thought would be to perhaps select a WCID from the unique table, count the occurence of that, select the next WCID from the unique table, count that and so on, however I'm not sure how to do it in PL/SQL. Perhaps select the WCID from the unique table into a cursor, loop through that and compare it with the original table and count the instances?
I hope this makes some sense, any help would be really appreciated
Thanks
Bill

Hi, Bill,
That sounds like a job for GROUP BY:
SELECT    wcid
,         COUNT (*)       AS num_found
FROM      table_x
GROUP BY wcid
ORDER BY wcid
I hope that answers your question.
If not, post CREATE TABLE and INSERT statements for a little sample data, and the results you want from that data.

How count the number of substrings in a string

Hi all,
I wonder if someone can help me and point out where I'm going wrong.
I want to count the number of substring starting with 'd' and ending with 'e' in a string supplied by a user.
currently my code is only counting the number of 'd' and 'e' characters in the string.
    public static void main (String [] args)
         Scanner input = new Scanner (System.in);
         String mstring;// string variable that the user will enter
         int i = 0; // upstep variable
         int count = 0; // int variable to count the number of substrings
         System.out.print("Enter a string: ");
         mstring = input.nextLine();
         while (i != mstring.length())
                   if(mstring.charAt(i)=='d'){count++;}
                   else if (mstring.charAt(i)=='e'){count++;}
                   i++;
         System.out.println("The total d e substrings is " + count);
}For example if the user enters adbedeaadffe the total number of substrings should be 5 but I'm getting 6 as the program is just counting the number of times it comes across 'd' and 'e'
Can anyone shed some light on this for me thanks.

Hi all I had a good few replies to this question but they seem to have been removed from the system for some reason.
I've changed my code slightly and it works but only for one round of the guard.
               Scanner input = new Scanner (System.in)
         String mstring;// main string variable that the user will enter
         int i = 0; // upstep variable
         int count = 0; // int variable to count the number of substrings
     System.out.print("Enter a string: ");
         mstring = input.nextLine();
         int y =mstring.length();
         while (i != mstring.length() && y !=0)
         if (mstring.charAt(i) == 'd' && mstring.charAt(y-1)=='e'){count = count+1;}
         y--;
         i++;
         System.out.println("The total a b substrings is " + (count));The above code will only print out one substring of adbedeaadffe but that is not what I need, I need to be able to print out all of the substrings.
Can anyone tell me where I'm going wrong.
Thanks in advance

A quick way to count the number of newlines '/n' in string of 200 chars

I am trying to establish the number of lines that a string will generate.
I can do this by counting the number of '/n' in the string. However my brute force method (shown below) is very slow.
Normally this would not be a problem on a 2800mhz Athlon (Standard) PC this takes < 1 second. However this code resides within a speed critical loop (not shown). The code shown below is a Achilles heal as far as the performance of this speed critical loop goes.
Can anyone suggest a faster way to count the number of �/n� (new lines) within a text string of around 50- 1000 chars, given that there may be 10 � 100 newline chars. Speed is a very important factor for this part of my program.
Thanks in advance
Andrew.
    int lineCount =0;
    String txt = this.getText();
    //loop throught text and count the carridge returns
    for (int i = 0; i < txt.length(); i++)
      char ch = txt.charAt(i);
      if (ch == '\n')
       lineCount ++;
    }//end forMessage was edited by:
scottie_uk
Message was edited by:
scottie_uk

Well, here is a C version. On my computer the Java version (reply 9 above) is slightly faster than C. YMMV. For stuff like this a compiler can be hard to beat even with assembler, as you need to do manual loop unrolling and method inlining which turn assembly into a maintenance nightmare.
// gcc -O6 -fomit-frame-pointer -funroll-loops -finline -o newlines.exe newlines.c
#include <stdio.h>
#include <string.h>
#if defined(__GNUC__) || defined(__unix__)
#include <time.h>
#include <sys/time.h>
#else
#include <windows.h>
#endif
#if defined(__GNUC__) || defined(__unix__)
typedef struct timeval TIMESTAMP;
void currentTime(struct timeval *time)
    gettimeofday(time, NULL);
int milliseconds(struct timeval *start, struct timeval *end)
    int usec = (end->tv_sec - start->tv_sec) * 1000000 +
     end->tv_usec - start->tv_usec;
    return (usec + 500) / 1000;
#else
typedef FILETIME TIMESTAMP;
void currentTime(FILETIME *time)
    GetSystemTimeAsFileTime(time);
int milliseconds(FILETIME *start, FILETIME *end)
    int usec = (end->dwHighDateTime - start->dwHighDateTime) * 1000000L +
     end->dwLowDateTime - start->dwLowDateTime;
    return (usec + 500) / 1000;
#endif
static int count(register char *txt)
    register int count = 0;
    register int c;
    while (c = *txt++)
     if (c == '\n')
         count++;
    return count;
static void doit(char *str)
    TIMESTAMP start, end;
    long time;
    register int n;
    int total = 0;
    currentTime(&start);
    for (n = 0; n < 1000000; n++)
     total += count(str);
    currentTime(&end);
    time = milliseconds(&start, &end);
    total *= 4;
    printf("time %ld, total %d\n", time, total);
    fflush(stdout);
int main(int argc, char **argv)
    char buf[1024];
    int n;
    for (n = 0; n < 256 / 4; n++)
     strcat(buf, "abc\n");
    for (n = 0; n < 5; n++)
     doit(buf);
}

How do I count the number of records returned in the CMIS query

How do I count the number of records returned in the query CMIS?
SELECT COUNT(*) FROM ora:t:IDC:GlobalProfile WHERE ora:p:xRegionDefinition = \'RD_PROJETOS_EXCLUSIVOS\''}
Euler Homero

Hi Euler,
interestingly enough, the reference guide for CMIS ( http://wiki.alfresco.com/wiki/CMIS_Query_Language ) that I found does not mention the COUNT function at all. On the other hand it states that: "The SELECT clause identifies which virtual columns to return in the result set. It can be either a comma-separated list of one or more queryNames of properties that are defined by queryable object types or * for all virtual columns."
There are, however, some other posts like e.g. http://alfrescoshare.wordpress.com/2010/01/20/count-the-total-number-of-documents-in-alfresco-using-sql/ which state that they could make it working.
Having asked in the WebCenter Portal forum, I assume that your content repository is WebCenter Content. The CMIS doc for the Content is available here: http://docs.oracle.com/cd/E23943_01/doc.1111/e15813.pdf (no COUNT there either). It does, however, mention explicitly that "CMIS queries return a Result Set where each Entry object will contain only the properties that were specified in the query.". This means your could rather investigate the Result Set. Note that there are also other means than CMIS how to get the requested result set (e.g. calling a search service directly via so-called RIDC).
In the given context I am also interested what your use case is. OOTB CMIS in WebCenter Portal is used, for instance, in Content Presenter, where it is content rather than "parameters" what's displayed.

Count the number of records

Hi all,
I have a BEx report that displays a list of empnos and their details based on a selection. Now I want to count the number of recrods that were selected and display the total number of records that were selected. How can this be done?
Thanks,
Sam

Dear all,
Thanks for the suggestions. Actially I had solved it in another way.
As suggested by Sharma, I had used a CKF as count, But i did not want the empnos to come as the rows with 1 displayed across each row.
I was trying to bring a summary in there.
So I had created a structure in Rows and in the structure created a new selection and in that selection included Employee.
This solved the problem that I was facing.
Thanks everyone for your time and effort!!
Cheers,
Sam

Count the number of records between two key values (BTREE)

How can I count the number of keys between two values?
I'm using python driver, and BTREE access method.
====>
ideally what I want is to average a whole time-series data set (the intervals can change) to a given number of points. The keys are the time stamps and the values are the data that needs to be averaged. I need to count the number of records between two time stamps so that I can divide that number by the number of points i need, and average the data. What is the best way to do this? Or should I just keep the intervals for the time stamp constant and use RECNO access method?
Thank you
(first post btw.. and why aren't there many people in stackoverflow who answer Berkeley DB questions?)

BDB is an embedded db and it does not have any internal counters or statistics that you could grap to use for this. You will need to do it manually.
You can create a cursor, grap the records you want, each time you get the next record you bump a counter.
If you are using RECNO, you can use a cursor to get the record number of the record (DB_GET_RECNO), and if all you data is in
sequentail records with no missing records you can figure out the total count by take last rec # - initial rec # + 1 to get a total count.
If you switch over to the SQL API, you can issue a SQL query to give you a count. Select count(*) Where .......
Since you have to grab the data anyway, then best may be to count records as you go along.
thanks
mike

Counting the number of detail records at target side...

Hi
I have a requirement where i need to caluclate the number of detail records formed and add it in trailer section...both at target side...
The output will look like this :
<Detail>
    <Field></Field>
</Detail>
<Detail>
    <Field></Field>
</Detail>
<Detail>
    <Field></Field>
</Detail>
<Trailer>
   <Count> 3 </Count>
</Trailer>
Basically, In trailer the Count field will contain the number of Detail records.
Can this be achived by a UDF?? Please help!!!

Actually teh detail node formation at target depends on a lot of nodes of source side...using Count I have already done it but due to lot of nodes @ source side it becomes kind of a mesh...
Hence I was trying to figure out a way in which if I can count the number of detail nodes at target side I can remove the present mapping...
Corect me if I am wrong but can't we declare this target detail node as global variable and then use 'Count' function to capture its occurence???
P.S : In XI 3.0 we used to have a tab to declare global variables but here in 7.1 we have Functions tab...could anybody tell me how to declare the variable in 7.1?

Count the number of instances a record appears in a string

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