Find no numbers characters in a string

HI, how i can find no numbers characters in a string
By example.
'45125454564#4545'.
I only want to know if the string contains no number characters or not, i dont want to know the character '#'.
Is there a function module for doing that?
Thanks

You can try:
NUMERIC_CHECK
OR
IF var CO '0123456789'.
ENDIF.

Similar Messages

Find Substring + 2 characters in a string

I have the following string: 1.0000000000000000E+01
I need to find subtring E+01 and delete it from the string.
problem is thet 01 can be anything and changes.
I need an expression that will find a subtring starting E+ and two characters that come after ie 01
<removed_by_moderator>
Edited by: Julius Bussche on Sep 15, 2009 10:31 AM

Hi,
Try the below code;
Data : str(100) type C value '1.0000000000000000E+01'.
Data : len type i.
len = strlen( str ) - 4.
str = str+0(len).
This will solve the problem.
>
Maksims Jegorovcevs wrote:
> Will award points pls help.
Do not offer points its against [Forum Rules|http://wiki.sdn.sap.com/wiki/x/FgQ].
Regards
Karthik D

Find non alpha characters in a string/email

HI,
I'm on 10g database.
I want to find out a list of email address where there are non alpha characters in an email .
Example
KEY1 EMAIL
1       [email protected]
2       [email protected]
3       AXIAP#[email protected] can i find out the emails with "#" and "-" characters only.
Thanks
Edited by: user527060 on Aug 14, 2009 8:15 AM

Hey Centinul, you have a DUP_VAL_ON_INDEX ;)
It's friday after all
SQL> with t as (
2 select 1 key1, '[email protected]' email from dual union all
3 select 2,'[email protected]' from dual union all
4 select 3,'AXIAP#[email protected]' from dual
5 )
6 select key1
7 ,      email
8 from   t
9 where length(regexp_replace(substr(email, 1, instr(email, '@') -1), '[[:alnum:]]')) is not null;
      KEY1 EMAIL
         1 [email protected]
         3 AXIAP#[email protected]
2 rows selected.http://www.oracle.com/technology/oramag/webcolumns/2003/techarticles/rischert_regexp_pt1.html
http://www.oracle.com/technology/oramag/webcolumns/2003/techarticles/rischert_regexp_pt2.html
However regexp consume CPU, why not just simply:
SQL> with t as (
2 select 1 key1, '[email protected]' email from dual union all
3 select 2,'[email protected]' from dual union all
4 select 3,'AXIAP#[email protected]' from dual
5 )
6 select key1
7 ,      email
8 from   t
9 where email like '%#%'
10 or     email like '%-%';
      KEY1 EMAIL
         1 [email protected]
         3 AXIAP#[email protected]
2 rows selected.if you're only interested in finding # or - occurences in an emailaddress?

How to find 2 different characters from a string

Hi All,
I would like to remove 'T' and 'Z' from "20110428T10:33:45Z" and convert the string in date/time. I was looking at instr function but that allows to search for one character only.
Thans for the help.

With 10g
SELECT REGEXP_REPLACE ('20110428T10:33:45Z', '[T|Z]') dt
FROM DUAL
DT
2011042810:33:45
1 row selected.For the conversion
SELECT TO_DATE (REGEXP_REPLACE ('20110428T10:33:45Z', '[T|Z]'),
                'YYYYMMDDHH24:MI:SS'
               ) dt
FROM DUAL
DT
2011-04-28 10:33:45
1 row selected.

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

How to find out if there are repeated characters in a string

hey guys
well i kinda have a little problem figuring out how to find out
if there are repeated characters in a string.
if anyone can help, would appreciate it.
thanks
milos

Try using the StringTokenizer class. if u already know which character to trace. this could do the job.
eg. String str = "here and there its everywhere";
StringTokenizer st = new StringTokenizer(str, "e");
int rep = st.countTokens();

Finding the total size of a string when number of characters are known.

Hi: I am having trouble inserting a field in a record that is CLOB. The number of characters in the string is 59298. Any idea how to know what is the size of this string? I am trying to figure out if I am exceeding the size of the string. The error I get is the following
ORA-06512: at "SYS.DBMS_LOB", line 789
ORA-06512: at line 1
Second exception: ORA-21560: argument 2 is null, invalid, or out of range
ORA-06512: at "SYS.DBMS_LOB", line 789
Thanks
Ray

Either your stacktrace is incomplete or you make something wrong by copy/paste. And please, by providing sql/plsql code, use the tags [ code ] [ / code ] or [ pre ] [ / pre ], otherwise it is very hard to read.
Here is small demonstration of what i mean by 3rd parameter:
SQL> CREATE TABLE T_LOB(ID NUMBER,MESSAGE CLOB)
2 /
Table created.
SQL> DECLARE
2 l_clob CLOB;
3 BEGIN
4 INSERT INTO t_lob VALUES(1,'Hello')
5 RETURNING message INTO l_clob;
6 dbms_lob.writeappend(l_clob,length(',world'));
7 END;
8 /
dbms_lob.writeappend(l_clob,length(',world'));
ERROR at line 6:
ORA-06550: line 6, column 1:
PLS-00306: wrong number or types of arguments in call to 'WRITEAPPEND'
ORA-06550: line 6, column 1:
PL/SQL: Statement ignored
SQL> DECLARE
2 l_clob CLOB;
3 BEGIN
4 INSERT INTO t_lob VALUES(1,'Hello')
5 RETURNING message INTO l_clob;
6 dbms_lob.writeappend(l_clob,length(',world'),',world');
7 END;
8 /
PL/SQL procedure successfully completed.
SQL> SELECT * FROM t_lob
2 /
ID MESSAGE
1 Hello,worldIn your examples, after length('some lengthy string...' there is no more parameters to dbms_lob.writeappend ( or i don't see any).
Best regards
Maxim

How do I drop the last 3 characters in a string?

Hello. I'm trying to find a way to drop the last 3 characters in a string. The length varies because it is a last name field. The problem is that when a customer has a suffix on their name, III for example, it is appended to the last name all in the last name field. If I use the ProperCase function, I end up with "Smith Iii". How can I tell it to drop the last 3 characters? I'll then append an UpperCase edition of the last 3 characters to that. Thank you!

Use the instring and right string functions then UpperCase

Need help in replacing special characters in a string

Hi,
please let me know the best way to replace all the special characters in a string with space.
other than alphabets and numbers
with regards.
sumanth.

please let me know the best way to replace all the special characters in a string with space.
other than alphabets and numbers
>
Sumanth Nag Kristam wrote:
> actually i need to replace hexa decimal char 0X1A in a string.... that is 'substitue' as per the chart
> any pointers....
>
> chk the link for the ASCII codes
> http://www.techonthenet.com/ascii/chart.php
But in Hexa decimal value there is no special characters?

Scanning values of characters in a string

Hey guys,
I'm trying to figure out a way I can scan the individual characters of a string to find their value [unicode value, if it's a number, punctuation etc] and I'm a bit stumped. I know I'll have to be using the chatAt method, but I can't think of how I could implement this to continuiously scan through a string.
If someone could help point me in the right direction it would be great.
Thanks in advance.

Do you know what a loop is?
http://java.sun.com/docs/books/tutorial/java/nutsandbolts/index.html
Do you know how to find out about methods of the String class such as toCharArray?
http://java.sun.com/javase/6/docs/api/java/lang/String.html

Replace multiple characters in a string

Hi, I have some string fields that contain special characters such as ô and û and I want to replace them with ō and ū, and their upper case equivalents as well. How do I write a single formula that would find any of these characters in a string and replace them?
Thanks,
Will

replace(replace(replace(replace(x,'ô','ō'),'û','ū'),'Ô','Ö'),'Û','Ü');
where x is the string field. I suggest using the unicodes rather than the actual character. I do not think that I have the correct uppercase characters. Please ensure that they are correct in your formula.

Number of characters in a string

Hi,
I need to find the number of characters in a string. For instance, i have a string type variable with the value "ABAP". How can i know how many letters "A" there are in that string? Is there any FM that does this?
Cheers,
Roberto

Hi,
you can make use of string operation FIND as:
DATA: text1 TYPE string VALUE `A`,
         text TYPE string,
      off TYPE i,
      moff TYPE i,
      mlen TYPE i.
off = 0.
WHILE sy-subrc = 0.
FIND text1 IN SECTION OFFSET off OF
       `ABAP`
       MATCH OFFSET moff
       MATCH LENGTH mlen.
IF sy-subrc = 0.
    WRITE / moff.
    off = moff + mlen.
ENDIF.
ENDWHILE.

Repeated characters in a string????

hey guys
well i kinda have a little problem figuring out how to find out
if there are repeated characters in a string.
if anyone can help, would appreciate it.
thanks
milos

And now a working version:
    public boolean hasRepeatedChars (String word) {
        boolean found = false;
        int i = 0;
        char lastChar = '\0';
        while ((i != word.length ()) && ! found) {
            char c = word.charAt (i);
            found = ((i != 0) && (c == lastChar));
            lastChar = c;
            i ++;
        return found;
    }

Removing characters from a String (concats, substrings, etc)

This has got me pretty stumped.
I'm creating a class which takes in a string and then returns the string without vowels and without even numbers. Vowels and even characters are considered "invalid." As an example:
Input: "hello 123"
Output: "hll 13"
The output is the "valid" form of the string.
Within my class, I have various methods set up, one of which determines if a particular character is "valid" or not, and it is working fine. My problem is that I can't figure out how to essentially "erase" the invalid characters from the string. I have been trying to work with substrings and the concat method, but I can't seem to get it to do what I want. I either get an OutOfBoundsException throw, or I get a cut-up string that just isn't quite right. One time I got the method to work with "hello" (it returned "hll") but when I modified the string the method stopped working again.
Here's what I have at the moment. It doesn't work properly for all strings, but it should give you an idea of where i'm going with it.
public String getValidString(){
          String valid = str;
          int start = 0;
          for(int i=0; i<=valid.length()-1; i++){
               if(isValid(valid.charAt(i))==false){
                         String sub1 = valid.substring(start, i);
                         while(isValid(valid.charAt(i))==false)
                              i++;
                         start=i;
                         String sub2 = valid.substring(i, valid.length()-1);
                         valid = sub1.concat(sub2);
          return valid;
     }So does anybody have any advice for me or know how I can get this to work?
Thanks a lot.

Thansk for the suggestions so far, but i'm not sure how to implement some of them into my program. I probably wasn't specific enough about what all I have.
I have two classes. One is NoVowelNoEven which contains the code for handling the string. The other is simply a test class, which has my main() method, used for running the program.
Here's my class and what's involved:
public class NoVowelNoEven {
//data fields
     private String str;
//constructors
     NoVowelNoEven(){
          str="hello";
     NoVowelNoEven(String string){
          str=string;
//methods
     public String getOriginal(){
          return str;
     public String getValidString(){
          String valid = str;
          int start = 0;
          for(int i=0; i<=valid.length()-1; i++){
               if(isValid(valid.charAt(i))==false){
                         String sub1 = valid.substring(start, i);
                         while(isValid(valid.charAt(i))==false)
                              i++;
                         start=i;
                         String sub2 = valid.substring(i, valid.length()-1);
                         valid = sub1.concat(sub2);
          return valid;
     public static int countValidChar(String string){
          int valid=string.length();
          for(int i=0; i<=string.length()-1; i++){
               if(isNumber(string.charAt(i))==false){
                    if((upperVowel(string.charAt(i))==true)||lowerVowel(string.charAt(i))){
                         valid--;}
               else if(even(string.charAt(i))==true)
                    valid--;
          return valid;
     private static boolean even(char num){
          boolean even=false;
          int test=(int)num-48;
          if(test%2==0)
               even=true;
          return even;
     private static boolean isNumber(char chara){
          boolean number=false;
          for(int i=0; i<=9; i++){
               if((int)chara-48==i){
                    number=true;
          return number;
     private static boolean isValid(char chara){
          boolean valid=true;
          if(isNumber(chara)==true){
               if(even(chara)==true)
                    valid=false;
          if((upperVowel(chara)==true)||(lowerVowel(chara)==true))
               valid=false;
          return valid;
}So in my test class i'd have something like this:
public class test {
public static void main(String[] args){
     NoVowelNoEven test1 = new NoVowelNoEven();
     NoVowelNoEven test2 = new NoVowelNoEven("hello 123");
     System.out.println(test1.getOriginal());
     //This prints hello   (default constructor string is "hello")
     System.out.println(test1.countValidChar(test1.getOriginal()));
     //This prints 3
     System.out.println(test1.getValidString());
     //This SHOULD print hll
}

Validating Characters in a String

Hey everyone first post here, I have a question about validating characters in a string. I have a program where u need to enter an item code and it must have 2 letters in the begining and 3 numbers at the end, all one word.
Now I already have an error test to see whether or not the string is 5 characters and all that but i have no idea where to start on checking whether or not each character meets the criteria, i've thought of checking each character individually but I thought there had to be an easier way.

yea right now i'm completley lost, i kind of
understand regex but i just have no idea how i would
type it out, Go step by step. You want a string that has five characters. The first
test could be this (let's call that string 'string' by lack of creativity):if (string == null || string.length() != 5) return false;This is the first test: if there is no string or if it doesn't contain five
characters, all bets are off. Here's the second test:for (int i= 0; i < 2; i++)
   if (!Character.isLetter(string.charAt(i)))
      return false;If the first two characters aren't letters the string is incorrect (according
to your specifications). Here's the last test:for (int i= 2; i < 5; i++)
   if (!Character.isDigit(string.charAt(i)))
      return false;If all those tests succeed, you simply do this:return true;All you have to do now is join these code fragments in a simple method
and you're in business. Comprendo, partner?
kind regards,
Jos

Find no numbers characters in a string

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