Regular expression - get longest number from string

I believe it is easy one but I can't get it.Lets say I have a string 'A1_1000' I want to substract the 1000 using regular expression. When I feed Match regular expression I get '1' which is not the longest number. I know other ways of doing that but I want clean solution in one step. Does anybody knows the right regular expression to accomplish that? Thanks!
LV 2011, Win7
Solved!
Go to Solution.

ceties wrote:
This is the best solution I was able to come with. I am just wondering if there is "smoother way" without the cycle.
Since multiple checks are required I would tend to beieve that we do have to loop through the possibilities. in this example
I start check at offset "0" into the string for a number. Provided i find a number I check if it is longer that any previous number I found and if so save the new longer number in the shift register.
Have fun!
Ben
Message Edited by Ben on 04-15-2009 09:23 AM
Ben Rayner
I am currently active on.. MainStream Preppers
Rayner's Ridge is under construction
Attachments:
Find_Longest.PNG ‏33 KB

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Hi,
I have a problem with an select.
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Regular expression - parse version number string

Hi,
I try to parse a string using regular expressions but id did not work correctly in all cases.
The string that i want to parse can have one of the following layout:
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I use the following code to parse the version number and extract the parts:
Pattern versionPattern = Pattern.compile("(\\d+)\\.{1}(\\d+)\\.{1}(\\d+)(?:\\.{1}(\\w+))?");
Matcher m = versionPattern.matcher(versionString);
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// assert that we matched every part
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heissm wrote:
Hi,
I try to parse a string using regular expressions but id did not work correctly in all cases.
The string that i want to parse can have one of the following layout:
3.4.5.v20090305 or
3.4.5The first three parts have to be integer values, the last part can be a free string (which is optional).
I use the following code to parse the version number and extract the parts:
Pattern versionPattern = Pattern.compile("(\\d+)\\.{1}(\\d+)\\.{1}(\\d+)(?:\\.{1}(\\w+))?");
Matcher m = versionPattern.matcher(versionString);
if (!m.find()) {
throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
// assert that we matched every part
// three groups (without qualifier) or four parts (with qualifier)
int groups = m.groupCount();
if (groups != 4) {
throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
// extract the parts
major = parseInt(m.group(1));
minor = parseInt(m.group(2));
micro = parseInt(m.group(3));
qualifier = m.group(4);The above regular expression works in all cases that i tested, except one.
The follwoing string is accepted as correct input: (but it shouldn't)
3.4.5a.v20090305And i get the result:
major = 3
minor = 4
micro = 5
qualifier = a.v20090305No, that can't be the output. Perhaps you have some old class files you're executing, because with the code you now posted, that can't be the result.
To verify this, execute this:
import java.util.regex.*;
public class Main {
    public static void main(String[] args) {
        String versionString = "3.4.5a.v20090305";
        Pattern versionPattern = Pattern.compile("(\\d+)\\.{1}(\\d+)\\.{1}(\\d+)(?:\\.{1}(\\w+))?");
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        // assert that we matched every part
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        int groups = m.groupCount();
        if (groups != 4) {
             throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
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        System.out.println("2 -> "+m.group(2));
        System.out.println("3 -> "+m.group(3));
        System.out.println("4 -> "+m.group(4));
}and you'll see that the following is the result:
1 -> 3
2 -> 4
3 -> 5
4 -> nullSome remarks about your pattern:
"(\\d+)\\.{1}(\\d+)\\.{1}(\\d+)(?:\\.{1}(\\w+))?"All the "{1}" can be omitted they don't add anything to your regex and are only cluttering it up. And you'll probably also want to "anchor" your regex using the start- and end-of-string-meta-characters, like this:
"^(\\d+)\\.(\\d+)\\.(\\d+)(?:\\.(\\w+))?$"

Cannot get regular expression to return true in String.matches()

Hi,
My String that I'm attempting to match a regular expression against is: value=='ORIG')
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Thanks.

The string doesn't contain what you think it contains, or you made a mistake in your implementation.
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OR ('|') in regular expressions (e.g. split a String into lines)

Which match gets used when you use OR ('|') to specify multiple possible matches in a regex, and there are multiple matches among the supplied patterns? The first one (in the order written) which matches? Or the one which matches the most characters?
To make this concrete, suppose that you want to split a String into lines, where the line delimiters are the same as the [line terminators used by Java regex|http://java.sun.com/javase/6/docs/api/java/util/regex/Pattern.html#lt] :
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     A line-separator character ('\u2028'), or
     A paragraph-separator character ('\u2029)
This problem has [been considered before|http://forums.sun.com/thread.jspa?forumID=4&threadID=464846] .
If we ignore the idiotic microsoft two char \r\n sequence, then no problem; the Java code would be:
String[] lines = s.split("[\\n\\r\\u0085\\u2028\\u2029]");How do we add support for \r\n? If we try
String[] lines = s.split("[\\n\\r\\u0085\\u2028\\u2029]|\\r\\n");which pattern of the compound (OR) regex gets used if both match? The
[\\n\\r\\u0085\\u2028\\u2029]or the
\\r\\n?
For instance, if the above code is called when
s = "a\r\nb";and if the first pattern
[\\n\\r\\u0085\\u2028\\u2029]is used for the match when the \r is encountered, then the tokens will be
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\\r\\n
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which is what you want.
On my particular box, using jdk 1.6.0_17, if I run this code
String s = "a\r\nb";
String[] lines = s.split("[\\n\\r\\u0085\\u2028\\u2029]|\\r\\n");
System.out.print(lines.length + " lines: ");
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System.out.println();
if (true) return;the answer that I get is
3 lines: "a" "" "b"So it seems like the first listed pattern is used, if it matches.
Therefore, to get the desired behavior, it seems like I should use
"\\r\\n|[\\n\\r\\u0085\\u2028\\u2029]"instead as the pattern, since that will ensure that the 2 char sequence is first tried for matches. Indeed, if change the above code to use this pattern, it generates the desired output
2 lines: "a" "b"But what has me worried is that I cannot find any documentation concerning this "first pattern of an OR" rule. This means that maybe the Java regex engine could change in the future, which is worrisome.
The only bulletproof way that I know of to do line splitting is the complicated regex
"(?:(?<=\\r)\\n)" + "|" + "(?:\\r(?!\\n))" + "|" + "(?:\\r\\n)" + "|" + "\\u0085" + "|" + "\\u2028" + "|" + "\\u2029"Here, I use negative lookbehind and lookahead in the first two patterns to guarantee that they never match on the end or start of a \r\n, but only on isolated \n and \r chars. Thus, no matter which order the patterns above are applied by the regex engine, it will work correctly. I also used non-capturing groups
(?:X)
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Is the above complicated regex the only reliable way to do line splitting?

bbatman wrote:
Which match gets used when you use OR ('|') to specify multiple possible matches in a regex, and there are multiple matches among the supplied patterns? The first one (in the order written) which matches? Or the one which matches the most characters?
The longest match wins, normally. Except for alternation (or) as can be read from the innocent sentence
The Pattern engine performs traditional NFA-based matching with ordered alternation as occurs in Perl 5.
in the javadocs. More information can be found in Friedl's book, the relevant page of which google books shows at
[http://books.google.de/books?id=GX3w_18-JegC&pg=PA175&lpg=PA175&dq=regular+expression+%22ordered+alternation%22&source=bl&ots=PHqgNmlnM-&sig=OcDjANZKl0VpJY0igVxkQ3LXplg&hl=de&ei=Dcg7S43NIcSi_AbX-83EDQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CA0Q6AEwAA#v=onepage&q=&f=false|http://books.google.de/books?id=GX3w_18-JegC&pg=PA175&lpg=PA175&dq=regular+expression+%22ordered+alternation%22&source=bl&ots=PHqgNmlnM-&sig=OcDjANZKl0VpJY0igVxkQ3LXplg&hl=de&ei=Dcg7S43NIcSi_AbX-83EDQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CA0Q6AEwAA#v=onepage&q=&f=false]
If this link does not survive, search google for
regular expression "ordered alternation"
My first hit went right into Friedl's book.
Harald.

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When I try to simply get a number for input into a program, and I use (char)Sysytem.in.read(),
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Hope this helps - although haven't tried out your code.
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How can I get some text from string like AB_CD_9.9.9_Sometext

How Can I get
only 9.9.9 from string "AB_CD_9.9.9_Sometext" with powershell and regular expression?
Thank you,
Lei

This works.
if('AB_CD_9.9.9_Sometext' -match '\d\.\d\.\d'){$matches[0]}else{'notfound'}
¯\_(ツ)_/¯

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Hi,
I would like to use the REGEX_LIKE to check a number with up to two digits and at least one decimal point:
Ex.
10.1
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This is what I have so far. Any help would be appreciated. Thanks.
if regexp_like(v_expr, '^(\d{0,2})+(\.[0-9]{1})?$') t

Hi,
Whenever you have a question, post a little sample data (CREATE TABLE and INSERT statements, relevant columns only) for all the tables involved, and the results you want from that data.
Explain, using specific examples, how you get those results from that data.
Always say what version of Oracle you're using (e.g. 11.2.0.2.0).
See the forum FAQ: https://forums.oracle.com/message/9362002
SammyStyles wrote:
Hi,
I would like to use the REGEX_LIKE to check a number with up to two digits and at least one decimal point:
Ex.
10.1
1.1
1
12
This is what I have so far. Any help would be appreciated. Thanks.
if regexp_like(v_expr, '^(\d{0,2})+(\.[0-9]{1})?$') t
Do you really mean "up to two digits", that is, 3 or more digits is unacceptable? What if there are 0 digits? (0 is less than 2.)
Do you really mean "at least one decimal point", that is, 2, 3, 4 or more decimal points are okay? Include some examples when you post the sample data and results.
It might be more efficient without regular expressions. For example
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Regular Expression for Invalid Number

Hi everyone,
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SQL> select * from v$version;
BANNER
Oracle Database 10g Enterprise Edition Release 10.2.0.4.0 - Prod
PL/SQL Release 10.2.0.4.0 - Production
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TNS for 32-bit Windows: Version 10.2.0.4.0 - Production
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RESIDENCE_DESC,
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J

Or use DML error logging:
create table t1
(col1 number);
exec dbms_errlog.create_error_log ('t1','t1_errors')
insert into t1
with t as
(select '1' col from dual union all
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   select '0.11' col from dual union all
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   select '1d' col from dual union all
   select '1e6' col from dual union all
   select '1e6.1' col from dual union all
   select '1e' col from dual
select col
from t
log errors into t1_errors
reject limit 20
col col1 for 999,999,999.99
select * from t1;
           COL1
           1.00
           1.10
            .11
            .11
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   1,000,000.00
col col1 for a30
select * from t1_errors;
ORA_ERR_NUMBER$ ORA_ERR_MESG$                  ORA_ERR_ROWID$       OR ORA_ERR_TAG$         COL1
           1722 ORA-01722: invalid number                           I                       1,1
           1722 ORA-01722: invalid number                           I                       11a
           1722 ORA-01722: invalid number                           I                       1d
           1722 ORA-01722: invalid number                           I                       1e6.1
           1722 ORA-01722: invalid number                           I                       1e

Regular expression to remove SIDs from list

Hey everyone so I have a script and in it I try to find the current user, as well as the last user. Currently I'm using a regular expression to throw the System account SIDs and other things like that out of the list. However this doesn't seem to be taking
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How to get page number from the PDF using Javascript

Hi,
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Here each pdf has a page number and in sequential order.
The task is to check whether all the pdfs are in sequential order (i.e i, ii, iii, 1, 2, 3). If any page is missing, the script should throw an error report.
To do this task, I am writing a Javascript to get the real page number from the PDF.
Can anybody help me how to get the page number from the PDF using Javascript.
Thanks,
Gopal

The "real" page number within a PDF is the count of the physical page starting at 0, zero.
pageNum numPages
The number printed on each page is the page label.
setPageLabels getPageLabel
You will have to open each PDF and your script would need to know the page label for that file. I would expect you would need to build a 2 dimensional array of the file names and the page label for the page within that array.

Regular expression - get longest number from string

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