Unique regular expression to check if a string contains letters and numbers

Hi all,
How can I verify if a string contains numbers AND letters using a regular expression only?
I can do that with using 3 different expressions ([0-9],[a-z],[A-Z]) but is there a unique regular expression to do that?
Thanks all

Darin.K wrote:
Missed the requirements:
single regex:
^([[:alpha:]]+[[:digit:]]+|[[:digit:]]+[[:alpha:]])[[:alnum:]]*$
You either have 1 or more digits followed by 1 or more letters or 1 or more letters followed by 1 or more digits. Once that is out of the way, the rest of the string must be all alphanumerics. The ^ and $ make sure the whole string is included in the match.
(I have not tested this at all, just typed it up hopefully I got all the brackets in the right place).
I think you just made my point. TWICE. While the lex class would be much more readable as a ring.... I know all my brackets are in the correct places and don't need to hope.
Jeff

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HI,
I am validating an EDI document like as follows
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Thanks.I had taken that into consideration.
But using regular expression I could say
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Check if a string contains another string

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Is there a way to differentiate string containing characters vs. numbers?

Hello, I am creating an application to read data from a serial port device. The problem is that the device, on startup, outputs a block of text containing characters describing the device. Shortly after, it outputs a command prompt "Command:" It is at this point that I want my VI to input the command to begin taking measurements.
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That makes it alot easier. In that case this will do the trick. You'll notice that the bottom case does not match even though it has valid vaules inside of it. The only thing that will match is if the entire string is of the form XX.XXXX. By the way, since we are only matching numbers now, you can remove the "ignore case" boolean as it no longer matters.
Message Edited by SiegeX on 11-24-2008 03:58 PM
Attachments:
regex.png ‏4 KB

Regular expression - get longest number from string

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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
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Have fun!
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Message Edited by Ben on 04-15-2009 09:23 AM
Ben Rayner
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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] :
     A newline (line feed) character ('\n'),
     A carriage-return character followed immediately by a newline character ("\r\n"),
     A standalone carriage-return character ('\r'),
     A next-line character ('\u0085'),
     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
"a", "", "b"
because there is an empty String between the \r and following \n. On the other hand, if the rule is use the pattern which matches the most characters, then the
\\r\\n
pattern will match that entire \r\n and the tokens will be
"a", "b"
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: ");
for (String line : lines) System.out.print(" \"" + line + "\"");
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)
to avoid memory wastage (since I am only interested in grouping, and not capturing).
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.

Using Regular Expressions to replace Quotes in Strings

I am writing a program that generates Java files and there are Strings that are used that contain Quotes. I want to use regular expressions to replace " with \" when it is written to the file. The code I was trying to use was:
String temp = "\"Hello\" i am a \"variable\"";
temp = temp.replaceAll("\"","\\\\\"");
however, this does not work and when i print out the code to the file the resulting code appears as:
String someVar = ""Hello" i am a "variable"";
and not as:
String someVar = "\"Hello\" i am a \"variable\"";
I am assumming my regular expression is wrong. If it is, could someone explain to me how to fix it so that it will work?
Thanks in advance.

Thanks, appearently I'm just doing something weird that I just need to look at a little bit harder.

Regular expression to check a value if it contains a specific word.

Hi All,
How can i check if a certain word exists in a value in regular expression ?
I have an attribute called Race. Race can contain the following:
White, Non-Hispanic
Black, Non-Hispanic
White, Non Hispanic
Black, Non Hispanic
White, NonHispanic
Non-Hispanic, white
Non Hispanic - black
What i want is to check if my value contains the word "NON" (NON can be at the beginning, middle or end), if it does, parse it and return it.
This is what I have, however I want to make sure it covers all cases and not missing anything else
select REGEXP_SUBSTR(UPPER(trim('Black, Non-Hispanic')), '[NON]+') from dual;Thanks in advance.

Rooney wrote:
Could you please explain what are the 2 ones's for ?The two 1 are not really needed for this. It is just taht the syntax requires those parameters when I add the fifth parameter.
http://docs.oracle.com/cd/E14072_01/server.112/e10592/functions148.htm
First 1 is where the search starts (same as in substr('Abc',1))
Second 1 is the number of occurences. Here meaning return the first occurence that was found. Replace it with 2 in my next example to see a (very slight) difference.
Also 'NON' alone will not cover all cases ?But you don't have non alone. You have regexp with non + upper. The 'i' replaces the upper. Also the output is slightly different. the 'i' version will return the same capitalization as it was found in the original. It depends a little what you want to achieve. And of cause INSTR will give the same info as your version. if the result is > 0 it means NON was found.
with testdata as (select 'White,Non-Hispanic' str from dual union all
                  select 'Non-White,nOn-Hispanic' str from dual union all
                  select 'White,Hispanic' str from dual
/* end of test data creation */
select str,
      REGEXP_SUBSTR(UPPER(TRIM(str)), 'NON') regexp1,
      REGEXP_SUBSTR(str, 'NON',1,1,'i') regexp2,
      instr(upper(str),'NON') instr
from testdata;
STR                    REGEXP1                REGEXP2                INSTR
White,Non-Hispanic     NON                    Non                        7
Non-White,Non-Hispanic NON                    Non                        1
White,Hispanic                                                           0

A regular expression to detect a blank string...

Anyone know how to write a regular expression that will detect a blank string?
I.e., in a webservice xsd I'm adding a restriction to stop the user specifying a blank string for an element in the webservice operation call.
But I can't figure out a regular expression that will detect an entirely blank string but that will on the other hand allow the string to contain blank spaces in the text.
So the restriction should not allow "" or " " to be specified but will allow "Joe Bloggs" to be specified.
I tried [^ ]* but this wont allow "Joe Bloggs" to pass.
Any ideas?
Thanks,
Ruairi.

Hi ruairiw,
there is a shortcut for the set of whitespace chars in Java. It is the Expression *\s* which is equal to *[ \t\n\f\r\u000b]*.
With this expression you can test whether a String consists only of whitespace chars.
Expamle:
String regex = "\\s*"; // the slash needs to be escaped
// same as String regex = "[ \t\n\f\r\u000b]";
System.out.println("".matches(regex)); // true
System.out.println(" ".matches(regex)); // true
System.out.println(" \r\n\t ".matches(regex)); // true
System.out.println("\n\nTom".matches(regex)); // false
System.out.println(" Tom Smith".matches(regex)); // falseBesh Wishes
esprimo

Regular Expression to Check number with at least one decimal point

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

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
WHERE   TRANSLATE ( str              -- nothing except digits and dots
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AND     str           LIKE '%.%'     -- at least 1 dot
AND     LENGTH ( REPLACE ( str       -- up to 2 digits
                )     <= 2

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')
My regular expression is: value=='ORIG'\\) The double backslashes are included as a delimiter for ')' which is a regular expression special character
However, when I call the String.matches() method for this regular expression it returns false. Where am I going wrong?
Thanks.

The string doesn't contain what you think it contains, or you made a mistake in your implementation.
public class Bar {
   public static void main(final String... args) {
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}

How check if a string contains a generic character?

I want know how i can check if this string
String mio= new String("Error 101");contains this substring
1xxwhere "xx" are two generic characters that follow "1".
Thanks
Message was edited by:
PremierITA

Use a regular expression.Pattern threeDigitNumberThatStartsWith1 = Pattern.compile(".*1\\d\\d.*");
Matcher matcher = pattern.matcher(mio);
if(matcher.matches()) {
// found it!
else {
// didn't find it
}More about pattern matching here: http://java.sun.com/javase/6/docs/api/java/util/regex/Pattern.html

Unique regular expression to check if a string contains letters and numbers

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