Replacing numbers in a string

Is it possible to look through a string find any 0's and replace them with a random Integer generated before?

Take a look at the methods in the String API and see
if you can find one that will replace all
occurrences of something within a string with
something else.
~@OP:
I was assuming you wanted to replace every zero with a different (pseudo) random number, in which case you can use replaceFirst(...) as I suggested.
If you want to replace every zero with the same (pseudo) random number, use yawmark's approach.

Similar Messages

Find and replace value in Delimited String

Hi All,
I have a requirement, where i need to find and replace values in delimited string.
For example, the string is "GL~1001~157747~FEB-13~CREDIT~A~N~USD~NULL~". The 4th column gives month and year. I need to replace it with previous month name. For example: "GL~1001~157747~JAN-13~CREDIT~A~N~USD~NULL~". I need to do same for last 12 months.
I thought of first devide the values and store it in variable and then after replacing it with required value, join it back.
I just wanted to know if there is any better way to do it?

for example (Assumption: the abbreviated month is the first occurance of 3 consecutive alphabetic charachters)
with testdata as (
select 'GL~1001~157747~FEB-13~CREDIT~A~N~USD~NULL~' str from dual
select
str
,regexp_substr(str, '[[:alpha:]]{3}') part
,to_date('01'||regexp_substr(str, '[[:alpha:]]{3}')||'2013', 'DDMONYYYY') part_date
,replace (str
         ,regexp_substr(str, '[[:alpha:]]{3}')
         ,to_char(add_months(to_date('01'||regexp_substr(str, '[[:alpha:]]{3}')||'2013', 'DDMONYYYY'),-1),'MON')
) res
from testdata
STR
PART
PART_DATE
RES
GL~1001~157747~FEB-13~CREDIT~A~N~USD~NULL~
FEB
02/01/2013
GL~1001~157747~JAN-13~CREDIT~A~N~USD~NULL~
with year included
with testdata as (
select 'GL~1001~157747~JAN-13~CREDIT~A~N~USD~NULL~' str from dual
select
str
,regexp_substr(str, '[[:alpha:]]{3}-\d{2}') part
,to_date(regexp_substr(str, '[[:alpha:]]{3}-\d{2}'), 'MON-YY') part_date
,replace (str
         ,regexp_substr(str, '[[:alpha:]]{3}-\d{2}')
         ,to_char(add_months(to_date(regexp_substr(str, '[[:alpha:]]{3}-\d{2}'), 'MON-YY'),-1),'MON-YY')
) res
from testdata
STR
PART
PART_DATE
RES
GL~1001~157747~JAN-13~CREDIT~A~N~USD~NULL~
JAN-13
01/01/2013
GL~1001~157747~DEC-12~CREDIT~A~N~USD~NULL~
Message was edited by: chris227 year included

How to replace numbers with text in tax return pdf using Adobe Acrobat X Pro

How do I replace numbers with text in tax return pdf using Adobe Acrobat X Pro? The tax return was created using CCH software. Thanks for your review.

Thanks Bill for your quick reply. CCH software is one of the major
suppliers of tax return software. I found an internal source that helped me
make the changes from numbers to text i.e. "$123,456" to "See Schedule O".
I am not sure if I am working in form or final text. Thanks again! Kelly

Searching for numbers in a string

I am trying to filter passwords stored in a string variable and ensure that they contain at least a number and one capital letter.
how do you search for numbers in a string and search if that string contains a character in uppercase?

function checkString(username){
var userfield = username;
var valid = '0123456789ABCDEFGHIJKLMN�OPQRSTUVWXYZ'; // define valid characters
if(!isValid(userfield.value,valid)){
alert("Invalid user name format");
userfield.value="";
userfield.select();
userfield.focus();
}else{
err = 0;
userfield = userfield.value;
function isValid(string,allowed) {
for (var i=0; i< string.length; i++) {
if (allowed.indexOf(string.charAt(i)) == -1)
return false;
return true;
}

Lua function to replace metatags in a string?

(Maybe this is a recipe request for the new cookbook site.)
I'd like a function that, given a string and a LrPhoto, will search and replace tags in the string with metadata from the photo. So, given a string 'The title of this photo is {{title}}', it would return a string where {{title}} has been replaced with the LrPhoto.getRawMetadata('title') (or maybe getFormattedMetadata). I'm sure it's been written more than once. Is there one out there to be shared before I write my own?
db

I spent the time on this myself and don't know why I bothered asking, it was so simple. Here's what I did. First, a general-purpose function to replace any token in a string with values from a table:
-- replace any string of characters in a string that are surrounded by curly
-- braces (e.g. {foo}) with the value in the table corresponding to 'foo'.
-- will replace all tokens in braces if there are more than one
function DFB.replaceTokens( tokenTable, str )
return str:gsub( '{(.-)}', function( token ) return tokenTable[token] end )
end
Then, if you want to do this with a photo's metadata, just pass the table that's returned from getRaw/FormattedMetadata:
line = DFB.replaceTokens( photo:getFormattedMetadata( nil ) , line )
So if line starts out with "my caption is {caption}" and the photo's caption is 'My Photo Caption', the result is "my caption is My Photo Caption".
I use this in a simple templating system that reads an HTML template from a file and does this for each line as it reads/writes it. Works well.

Replacing 'word' in a string

Hi! I want to replace Strings in a String, but if the the substring which has to be replaced is not a word, then i don't want it to be replaced. Eg.: I want to replace Words for other strings, where words are not substring of other Words :)
Just want to ask if someone has made this before, and can help me out
cu
athalay

But, my original question wasn't this. I
want to replace words(string) for other strings in
sentences(string). I hope it's clearer. So for example
I have a sentence: This is an applepie. And if I want
to replace apple for pear, then do not replace this
sentence for pearpie, just remain it, because there's
no "apple" word in this sentence. I hope it's clearer.
My main problem is, that handling the First word of
the sentence. It easy to replace " "+word+" " for what
I want, but there are separatorchars like , . etc :)Search for the word you want to replace. Once you have found it, you will need to do a series of other checks...
1) is it at the beginning or end of the sentence? (then you need to look at the character before or after the word depending on location in entire String to see if it is whitespace or separator character)
2) is it surrounded by whitespace or separator characters on both sides? (get the surrounding characters and compare to a list of acceptable characters to make this a word, ie " ", ",", ";"...
Only replace words that pass the conditions (beginning of sentence and followed by whitespace or separator, end of sentence and preceeded by whitespace or separator, or middle of sentence and surrounded by whitespace or separators).

How to use replace(CharacterSequence, CharacterSequence) in String class

Need simple java program using the following
how to use replace(CharacterSequence, CharacterSequence) in String class
Kindly help

From http://java.sun.com/j2se/1.5.0/docs/api/java/lang/String.html#replace(java.lang.CharSequence,%20java.lang.CharSequence)
replace
public String replace(CharSequence target,
CharSequence replacement)Replaces each substring of this string that matches the literal target sequence with the specified literal replacement sequence. The replacement proceeds from the beginning of the string to the end, for example, replacing "aa" with "b" in the string "aaa" will result in "ba" rather than "ab".
Parameters:
target - The sequence of char values to be replaced
replacement - The replacement sequence of char values
Returns:
The resulting string
Throws:
NullPointerException - if target or replacement is null.
Since:
1.5

CS3: grep search to replace part of a string

Can I do a grep search for a string of text but only replace a portion of the string while leaving the remainder intact?
iMac G5, OS 10.5.5

In addition to Eric:
Find "some search string", Replace "some replacement string " also replaces part of the string, not touching anything...
But the GREP search has two advantages: First, if you define formatting in the Replace field (italics, superscript, a character style), only the 'search'/'replacement' part will be affected, rather than the entire found string (as in regular search).
The second big advantage is you can use any regular GREP single character wildcard in the 'left' and 'right' (matching-but-not-marking) parts. A few useful examples: '\d' (digit), '\l' and '\u' (lowercase and uppercase), '.' (any character), or even '[a-f]' (lowercase 'a' to 'f'). The only limitation is that you cannot use the once, zero-or-more, or once-or-more modifiers -- the strings must have a fixed length. FYI, those three modifiers are, respectively, ?, *, and +.
This restriction does not apply to the 'middle' string, the one you are going to replace anyway -- use whatever GREP you want.

Replace frist in html string

Hi Guys,
I have an html string defined in property file.<html>,,,,,,<body>,,,,,blah blah $1 blah blah .........</html>
I have html and eg $1, $2 so on.
I want to replace $1 with another string value in the program, I tried str.replaceFirst("\\$1",string);
but didnt work,
please help me replcing it,
thanks,

DrClap wrote:
It didn't work because that method takes a regular expression as its first parameter. (The API documentation does mention that fact.) The regular expression you passed it ("$1") ...The OP did escape the $, but \\ get eaten by the forum software as a new-line char.
@OP, the replaceFirst(...) returns a new String (and does not change the original String!), so you will have to do:
String s = "<html>,,,,,,<body>,,,,,blah blah $1 blah blah .........</html>";
s = s.replaceFirst("\\$1", "XYZ");

Find & replace part of a string in Numbers using do shell script in AppleScript

Hello,
I would like to set a search-pattern with a wildcard in Applescript to find - for example - the pattern 'Table 1::$*$4' for use in a 'Search & Replace script'
The dollar signs '$' seem to be a bit of problem (refers to fixed values in Numbers & to variables in Shell ...)
Could anyone hand me a solution to this problem?
The end-goal - for now - would be to change the reference to a row-number in a lot of cells (number '4' in the pattern above should finally be replaced by 5, 6, 7, ...)
Thx.

Hi,
Here's how to do that:
try
    tell application "Numbers" to tell front document to tell active sheet
        tell (first table whose selection range's class is range)
            set sr to selection range
            set f to text returned of (display dialog "Find this in selected cells in Numbers " default answer "" with title "Find-Replace Step 1" buttons {"Cancel", "Next"})
            if f = "" then return
            set r to text returned of (display dialog "Replace '" & f & "' with " default answer f with title "Find-Replace Step 2")
            set {f, r} to my escapeForSED(f, r) -- escape some chars, create back reference for sed
            set tc to count cells of sr
            tell sr to repeat with i from 1 to tc
                tell (cell i) to try
                    set oVal to formula
                    if oVal is not missing value then set value to (my find_replace(oVal, f, r))
                end try
            end repeat
        end tell
    end tell
on error number n
    if n = -128 then return
    display dialog "Did you select cells?" buttons {"cancel"} with title "Oops!"
end try
on find_replace(t, f, r)
    do shell script "/usr/bin/sed 's~" & f & "~" & r & "~g' <<< " & (quoted form of t)
end find_replace
on escapeForSED(f, r)
    set tid to text item delimiters
    set text item delimiters to "*" -- the wildcard
    set tc1 to count (text items of f)
    set tc2 to count (text items of r)
    set text item delimiters to tid
    if (tc1 - tc2) < 0 then
        display alert "The number of wildcard in the replacement string must be equal or less than the number of wildcard in the search string."
        error -128
    end if
    -- escape search string, and create back reference for each wildcard (the wildcard is a dot in sed) --> \$.\$
    set f to do shell script "/usr/bin/sed -e 's/[]~$.^|[]/\\\\&/g;s/\\*/\\\$.\\\$/g' <<<" & quoted form of f
    -- escape the replacement string, Perl replace wildcard by two backslash and an incremented integer, to get the back reference --> \\1 \\2
    return {f, (do shell script "/usr/bin/sed -e 's/[]~$.^|[]/\\\\&/g' | /usr/bin/perl -pe '$n=1;s/\\*/\"\\\\\" . $n++/ge'<<<" & (quoted form of r))}
end escapeForSED
For what you want to do, you must have the wildcard in the same position in both string. --> find "Table 1::$*$3", replace "Table 1::$*$4"
Important, you can use no wildcard in both (the search string and the replacement string) or you can use any wildcard in the search string with no wildcard in the replacement string).
But, the number of wildcard in the replacement string must be equal or less than the number of wildcard in the search string.

How to compare numbers in a string variable?

I have a powershell script that is grabbing the output of a website and storing it into a string variable. What I would like to do is check a line in the output which contains two numbers - the number of emails sent out of the total number of emails.
I have not come up with a way to do this that works. I was thinking of maybe using a regex, but there are other numbers I don't care about that keep getting caught.
Here is a sample of what is contained in the string variable:
Email ID: 46862270 : Succeded. Time Taken: 00:00:00.0924511
Email ID: 46862272 : Succeded. Time Taken: 00:00:00.0493887
-- Sent 250 of 250 emails successfully. Total Dispatch Time: 00:00:08.1406995
--State Written to DB. Time Taken for DB Write: 00:00:00.4611053
Here is another sample:
Email ID: 46865840 : Succeded. Time Taken: 00:00:00.9333887
-- Sent 17 of 17 emails successfully. Total Dispatch Time: 00:00:01.2416905
--State Written to DB. Time Taken for DB Write: 00:00:00.0402957
So if I have the above contained in a variable called $httpResponse how can I find only the line that starts with -- Sent
and then subtract the first number from the second number?
So for example if I have this:
Email ID: 46865840 : Succeded. Time Taken: 00:00:00.9333887
-- Sent 207 of 217 emails successfully. Total Dispatch Time: 00:00:01.2416905
--State Written to DB. Time Taken for DB Write: 00:00:00.0402957
I would want the check to return 10. Is this even possible?
Any help is appreciated!

Sure.
$String = @'
Email ID: 46865840 : Succeded. Time Taken: 00:00:00.9333887
-- Sent 207 of 217 emails successfully. Total Dispatch Time: 00:00:01.2416905
--State Written to DB. Time Taken for DB Write: 00:00:00.0402957
if ( $string -match 'Sent (\d+) of (\d+) emails .+' )
{ $FailedEmailCount = [int]$matches[2] - $matches[1] }
$FailedEmailCount
10
Each set of parens will designate a capture group, and those can be referenced from $Matches after a successful -match.
Just remember that they will all be strings, and to do a math operation you have to cast at least the one on the LH side to a numeric type first.
[string](0..33|%{[char][int](46+("686552495351636652556262185355647068516270555358646562655775 0645570").substring(($_*2),2))})-replace " "

Padding numbers within a string

Hi Folks,
I have some strings which are all a mix of letters and numbers - no particular rules, but no special characters.
For ordering purposes, I would like to be able to lpad each number string to 5 digits (in reality there will never be more than 3 digits in a row)
For example:
XX01 -> XX00001
XX100 -> XX00100
XX1YY10ZZ1 ->XX00001YY00010ZZ00001
Any thoughts on how this can be achieved?
I am currently trying to get something working with REGEXP_REPLACE but cant quite get it working:
select pc.pcode,REGEXP_REPLACE(pc.pcode, '([[:digit:]]+)',lpad('\1',5,0))
from test_tab pc
order by 2
The actual/desired results are as follows:
ORIG DESIRED RESULT ACTUAL RESULT
=== =========== ==========
XX01 XX00001 XX00001 (correct)
XX10 XX00010 XX00010 (correct)
XX100 XX00100 XX000100 (incorrect)
XX1YY10ZZ1 XX00001YY00010ZZ00001 XX0001YY00010ZZ0001 (incorrect)
Thanks for any input you may have!
Matt
Edited by: Matthew Morgan on Jan 10, 2013 7:59 PM
Edited by: Matthew Morgan on Jan 10, 2013 7:59 PM
Edited by: Matthew Morgan on Jan 10, 2013 8:00 PM

Hi, Matt,
Matthew Morgan wrote:
... I am currently trying to get something working with REGEXP_REPLACE but cant quite get it working:
select pc.pcode,REGEXP_REPLACE(pc.pcode, '([[:digit:]]+)',lpad('\1',5,0))
from test_tab pc
order by 2
/In case you're wondering why that didn't work:
The arguments to REGEXP_REPLACE can be any kind of expressions of the right datatype. In particular, the 3rd argument is supposed to be a string, so it's perfectly okay to use a function that returns a string, such as
LPAD ('\1', 5, '0')in that place. (The 3rd argument to LPAD is supposed to be string. You passed a NUMBER instead, but Oracle implicitly converted that to a string for you, and no harm was done this time.)
When you pass an expression like that as an argument to REGEXP_REPLACE, the expression is evaluated first, and then it's results are passed back to REGEXP_REPLACE. What does
LPAD ('\1', 5, '0')evaluated to? It's the 5-character string:
000\1'\1' has no special meaning to LPAD; it's just a 2-character string. '\1' has special meaning in REGEXP_REPLACE, but this is not being done in the context of REGEXP_REPLACE, in fact, this has to be done before REGEXP_REPLACE can do its job. So, regardless of what's in pcode, what you wrote is equvalent to
select    pc.pcode
,       REGEXP_REPLACE ( pc.pcode
                   , '([[:digit:]]+)'
                   , '000\1'
from      test_tab pc
order by 2which adds exactly 3 0's before every number, regardless of how many digits are in that number. So when you happened to have 2-digit numbers (as in 'XX01') it happened to return the results you wanted.
chris227 wrote:
A more cautious version of frank solution could be to replace only the kind of digits added previously
SELECT       str
,       REGEXP_REPLACE ( REGEXP_REPLACE ( str
                                   , '(\d+)'
                           , '0000\1'
                   , '0+(\d{5})' --changed \d to 0
                   , '\1'
                   )     AS padded_str
FROM       str
ORDER BY padded_str
Good idea!
Edited by: Frank Kulash on Jan 11, 2013 10:28 AM

GREP search and replace: numbers and commas

I have a list of numbers 1000, 15000, 250500, etc. and I would like to find only numbers that have 5 or more digitals, and insert a comma, like the following:
1000 no change
15000 replace with 15,000
250500 replace with 250,500
TIA

One note, the replace patterns in ID need to use $, so that would be $1,$2 - otherwise I think that will work fine.
You can also use curly brackets to indicate the number of possible occurrences of a pattern you're looking for.
(\s\d{1,3})(\d{3}\s)
Means find a space, 1 to 3 digits followed by 3 more digits and another space.
One little note, if you were to put a bunch of number separated by single spaces in a string, this would skip alternate ones, because the trailing space of one number would be considered part of the find pattern for that match, and so would not be available as the leading space of the next number. (That may be very badly phrased, sorry.) In that case I think you'd have to break those spaces out to lookaheads/lookbehinds, where you include them in the find logic without making them part of the matched string. (Which may have made this response even less intelligible, but are at least terms to look at in the documentation.)

How to replace special characters in string.

Hello,
I want to replace special characters such as , or ; with any other character or " ".I find out there is no such function is java for this.There is only replace() but it accepts only chars.If anybody know how to do this?.
Thanks,

Hello,
I want to replace special characters such as , or ;
with any other character or " ".I find out there is no
such function is java for this.There is only replace()
but it accepts only chars.If anybody know how to do
this?.
Thanks,Can't you just do the following?
public class Test
     public static void main(String[] args)
     String testString = "Hi, there?";
     System.out.println(testString.replace(',',' '));
}

How can I get the last 18 numbers from this string

Hi,
I'm querying this field from a table. How can I only get the last 18 numbers:
The numbers look like this:
500000000818118

If it is always the last 18 numbers just use the Right() function. Examples:
Right("500000000818118", 18)
or
<cfset mystring = "500000000818118">
<cfset mynewstring = Right(mystring, 18)>
or
#Right(yourquerycolumn, 18)#
etc.
Edit:
Well I just noticed that your example string is only 15 characters long. Perhaps I mis-understood your question?
Message was edited by: Miguel-F

Replacing numbers in a string

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