Replace part of a string

I want to replace a part of a string value with another in table column(VARCHAR). For example \\server01\data\doc\123.eps with \\server02\data\doc\123.eps . How to replace the server01 part with server02?

Use REPLACE.
Cheers, APC
SQL> select replace('\\server01\data\doc\123.eps', 'server01', 'server02')
2 from dual
3 /
REPLACE('\\SERVER01\DATA\DO
\\server02\data\doc\123.eps
SQL>

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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.

How to replace part of a String with another String :s ?

Got a String
" Team_1/Team_2/Team_3/ "
And I want to be able to rename part of the string (they are seperated by '/'), for example I want to rename "Team_3" to "Team C", and "Team_1" to "Team A" while retaining the '/' character how would I do this?
I have tried using String.replace(oldValue, newValue); but this doesnt work, any ideas?

What do you mean that it doesn't work? You do know that the method returns the modified string? The actual string that you invoke the method on is not altered.
/Kaj

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
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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.

Replacing a part of a String with a new String

Hi everybody,
is there a option or a method to replace a part of a String with a String???
I only found the method "replace", but with this method I only can replace a char of the String. I don't need to replace only a char of a String, I have to replace a part of a String.
e.g.:
String str = "Hello you nice world!";
str.replace("nice","wonderfull");   // this won't work, because I can't replace a String with the method "replace"
                                    // with this method I'm only able to replace charsDoes anyone know some method like I need???
Thanks for your time on answering my question!!
king regards
IceCube-D

do check java 1.4 api, I think there is a method in it, however for jdk1.3 you can use
private static String replace(String str, String word,String word2) {
     if(str==null || word==null || word2 == null ||
           word.equals("") || word2.equals("") || str.equals("")) {
          return str;
     StringBuffer buff = new StringBuffer(str);
     int lastPosition = 0;
     while(lastPosition>-1) {
          int startIndex = str.indexOf(word,lastPosition);
          if(startIndex==-1) {
               break;
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          buff.delete(startIndex,startIndex+len);
          char[] charArray = word2.toCharArray();
          buff.insert(startIndex,charArray);
          str = buff.toString();
          int len2 = startIndex+word2.length();
          lastPosition = str.indexOf(word,len2);
     return buff.toString();

How to used REGEXP_REPLACE for replace part of string ?

hi
How can i replace part of string as following , i want to replace space in date by "-"
SELECT
REGEXP_REPLACE(upper('Daivd bought stuff by 2000 USD on 12 Sep 2012 from KL and left kl on 20 Sep 2012'),
'[0-9]{1,2}[^0-9](JAN|FEB|MAR|APR|JUN|JUL|AUG|SEP|OCT|NOV|DEC)[^0-9][0-9]{4}',
' ','-') "REGEXP_REPLACE"
FROM DUAL;
the output will be like this
Daivd bought stuff by 2000 USD on 12-Sep-2012 from KL and left kl on 20-Sep-2012
regards

I thought the questions is answered.
Your code will not work, because the alternate expression applies only to the four digit year and the month.
If you want to recognize both date formats with one expressions, you have to group the complete expressions.
The disadvantage of this would be, that the backreferences would not work in the same way because you would have more groups.
I advice to use two separate regular expressions for this task.
Take a look at the following example if you simply want to fill the gaps with -:
with yourtable as
select 'Daivd bought stuff by 2000 USD on 12 Sep 2012 from KL and left kl on 20 Sep 2012' text from dual union all
select 'Daivd bought stuff by 2000 USD on Sep, 20 2012 from KL and left kl on Sep, 20 2012' text from dual
SELECT
REGEXP_REPLACE(
REGEXP_REPLACE(text,
'([0-9]{1,2}) (JAN|FEB|MAR|APR|JUN|JUL|AUG|SEP|OCT|NOV|DEC) ([0-9]{4})','\1-\2-\3',1,0,'i'),
'(JAN|FEB|MAR|APR|JUN|JUL|AUG|SEP|OCT|NOV|DEC), ([0-9]{1,2}) ([0-9]{4})','\1-\2-\3',1,0,'i') regexp_replace
FROM yourtable;If you want same output-format for both date formats you could use this:
with yourtable as
select 'Daivd bought stuff by 2000 USD on 12 Sep 2012 from KL and left kl on 20 Sep 2012' text from dual union all
select 'Daivd bought stuff by 2000 USD on Sep, 20 2012 from KL and left kl on Sep, 20 2012' text from dual
SELECT
REGEXP_REPLACE(
REGEXP_REPLACE(text,
'([0-9]{1,2}) (JAN|FEB|MAR|APR|JUN|JUL|AUG|SEP|OCT|NOV|DEC) ([0-9]{4})','\1-\2-\3',1,0,'i'),
'(JAN|FEB|MAR|APR|JUN|JUL|AUG|SEP|OCT|NOV|DEC), ([0-9]{1,2}) ([0-9]{4})','\2-\1-\3',1,0,'i') regexp_replace
FROM yourtable;Edited by: hm on 25.09.2012 22:00

REPLACE part of string

Hi.
I have a string column in a DB where it's values contain the following midway through the string ([DOCUMENTGUID] is a uniqueidentifier that is different for each row):
<a href="../downloadDoc.aspx?dg=[DOCUMENTGUID]" target="_blank">
I would like to replace this part of the string with a different piece of text.
I know that I should be using PATINDEX and REPLACE functions but I don't know how to utilise.
How can I go about this please?
Thanks in advance.

Assuming the string you gave is consistent you can use this
SELECT STUFF(YourColumn,PATINDEX('%<a href="../downloadDoc.aspx?dg=%',YourCol),PATINDEX('%target="_blank">%',YourCol)- PATINDEX('%<a href="../downloadDoc.aspx?dg=%',YourCol) +1 ,'<your new text')
FROM table
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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

Get a part of the string

Hi SAP gurus
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xsl:value-of select="concat('A.B.N. ',$vABNNameValue)"/
this is the part of the xsl i am trying to change the above statement in .
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please advice its so urgent i have a production issue
i know there is a function substring
thanks

Have a look at this
http://www.w3schools.com/xpath/xpath_functions.asp#string
i guess this might be useful to you :
<b>replace(string,pattern,replace)      </b>
Returns a string that is created by replacing the given pattern with the replace argument
Example: replace("Bella Italia", "l", "")
Result: 'Bea Itaia'
in your case
<b>replace("someA.B.Nssss","A.B.N","")</b>
I think it would work. try this....
PS: award points if answer helps you.
thanks,
Pooja

T400 case number and replacement part?

I have a clients T400 that has come out of warranty. His 160GB HD died. I called Lenovo first to make sure the machine was out of warranty, the tech advise me that is way and gave me the HD FRU number and the phone number to call to get a replacement.
My colleague called IBM and gave them the part number, they said the FRU doesn't exsist and got us the correct part number but said unless we had a case number from Lenovo the drive would be sold AS-IS. So if we bought it and for someone reason it didn't work they would not warranty or exchange the drive.
So my colleague called Lenovo back who said "We don't give case numbers out for machines out of warranty". It seems that IBM thinks differently. Lenovo also said there some way of creating a case on their website which will send me an email with the correct replacement part number...do you think I could find what they are talking about??? NO!
Has anyone had to deal this mayhem? It's like the left hand doesn't know what the right hand is doing....Plus it makes not sense to not properly help people who have OOW machines that are still fairly new.
I have thought of maybe getting the one of the big bosses at IBM on the phone to fight it out with the boss at Lenovo!
Andrew
Xbase Technologies Corp.

hey xbasetech,
could you pm to me the following and i will see what i can do for you :
Name:
Country:
Mobile:
Email:
MTM [machine type model]:
(To locate MTM - http://support.lenovo.com/en_US/FindProductNumber.page#find)
S/N:
Date of Purchase:
Case/Order Number : (if any)
Screenshot of Error(if applicable) : (upload it to a hosting site and paste the link here)
Location of unit : Home / Repair Center (delete where appropriate)
Description of issue :
WW Social Media
Important Note: If you need help, post your question in the forum, and include your system type, model number and OS. Do not post your serial number.
Did someone help you today? Press the star on the left to thank them with a Kudo!
If you find a post helpful and it answers your question, please mark it as an "Accepted Solution"!
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How to send a private message? --> Check out this article.

Converting part of the string to a date and subtract with sysdate.

HINT! In order solve this you must know how the pnr is assembled. Study this:
650323-5510, we only need the first six characters. They inform us about when the person (car owner) was born. In this case it is 23 Mars 1965. You have to use several oracle built-in-functions to solve this. Hint! Begin by converting part of the string to a date and subtract with sysdate.
select to_char(to_date(cast(pnr,'YYMMDDMM'))) from car_owner;
please what am i doing wrong. i need the result to be something like this
Hans, Rosenboll, 59,6 years.

Hi.
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<code>
SELECT months_between(trunc(SYSDATE),
to_date('19' || substr('650323-5510',
1,
6),
'YYYYMMDD')) / 12 years_old
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</code>
Suppose you are expecting the age of the car owner as a result above code will give you that. One again notice the '19' I appended.
Best regards.

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This question was solved.
View Solution.

Hi:
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Replacement part number for (MCS7828I4-K9-BE7)

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Hi Samson,
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Call the Apple store and find out.
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Replace part of a string

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