Excel : script to identify text in one column (based on regex) and paste it in the next column once found
I have an excel file with more than 10,000 columns of text like this(example):
Identified a potential Security Vulnerability CVE-2018-1231
Is there anyway i can identify the columns having CVE
and write the whole CVE-2018-1231
in the next column by writing a regex or formula?
excel vba excel-vba excel-formula excel-2010
add a comment |
I have an excel file with more than 10,000 columns of text like this(example):
Identified a potential Security Vulnerability CVE-2018-1231
Is there anyway i can identify the columns having CVE
and write the whole CVE-2018-1231
in the next column by writing a regex or formula?
excel vba excel-vba excel-formula excel-2010
4
Yes there are several ways of doing this. Why don't you go and do some research, try something and then post back when you have a specific problem?
– SJR
Nov 13 '18 at 12:21
1
Hi @SJR, thanks for your response. I did some research and found that it can be done using ISNUMBER and SEARCH(example): =IF(ISNUMBER(SEARCH("apple",A2)),"Apple", What i am not sure about is how do i use this in numbers after "CVE-XXXX-XXXX"?
– Hammad Qureshi
Nov 13 '18 at 12:54
Yes I don't think that will work for your situation although you can use wildcards in the SEARCH formula (but doesn't distinguish numbers from anything else). I think either the answer below or Regexp is probably the way to go here.
– SJR
Nov 13 '18 at 12:58
add a comment |
I have an excel file with more than 10,000 columns of text like this(example):
Identified a potential Security Vulnerability CVE-2018-1231
Is there anyway i can identify the columns having CVE
and write the whole CVE-2018-1231
in the next column by writing a regex or formula?
excel vba excel-vba excel-formula excel-2010
I have an excel file with more than 10,000 columns of text like this(example):
Identified a potential Security Vulnerability CVE-2018-1231
Is there anyway i can identify the columns having CVE
and write the whole CVE-2018-1231
in the next column by writing a regex or formula?
excel vba excel-vba excel-formula excel-2010
excel vba excel-vba excel-formula excel-2010
edited Nov 13 '18 at 12:56
Pᴇʜ
20.5k42650
20.5k42650
asked Nov 13 '18 at 12:16
Hammad QureshiHammad Qureshi
1
1
4
Yes there are several ways of doing this. Why don't you go and do some research, try something and then post back when you have a specific problem?
– SJR
Nov 13 '18 at 12:21
1
Hi @SJR, thanks for your response. I did some research and found that it can be done using ISNUMBER and SEARCH(example): =IF(ISNUMBER(SEARCH("apple",A2)),"Apple", What i am not sure about is how do i use this in numbers after "CVE-XXXX-XXXX"?
– Hammad Qureshi
Nov 13 '18 at 12:54
Yes I don't think that will work for your situation although you can use wildcards in the SEARCH formula (but doesn't distinguish numbers from anything else). I think either the answer below or Regexp is probably the way to go here.
– SJR
Nov 13 '18 at 12:58
add a comment |
4
Yes there are several ways of doing this. Why don't you go and do some research, try something and then post back when you have a specific problem?
– SJR
Nov 13 '18 at 12:21
1
Hi @SJR, thanks for your response. I did some research and found that it can be done using ISNUMBER and SEARCH(example): =IF(ISNUMBER(SEARCH("apple",A2)),"Apple", What i am not sure about is how do i use this in numbers after "CVE-XXXX-XXXX"?
– Hammad Qureshi
Nov 13 '18 at 12:54
Yes I don't think that will work for your situation although you can use wildcards in the SEARCH formula (but doesn't distinguish numbers from anything else). I think either the answer below or Regexp is probably the way to go here.
– SJR
Nov 13 '18 at 12:58
4
4
Yes there are several ways of doing this. Why don't you go and do some research, try something and then post back when you have a specific problem?
– SJR
Nov 13 '18 at 12:21
Yes there are several ways of doing this. Why don't you go and do some research, try something and then post back when you have a specific problem?
– SJR
Nov 13 '18 at 12:21
1
1
Hi @SJR, thanks for your response. I did some research and found that it can be done using ISNUMBER and SEARCH(example): =IF(ISNUMBER(SEARCH("apple",A2)),"Apple", What i am not sure about is how do i use this in numbers after "CVE-XXXX-XXXX"?
– Hammad Qureshi
Nov 13 '18 at 12:54
Hi @SJR, thanks for your response. I did some research and found that it can be done using ISNUMBER and SEARCH(example): =IF(ISNUMBER(SEARCH("apple",A2)),"Apple", What i am not sure about is how do i use this in numbers after "CVE-XXXX-XXXX"?
– Hammad Qureshi
Nov 13 '18 at 12:54
Yes I don't think that will work for your situation although you can use wildcards in the SEARCH formula (but doesn't distinguish numbers from anything else). I think either the answer below or Regexp is probably the way to go here.
– SJR
Nov 13 '18 at 12:58
Yes I don't think that will work for your situation although you can use wildcards in the SEARCH formula (but doesn't distinguish numbers from anything else). I think either the answer below or Regexp is probably the way to go here.
– SJR
Nov 13 '18 at 12:58
add a comment |
2 Answers
2
active
oldest
votes
for cell A1 containing your example:
=if(isnumber(search("CVE";A1));right(A1;13);"")
with , as delimiter
=if(isnumber(search("CVE",A1)),right(A1,13),"")
Hi, Many thanks for your response. I tried using this but didn't work. please have a look here.
– Hammad Qureshi
Nov 13 '18 at 12:34
1
possibly due to the fact that you are using , as delimiter instead of ; like I do. Change all ; to , and try again
– Lambik
Nov 13 '18 at 12:41
2
The formulas are missing a closing parentheses (and therefore cannot work).
– Pᴇʜ
Nov 13 '18 at 13:02
corrected the missing parentheses, thanks for the remark
– Lambik
Nov 13 '18 at 14:55
add a comment |
Use this formula in column B (assuming column A contains the data)
=IFERROR(MID(A1,SEARCH("CVE",A1),13),"")
This will even work if the CVE
is not necessarily in the end. It will work for both examples below:
Column A | Column B
Identified a potential Security Vulnerability CVE-2018-1231 | CVE-2018-1231
Identified a potential Security Vulnerability CVE-2018-1232 aeuia e | CVE-2018-1232
If your CVE number is not constantly 13 characters you must use:
=IFERROR(MID(A1,SEARCH("CVE",A1),IFERROR(SEARCH(")",A1,SEARCH("CVE",A1)+1),IFERROR(SEARCH(" ",A1,SEARCH("CVE",A1)+1),LEN(A1)))-SEARCH("CVE",A1)),"")
This formula will cut out the CVE until a closing parentheses or a space comes or until the end of the text.
Hi, many thanks for your answer. You almost made it work. :)
– Hammad Qureshi
Nov 13 '18 at 14:05
If it solved your issue please mark it as solution: Accepting Answers: How does it work?
– Pᴇʜ
Nov 13 '18 at 14:06
Just one last question. We are using 13 digits string in output, and considering "1231" (the last part) will only be 4 digits, when i have 5 digit long string in the last part and change the "13" digit limit to 14, it takes one extra character in the end f our digit as well. HERES an example. Can you suggested something to remediate this?
– Hammad Qureshi
Nov 13 '18 at 14:14
As you will see, the last ")" in B1 is useless.
– Hammad Qureshi
Nov 13 '18 at 14:15
@HammadQureshi see my edit.
– Pᴇʜ
Nov 13 '18 at 14:23
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
for cell A1 containing your example:
=if(isnumber(search("CVE";A1));right(A1;13);"")
with , as delimiter
=if(isnumber(search("CVE",A1)),right(A1,13),"")
Hi, Many thanks for your response. I tried using this but didn't work. please have a look here.
– Hammad Qureshi
Nov 13 '18 at 12:34
1
possibly due to the fact that you are using , as delimiter instead of ; like I do. Change all ; to , and try again
– Lambik
Nov 13 '18 at 12:41
2
The formulas are missing a closing parentheses (and therefore cannot work).
– Pᴇʜ
Nov 13 '18 at 13:02
corrected the missing parentheses, thanks for the remark
– Lambik
Nov 13 '18 at 14:55
add a comment |
for cell A1 containing your example:
=if(isnumber(search("CVE";A1));right(A1;13);"")
with , as delimiter
=if(isnumber(search("CVE",A1)),right(A1,13),"")
Hi, Many thanks for your response. I tried using this but didn't work. please have a look here.
– Hammad Qureshi
Nov 13 '18 at 12:34
1
possibly due to the fact that you are using , as delimiter instead of ; like I do. Change all ; to , and try again
– Lambik
Nov 13 '18 at 12:41
2
The formulas are missing a closing parentheses (and therefore cannot work).
– Pᴇʜ
Nov 13 '18 at 13:02
corrected the missing parentheses, thanks for the remark
– Lambik
Nov 13 '18 at 14:55
add a comment |
for cell A1 containing your example:
=if(isnumber(search("CVE";A1));right(A1;13);"")
with , as delimiter
=if(isnumber(search("CVE",A1)),right(A1,13),"")
for cell A1 containing your example:
=if(isnumber(search("CVE";A1));right(A1;13);"")
with , as delimiter
=if(isnumber(search("CVE",A1)),right(A1,13),"")
edited Nov 13 '18 at 14:55
answered Nov 13 '18 at 12:24
LambikLambik
453410
453410
Hi, Many thanks for your response. I tried using this but didn't work. please have a look here.
– Hammad Qureshi
Nov 13 '18 at 12:34
1
possibly due to the fact that you are using , as delimiter instead of ; like I do. Change all ; to , and try again
– Lambik
Nov 13 '18 at 12:41
2
The formulas are missing a closing parentheses (and therefore cannot work).
– Pᴇʜ
Nov 13 '18 at 13:02
corrected the missing parentheses, thanks for the remark
– Lambik
Nov 13 '18 at 14:55
add a comment |
Hi, Many thanks for your response. I tried using this but didn't work. please have a look here.
– Hammad Qureshi
Nov 13 '18 at 12:34
1
possibly due to the fact that you are using , as delimiter instead of ; like I do. Change all ; to , and try again
– Lambik
Nov 13 '18 at 12:41
2
The formulas are missing a closing parentheses (and therefore cannot work).
– Pᴇʜ
Nov 13 '18 at 13:02
corrected the missing parentheses, thanks for the remark
– Lambik
Nov 13 '18 at 14:55
Hi, Many thanks for your response. I tried using this but didn't work. please have a look here.
– Hammad Qureshi
Nov 13 '18 at 12:34
Hi, Many thanks for your response. I tried using this but didn't work. please have a look here.
– Hammad Qureshi
Nov 13 '18 at 12:34
1
1
possibly due to the fact that you are using , as delimiter instead of ; like I do. Change all ; to , and try again
– Lambik
Nov 13 '18 at 12:41
possibly due to the fact that you are using , as delimiter instead of ; like I do. Change all ; to , and try again
– Lambik
Nov 13 '18 at 12:41
2
2
The formulas are missing a closing parentheses (and therefore cannot work).
– Pᴇʜ
Nov 13 '18 at 13:02
The formulas are missing a closing parentheses (and therefore cannot work).
– Pᴇʜ
Nov 13 '18 at 13:02
corrected the missing parentheses, thanks for the remark
– Lambik
Nov 13 '18 at 14:55
corrected the missing parentheses, thanks for the remark
– Lambik
Nov 13 '18 at 14:55
add a comment |
Use this formula in column B (assuming column A contains the data)
=IFERROR(MID(A1,SEARCH("CVE",A1),13),"")
This will even work if the CVE
is not necessarily in the end. It will work for both examples below:
Column A | Column B
Identified a potential Security Vulnerability CVE-2018-1231 | CVE-2018-1231
Identified a potential Security Vulnerability CVE-2018-1232 aeuia e | CVE-2018-1232
If your CVE number is not constantly 13 characters you must use:
=IFERROR(MID(A1,SEARCH("CVE",A1),IFERROR(SEARCH(")",A1,SEARCH("CVE",A1)+1),IFERROR(SEARCH(" ",A1,SEARCH("CVE",A1)+1),LEN(A1)))-SEARCH("CVE",A1)),"")
This formula will cut out the CVE until a closing parentheses or a space comes or until the end of the text.
Hi, many thanks for your answer. You almost made it work. :)
– Hammad Qureshi
Nov 13 '18 at 14:05
If it solved your issue please mark it as solution: Accepting Answers: How does it work?
– Pᴇʜ
Nov 13 '18 at 14:06
Just one last question. We are using 13 digits string in output, and considering "1231" (the last part) will only be 4 digits, when i have 5 digit long string in the last part and change the "13" digit limit to 14, it takes one extra character in the end f our digit as well. HERES an example. Can you suggested something to remediate this?
– Hammad Qureshi
Nov 13 '18 at 14:14
As you will see, the last ")" in B1 is useless.
– Hammad Qureshi
Nov 13 '18 at 14:15
@HammadQureshi see my edit.
– Pᴇʜ
Nov 13 '18 at 14:23
add a comment |
Use this formula in column B (assuming column A contains the data)
=IFERROR(MID(A1,SEARCH("CVE",A1),13),"")
This will even work if the CVE
is not necessarily in the end. It will work for both examples below:
Column A | Column B
Identified a potential Security Vulnerability CVE-2018-1231 | CVE-2018-1231
Identified a potential Security Vulnerability CVE-2018-1232 aeuia e | CVE-2018-1232
If your CVE number is not constantly 13 characters you must use:
=IFERROR(MID(A1,SEARCH("CVE",A1),IFERROR(SEARCH(")",A1,SEARCH("CVE",A1)+1),IFERROR(SEARCH(" ",A1,SEARCH("CVE",A1)+1),LEN(A1)))-SEARCH("CVE",A1)),"")
This formula will cut out the CVE until a closing parentheses or a space comes or until the end of the text.
Hi, many thanks for your answer. You almost made it work. :)
– Hammad Qureshi
Nov 13 '18 at 14:05
If it solved your issue please mark it as solution: Accepting Answers: How does it work?
– Pᴇʜ
Nov 13 '18 at 14:06
Just one last question. We are using 13 digits string in output, and considering "1231" (the last part) will only be 4 digits, when i have 5 digit long string in the last part and change the "13" digit limit to 14, it takes one extra character in the end f our digit as well. HERES an example. Can you suggested something to remediate this?
– Hammad Qureshi
Nov 13 '18 at 14:14
As you will see, the last ")" in B1 is useless.
– Hammad Qureshi
Nov 13 '18 at 14:15
@HammadQureshi see my edit.
– Pᴇʜ
Nov 13 '18 at 14:23
add a comment |
Use this formula in column B (assuming column A contains the data)
=IFERROR(MID(A1,SEARCH("CVE",A1),13),"")
This will even work if the CVE
is not necessarily in the end. It will work for both examples below:
Column A | Column B
Identified a potential Security Vulnerability CVE-2018-1231 | CVE-2018-1231
Identified a potential Security Vulnerability CVE-2018-1232 aeuia e | CVE-2018-1232
If your CVE number is not constantly 13 characters you must use:
=IFERROR(MID(A1,SEARCH("CVE",A1),IFERROR(SEARCH(")",A1,SEARCH("CVE",A1)+1),IFERROR(SEARCH(" ",A1,SEARCH("CVE",A1)+1),LEN(A1)))-SEARCH("CVE",A1)),"")
This formula will cut out the CVE until a closing parentheses or a space comes or until the end of the text.
Use this formula in column B (assuming column A contains the data)
=IFERROR(MID(A1,SEARCH("CVE",A1),13),"")
This will even work if the CVE
is not necessarily in the end. It will work for both examples below:
Column A | Column B
Identified a potential Security Vulnerability CVE-2018-1231 | CVE-2018-1231
Identified a potential Security Vulnerability CVE-2018-1232 aeuia e | CVE-2018-1232
If your CVE number is not constantly 13 characters you must use:
=IFERROR(MID(A1,SEARCH("CVE",A1),IFERROR(SEARCH(")",A1,SEARCH("CVE",A1)+1),IFERROR(SEARCH(" ",A1,SEARCH("CVE",A1)+1),LEN(A1)))-SEARCH("CVE",A1)),"")
This formula will cut out the CVE until a closing parentheses or a space comes or until the end of the text.
edited Nov 13 '18 at 14:55
answered Nov 13 '18 at 13:04
PᴇʜPᴇʜ
20.5k42650
20.5k42650
Hi, many thanks for your answer. You almost made it work. :)
– Hammad Qureshi
Nov 13 '18 at 14:05
If it solved your issue please mark it as solution: Accepting Answers: How does it work?
– Pᴇʜ
Nov 13 '18 at 14:06
Just one last question. We are using 13 digits string in output, and considering "1231" (the last part) will only be 4 digits, when i have 5 digit long string in the last part and change the "13" digit limit to 14, it takes one extra character in the end f our digit as well. HERES an example. Can you suggested something to remediate this?
– Hammad Qureshi
Nov 13 '18 at 14:14
As you will see, the last ")" in B1 is useless.
– Hammad Qureshi
Nov 13 '18 at 14:15
@HammadQureshi see my edit.
– Pᴇʜ
Nov 13 '18 at 14:23
add a comment |
Hi, many thanks for your answer. You almost made it work. :)
– Hammad Qureshi
Nov 13 '18 at 14:05
If it solved your issue please mark it as solution: Accepting Answers: How does it work?
– Pᴇʜ
Nov 13 '18 at 14:06
Just one last question. We are using 13 digits string in output, and considering "1231" (the last part) will only be 4 digits, when i have 5 digit long string in the last part and change the "13" digit limit to 14, it takes one extra character in the end f our digit as well. HERES an example. Can you suggested something to remediate this?
– Hammad Qureshi
Nov 13 '18 at 14:14
As you will see, the last ")" in B1 is useless.
– Hammad Qureshi
Nov 13 '18 at 14:15
@HammadQureshi see my edit.
– Pᴇʜ
Nov 13 '18 at 14:23
Hi, many thanks for your answer. You almost made it work. :)
– Hammad Qureshi
Nov 13 '18 at 14:05
Hi, many thanks for your answer. You almost made it work. :)
– Hammad Qureshi
Nov 13 '18 at 14:05
If it solved your issue please mark it as solution: Accepting Answers: How does it work?
– Pᴇʜ
Nov 13 '18 at 14:06
If it solved your issue please mark it as solution: Accepting Answers: How does it work?
– Pᴇʜ
Nov 13 '18 at 14:06
Just one last question. We are using 13 digits string in output, and considering "1231" (the last part) will only be 4 digits, when i have 5 digit long string in the last part and change the "13" digit limit to 14, it takes one extra character in the end f our digit as well. HERES an example. Can you suggested something to remediate this?
– Hammad Qureshi
Nov 13 '18 at 14:14
Just one last question. We are using 13 digits string in output, and considering "1231" (the last part) will only be 4 digits, when i have 5 digit long string in the last part and change the "13" digit limit to 14, it takes one extra character in the end f our digit as well. HERES an example. Can you suggested something to remediate this?
– Hammad Qureshi
Nov 13 '18 at 14:14
As you will see, the last ")" in B1 is useless.
– Hammad Qureshi
Nov 13 '18 at 14:15
As you will see, the last ")" in B1 is useless.
– Hammad Qureshi
Nov 13 '18 at 14:15
@HammadQureshi see my edit.
– Pᴇʜ
Nov 13 '18 at 14:23
@HammadQureshi see my edit.
– Pᴇʜ
Nov 13 '18 at 14:23
add a comment |
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4
Yes there are several ways of doing this. Why don't you go and do some research, try something and then post back when you have a specific problem?
– SJR
Nov 13 '18 at 12:21
1
Hi @SJR, thanks for your response. I did some research and found that it can be done using ISNUMBER and SEARCH(example): =IF(ISNUMBER(SEARCH("apple",A2)),"Apple", What i am not sure about is how do i use this in numbers after "CVE-XXXX-XXXX"?
– Hammad Qureshi
Nov 13 '18 at 12:54
Yes I don't think that will work for your situation although you can use wildcards in the SEARCH formula (but doesn't distinguish numbers from anything else). I think either the answer below or Regexp is probably the way to go here.
– SJR
Nov 13 '18 at 12:58