Using findstr /v on a string enclosed by quotation marks

[gunitinug]

I echo hi logan^^ to terminal. I want to remove ^^ at the end
The following works as intended...

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C:\Users\Logan>echo hi logan^^ | findstr /v "^^^^$"
hi logan

But if I echo "hi logan^^" it doesnt work as I intended...

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C:\Users\Logan>echo "hi logan^^" | findstr /v "^^^^""$"
"hi logan^^"

I want it to return

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"hi logan

THX

[penpen]

You cannot use findstr to eliminate characters within a single line.
The reason why the CIRCUMFLEX ACCENT ('^') is disappearing is because it is an escape character, that disappears when a command is interpreted (escaping the next character consuming the escape character):
First the two accents ("^^") are reduced to one ("^") when interpreting the complete line creating the pipe and then executing the single commands (on the left and right), and
then the single one disappears when interpreting the first command.
Because both accent characters were consumed when escaping characters, no circumflex accent passed through the pipe.

You could test it by creating a file "test.txt":

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hi logan^

Then just use this command:

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type test.txt | findstr /v "^^^^$"

The accents seem to appear when using doublequotes, because doublequotes escape all characters between a matching pair, so the accents are not treated as escape characters.

Beside this the circumflex accent also is a special character in findstr, which simply means "start of line".
Actually findstr doesn't display lines with four "starts of line" and one "end of line" - which matches any line because no line could have two successive line starts without a line end.


penpen

[Squashman]

If you need inline stream editing then look at using JREPL.

You could also download a version of SED that works for Windows as well.

Another popular find and replace utility is FART. But JREPL is much more powerful.

[gunitinug]

I understand... I got confused... Now I understand findstr works on lines.

But I have another question about findstr /v.

I have persons.txt

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person1^
person1
person2^
person2

Now I try

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C:\Users\CMY\Desktop\test batch scripts\findstr\v switch>type persons.txt | findstr /v "^^$"
person1^
person1
person2^
person2

I was expecting

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person1
person2

These work as expected

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C:\Users\CMY\Desktop\test batch scripts\findstr\v switch>type persons.txt | findstr /r "[1-9]$"
person1
person2

C:\Users\CMY\Desktop\test batch scripts\findstr\v switch>type persons.txt | findstr /v "[1-9]$"
person1^
person2^

THX

[Squashman]

Read the help file. It should be evident what the /V switch does

[gunitinug]

Read the help file. It should be evident what the /V switch does

/V Print only lines that do NOT contain a match.

I already read it.

I also looked at http://www.robvanderwoude.com/escapechars.php
for escaping ^ character.

[gunitinug]

I got it

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C:\Users\CMY\Desktop\test batch scripts\findstr\v switch>type persons.txt | findstr /v "\^"
person1
person2

C:\Users\CMY\Desktop\test batch scripts\findstr\v switch>type persons.txt | findstr /r "\^"
person1^
person2^

Used \ to escape ^ character........ on the link it used ^^ which didn't work in this case.

[Squashman]

As you were told by Penpen, the caret is a special character with the findstr command and again if you read the help for the command it literally tells you to use the \ to escape characters in your search string.

[pieh-ejdsch]

Normally it is to use:

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findstr /veLc:"^"