How to prevent variable substitution in SET command

[John_L.]

I am trying to do the following:

Code: [Select]

SET I=http://www.someURL.com/My%20Webpage%28Version%20B%29.html
in a batch file. However, the second parameter to the file (%2) is being substituted in the above.

How do I prevent variable substitution in the above example?

Thanks in advance for your help,

John L.

[Lemonilla]

SET I=http://www.someURL.com/My%%%20Webpage%%%28Version%%%20B%%%29.html

you have to use the escape the %, otherwise it thinks you are referring to a veritable

for most charictors (>, |, &) you need to use ^ but with % you must use another % or surround it with %'s

[foxidrive]

you have to use the escape the %, otherwise it thinks you are referring to a veritable

for most charictors (>, |, &) you need to use ^ but with % you must use another %

That's correct.

or surround it with %'s

That bit is wrong.   Just double the percent signs.

Code: [Select]

SET I=http://www.someURL.com/My%%20Webpage%%28Version%%20B%%29.html

[Lemonilla]

interesting, triple % has always worked for me, but maybe only in certain instances.

[Salmon Trout]

interesting, triple % has always worked for me, but maybe only in certain instances.

What circumstances?

[Lemonilla]

What circumstances?

Everything I've ever written, would you like me to post some examples? I'll have to go digging.

[Salmon Trout]

would you like me to post some examples?

Yes please.

[Salmon Trout]

Here, I'll do it for you.

Code: [Select]

@echo off
echo Passed parameters:
echo Parameter 1 %1
echo Parameter 2 %2
echo.
echo Attempting to escape percent signs in string:
echo.
echo 1. No escaping
SET I=http://www.someURL.com/My%20Webpage%28Version%20B%29.html
echo %I%
echo.
echo 2. Replace one percent sign with two
SET I=http://www.someURL.com/My%%20Webpage%%28Version%%20B%%29.html
echo %I%
echo.
echo 3. Replace one percent sign with three
SET I=http://www.someURL.com/My%%%20Webpage%%%28Version%%%20B%%%29.html
echo %I%
echo.
echo 4. Replace one percent sign with four
SET I=http://www.someURL.com/My%%%%20Webpage%%%%28Version%%%%20B%%%%29.html
echo %I%
echo.

1. With parameters passed

Code: [Select]

Passed parameters:
Parameter 1 cat
Parameter 2 dog

Attempting to escape percent signs in string:

1. No escaping
http://www.someURL.com/Mydog0Webpagedog8Versiondog0Bdog9.html

2. Replace one percent sign with two
http://www.someURL.com/My%20Webpage%28Version%20B%29.html

3. Replace one percent sign with three
http://www.someURL.com/My%dog0Webpage%dog8Version%dog0B%dog9.html

4. Replace one percent sign with four
http://www.someURL.com/My%%20Webpage%%28Version%%20B%%29.html

With no parameters passed

Code: [Select]

Passed parameters:
Parameter 1
Parameter 2

Attempting to escape percent signs in string:

1. No escaping
http://www.someURL.com/My0Webpage8Version0B9.html

2. Replace one percent sign with two
http://www.someURL.com/My%20Webpage%28Version%20B%29.html

3. Replace one percent sign with three
http://www.someURL.com/My%0Webpage%8Version%0B%9.html

4. Replace one percent sign with four
http://www.someURL.com/My%%20Webpage%%28Version%%20B%%29.html

If you study the above you will see that to get ONE percent sign in the URL string using SET, you need exactly TWO percent signs in the code.

[Salmon Trout]

Put simply, 2N percents in the code are replaced by N percents in the displayed result, and 2N+1 percents are replaced by N+1 percents.