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This article is about how DOS intercepts strings. The editor thinks it is very practical, so share it with you as a reference and follow the editor to have a look.

To illustrate the problem of intercepting strings in variables, the following example has been given:

The code is as follows:

@ echo off

Set str=123456789

The first character of echo is:% str:~0,1%

The first two characters of echo are:% str:~0,2%

The first five characters of echo are:% str:~0,5%

The string after the last character is removed by echo is:% str:~0,-1%

The string after the last three characters are removed by echo is:% str:~0,-3%

The fourth character of echo is:% str:~3,1%

The 4th and the following 3 characters of echo are:% str:~3,4%

The last character of echo is:% str:~-1%

The last character of echo is:% str:~-1,1%

The last character of echo is:% str:~-1,2%

The penultimate character of echo is:% str:~-4,1%

The penultimate character of echo and the characters after it are:% str:~-4%

The penultimate character of echo and the 1 character after it are:% str:~-4,2%

The penultimate character of echo and the following two characters are:% str:~-4,3%

Pause

In order to illustrate this problem, I take the batch processing characters here and give a further explanation, hoping to enlighten the novice.

It is as follows:

Echo var:~n,k%

Here we give a description of each parameter: "% var", that is, the string from which we want to intercept characters. "~" take the word

Sign sign (as I understand it), "n" we understand it as a pointer, "k" we understand it as an offset address. (note

Pointers and offset addresses are counted from zero)

Let's use the example of the namejm moderator to illustrate:

The code is as follows:

@ echo off

Set str=123456789

Rem defines a str string as 123456789

The first character of echo is:% str:~0,1%

The rem pointer is 0 and the offset address is 1, that is, starting with bit 0, take 1 bit

The first two characters of echo are:% str:~0,2%

The rem pointer is 0 and the offset address is 2, that is, starting with bit 0, take 2 bits

The first five characters of echo are:% str:~0,5%

The rem pointer is 0 and the offset address is 5, that is, starting with bit 0, take 5 bits

The string after the last character is removed by echo is:% str:~0,-1%

Rem when "k" is negative, we can understand it like this: take all the characters following it from the beginning of the pointer, and then subtract the following "abs (k) bit".. So we can explain this sentence as follows: take all its characters from bit 0.

Is: 123456789 and then subtracts the abs (k) bit from behind, so the final result is: 12345678

The string after the last three characters are removed by echo is:% str:~0,-3%

Rem this sentence is explained as ↑ above.

The last character of echo is:% str:~-1%

The rem parameters "n," and "k" can both be the default, and the default "n," can be understood as taking all of them from the abs (k) bit.

The penultimate character of echo and the characters after it are:% str:~-4%

Rem interpretation is the same as ↑ above

The last character of echo is:% str:~-1,1%

When rem n is negative, it means that characters are intercepted from behind and k bits are taken (n should be counted from 1 in this case)

The last character of echo is:% str:~-1,2%

Rem interpretation is the same as ↑ above

The penultimate character of echo is:% str:~-4,1%

Rem interpretation is the same as ↑ above

The penultimate character of echo and the 1 character after it are:% str:~-4,2%

Rem interpretation is the same as ↑ above

The penultimate character of echo and the following two characters are:% str:~-4,3%

Rem interpretation is the same as ↑ above

Pause

Thank you for reading! This is the end of the article on "how to intercept strings in DOS". I hope the above content can be of some help to you, so that you can learn more knowledge. if you think the article is good, you can share it for more people to see!

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