bash


Remove the first character from a bash variable

The following method, removes the first character from a bash variable by using a regular expression that matches any character.


VAR=${VAR#?};

The functionality ${string#substring} deletes shortest match of $substring from the front of $string.
The question mark ? matches zero or one of the previous regular expression. It is generally used for matching single characters.
So by using ? as our $substring we instruct bash to match any character at the beginning of the variable and remove it.

Example


$ VAR="Banana";
$ VAR=${VAR#?};
$ echo $VAR;
anana


Linux Bash: How to print leading zeroes on a variable

In order to print the leading zeros to a variable you need to use the function printf as follows, where %03d states that the number should consist of minimum 3 digits and thus it will put the missing leading zeros:

 printf %03d $counter

The following example renames a file that it has the following format for a filename: number.anything and adds the leading zeros in order to make it easier while sorting multiple files.
The name is contained in the variable $a.
If for example we had the file 11.txt  it will become 0011.txt

mv $a `printf %04d.%s ${a%.*} ${a##*.}`

 


Linux Bash: How to find a string inside C/C++ Source and Header Files

Just issue the following in the folder of the source codes and replace ‘FIND ME’ with the string you want to query:

find -O3 . -regex '.*\.\(c\|cpp\|h\)

You can change the contents of the regular expression to search in different file extensions (file types).  -exec grep 'FIND ME!' -sl '{}' \;

You can change the contents of the regular expression to search in different file extensions (file types).