Unzip to a folder with the same name as the file (without the .zip extension)

I use unar for this; by default, if an archive contains more than one top-level file or directory, it creates a directory to store the extracted contents, named after the archive in the way you describe:

unar foo.zip

You can force the creation of a directory in all cases with the -d option:

unar -d foo.zip

Alternatively, a function can do this with unzip:

unzd() {
    if [[ $# != 1 ]]; then echo I need a single argument, the name of the archive to extract; return 1; fi
    target="${1%.zip}"
    unzip "$1" -d "${target##*/}"
}

The

target=${1%.zip}

line removes the .zip extension, with no regard for anything else (so foo.zip becomes foo, and ~/foo.zip becomes ~/foo). The

${target##*/}

parameter expansion removes anything up to the last /, so ~/foo becomes foo. This means that the function extracts any .zip file to a directory named after it, in the current directory. Use unzip $1 -d "${target}" if you want to extract the archive to a directory alongside it instead.

Use unzip -d exdir zipfile.zip to extract a zipfile into a particular directory. In principle from reading your post literally you could write a function like this:

unzip_d () {
    unzip -d "$1" "$1"
}

Since you want the .zip extension removed though, you can use special variable syntax to do that:

unzip_d () {
    zipfile="$1"
    zipdir=${1%.zip}
    unzip -d "$zipdir" "$zipfile"
}