Get the next element in $@

I’m trying to parse args like I do in C:

if (str[i] == something && str[i + 1] == something) {
      //do somthing
}

I can’t figure out how to do the str[i + 1]. Here’s what I tried:

for a in $@ ; do
        if [ $a == "create" ] && [sed 's/table/']; then
            echo "create database"
        fi
done

I’m trying to to something if I find create and right after create if there’s database.

>Solution :

In bash, we can use indirect variables to do the same thing:

for ((i = 1, j = 2; i < $#; i++, j++)); do
    if [[ ${!i} == "create" && ${!j} == "database" ]]; then
        echo "create database
        break
    fi
done

This technique does not consume the arguments like shift does.

Ref: 4th paragraph of 3.5.3 Shell Parameter Expansion

A couple of points about your code:

for a in $@ ; do
        if [ $a == "create" ] && [sed 's/table/']; then
            echo "create database"
        fi
done

1. Always quote "$@" — that’s the only way to keep "arguments with whitespace" together.
2. in bash, use [[...]] instead of [...] — the double bracket form is more safe regarding unquoted variables.
3. the brackets are not mere syntax, [ is a command. For external commands, don’t use brackets:

   if [[ $a == "create" ]] && sed 's/table/other/' some_file; then
   

   Carefully read help if at a bash prompt.