Get the index of an item by value in a redis list

I don't know the nodejs client details for this, but the following is an implementation of a very simple indexOf command in lua.

In a my file indexof.lua i have the following code:

local key = KEYS[1]
local obj = ARGV[1]
local items = redis.call('lrange', key, 0, -1)
for i=1,#items do
    if items[i] == obj then
        return i - 1
    end
end 
return -1

Lets push a few values to a mylist.

> rpush mylist foo bar baz qux
(integer) 4

We can use the lua script to find the index of any value within the list. The command is O(N).

$ redis-cli --eval indexof.lua mylist , bar
(integer) 1

index of bar was 1

> lindex mylist 1
"bar"

index of nil is -1

$ redis-cli --eval indexof.lua mylist , nil
(integer) -1

Look at the http://redis.io/commands/eval further documentation on EVAL command.

Use sorted sets to implement a queue.

Add members and use timestamp as score.

> ZADD queue 1326990501 foo 1326990502 bar 1326990503 baz 1326990504 qux
(integer) 4

You can return members in FIFO and LIFO order by the use of ZRANGE and ZREVRANGE respectively.

FIFO:

> ZRANGE queue 0 0
"foo"

LIFO:

> ZREVRANGE queue 0 0
"qux"

To find the index of a member use ZRANK. ZRANK op is O(log(N))

> ZRANK queue bar
(integer) 1