How to split a delimited string into an array in awk?

Question

How to split the string when it contains pipe symbols | in it. I want to split them to be in array.

I tried

echo \"12:23:11\" | awk \'{s         


        

Answer 1

To split a string to an array in awk we use the function split():

 awk '{split($0, a, ":")}'
 #           ^^  ^  ^^^
 #            |  |   |
 #       string  |   delimiter
 #               |
 #               array to store the pieces

---

If no separator is given, it uses the FS, which defaults to the space:

$ awk '{split($0, a); print a[2]}' <<< "a:b c:d e"
c:d

We can give a separator, for example ::

$ awk '{split($0, a, ":"); print a[2]}' <<< "a:b c:d e"
b c

Which is equivalent to setting it through the FS:

$ awk -F: '{split($0, a); print a[1]}' <<< "a:b c:d e"
b c

In gawk you can also provide the separator as a regexp:

$ awk '{split($0, a, ":*"); print a[2]}' <<< "a:::b c::d e" #note multiple :
b c

And even see what the delimiter was on every step by using its fourth parameter:

$ awk '{split($0, a, ":*", sep); print a[2]; print sep[1]}' <<< "a:::b c::d e"
b c
:::

Let's quote the man page of GNU awk:

split(string, array [, fieldsep [, seps ] ])

Divide string into pieces separated by fieldsep and store the pieces in array and the separator strings in the seps array. The first piece is stored in array[1], the second piece in array[2], and so forth. The string value of the third argument, fieldsep, is a regexp describing where to split string (much as FS can be a regexp describing where to split input records). If fieldsep is omitted, the value of FS is used. split() returns the number of elements created. seps is a gawk extension, with seps[i] being the separator string between array[i] and array[i+1]. If fieldsep is a single space, then any leading whitespace goes into seps[0] and any trailing whitespace goes into seps[n], where n is the return value of split() (i.e., the number of elements in array).