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Top Forums Shell Programming and Scripting Replacing a variable in a loop Post 302740575 by michaelrozar17 on Thursday 6th of December 2012 11:07:47 AM
Old 12-06-2012
Replace with the below sed in the loop and try..
Code:
sed 's/k/'$i'/g' inputfile

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lreplace(n)						       Tcl Built-In Commands						       lreplace(n)

__________________________________________________________________________________________________________________________________________________

NAME
lreplace - Replace elements in a list with new elements SYNOPSIS
lreplace list first last ?element element ...? _________________________________________________________________ DESCRIPTION
lreplace returns a new list formed by replacing one or more elements of list with the element arguments. first and last are index values | specifying the first and last elements of the range to replace. The index values first and last are interpreted the same as index values | for the command string index, supporting simple index arithmetic and indices relative to the end of the list. 0 refers to the first ele- | ment of the list, and end refers to the last element of the list. If list is empty, then first and last are ignored. If first is less than zero, it is considered to refer to before the first element of the list. For non-empty lists, the element indicated by first must exist or first must indicate before the start of the list. If last is less than first, then any specified elements will be inserted into the list at the point specified by first with no elements being deleted. The element arguments specify zero or more new arguments to be added to the list in place of those that were deleted. Each element argu- ment will become a separate element of the list. If no element arguments are specified, then the elements between first and last are sim- ply deleted. If list is empty, any element arguments are added to the end of the list. EXAMPLES
Replacing an element of a list with another: % lreplace {a b c d e} 1 1 foo a foo c d e Replacing two elements of a list with three: % lreplace {a b c d e} 1 2 three more elements a three more elements d e Deleting the last element from a list in a variable: % set var {a b c d e} a b c d e % set var [lreplace $var end end] a b c d A procedure to delete a given element from a list: proc lremove {listVariable value} { upvar 1 $listVariable var set idx [lsearch -exact $var $value] set var [lreplace $var $idx $idx] } SEE ALSO
list(n), lappend(n), lindex(n), linsert(n), llength(n), lsearch(n), lset(n), lrange(n), lsort(n), string(n) | KEYWORDS
element, list, replace Tcl 7.4 lreplace(n)
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