help me

rinku · #1 (**permalink**) 05-28-2007

how can I print A to Z using for loop or If condition.
Exmp:
I have one file name 098A. I run one script which check that file name
Now every time one script run the file name(098A) will change and it will look like 098B then 098C.... the atlast 098Z.
please help me.

matrixmadhan · #2 (**permalink**) 05-28-2007

Code:

awk '{ for(i=65; i<=90; i++) { printf "%c\n", i }; exit }' /bin/ls

rinku · #3 (**permalink**) 05-28-2007

Thanx..
If I want to store The value of "%c" in one variable without print this then what is the command. please tell me as earli as possible.

matrixmadhan · #4 (**permalink**) 05-28-2007

Quote:

please tell me as earli as possible.

This is not encouraged.

Looking for something like this,

Just to store,

Code:

awk '{ for(i=65; i<=90; i++) {  arr[i]=i } }' /bin/ls

To store and display

Code:

awk '{ for(i=65; i<=90; i++) {  arr[i]=i } }END{ for ( i in arr ) { printf "%d %c\n", i, arr[i] } }' /bin/ls

rinku · #5 (**permalink**) 05-28-2007

awk '{ for(i=65; i<=90; i++) { arr[i]=i } }END{ for ( i in arr ) { printf "%d %c\n", i, arr[i] } }' /bin/ls

output of this is::::
70 F
71 G
72 H
73 I
74 J
75 K
76 L
77 M
78 N
79 O
90 Z
65 A
66 B
67 C
68 D
69 E
80 P
81 Q
82 R
83 S
84 T
85 U
86 V
87 W
88 X
89 Y

Why it is not giving in proper way. can I get A to Z serially.

matrixmadhan · #6 (**permalink**) 05-28-2007

Code:

awk '{ for(i=65; i<=90; i++) { arr[i]=i } }END{ for ( i in arr ) { printf "%d %c\n", i, arr[i] } }' /bin/ls | sort -k1

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