如何遍历 Bash 中的所有 ASCII 字符？
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原文地址: http://stackoverflow.com/questions/13127950/
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How to iterate through all ASCII characters in Bash?
提问by Shamdor
I know how to iterate through alphabets:
我知道如何遍历字母表：
for c in {a..z}; do ...; done
But I can't figure out how to iterate through all ASCII characters. Does anyone know how?
但我不知道如何遍历所有 ASCII 字符。有谁知道怎么做？
回答by Milan
What you can do is to iterate from 0 to 127 and then convert the decimal value to its ASCII value(or back).
您可以做的是从 0 到 127 迭代，然后将十进制值转换为其 ASCII 值（或返回）。
You can use thesefunctions to do it:
您可以使用这些功能来做到这一点：
# POSIX
# chr() - converts decimal value to its ASCII character representation
# ord() - converts ASCII character to its decimal value
chr() {
[ -lt 256 ] || return 1
printf \$(printf '%03o' )
}
# Another version doing the octal conversion with arithmetic
# faster as it avoids a subshell
chr () {
[ -lt 256 ] || return 1
printf \$((/64*100+%64/8*10+%8))
}
# Another version using a temporary variable to avoid subshell.
# This one requires bash 3.1.
chr() {
local tmp
[ -lt 256 ] || return 1
printf -v tmp '%03o' ""
printf \"$tmp"
}
ord() {
LC_CTYPE=C printf '%d' "'"
}
# hex() - converts ASCII character to a hexadecimal value
# unhex() - converts a hexadecimal value to an ASCII character
hex() {
LC_CTYPE=C printf '%x' "'"
}
unhex() {
printf \x""
}
# examples:
chr $(ord A) # -> A
ord $(chr 65) # -> 65
回答by mata
A possibility using only echos octal escape sequences:
仅使用echos 个八进制转义序列的可能性：
for n in {0..7}{0..7}{0..7}; do echo -ne "\0$n"; done
回答by Andrew Clark
Here is what I came up with for a one-liner taking some pieces from sampson-chen's and mata's answers:
这是我从 sampson-chen 和 mata 的答案中提取一些片段的单行文章：
for n in {0..127}; do awk '{ printf("%c", for n in {0..127}; do echo $n; done | awk '{ printf("%c", echo "65" | awk '{ printf("%c", A
); }'
); }'
); }' <<< $n; done
Or alternatively:
或者：
# ascii for A starts at 65:
ascii=65
index=1
total=26
while [[ $total -ge $index ]]
do
letter=$(echo "$ascii" | awk '{ printf("%c", awk 'function utf32(i) {printf("%c%c%c%c",i%0x100,i/0x100%0x100,i/0x10000%0x100,i/0x1000000) } BEGIN{for(i=0;i<0x110000;i++){utf32(i);utf32(0xa)}}' | iconv --from-code=utf32 --to-code=utf8 | grep -a '[[:print:]]'
); }')
echo "The $index'th letter is $letter"
# Increment the index counter as well as the ascii counter
index=$((index+1))
ascii=$((ascii+1))
done
回答by sampson-chen
Here's how you print an integer as its corresponding ascii character with awk:
以下是将整数打印为其对应的 ascii 字符的方法awk：
which will print:
这将打印：
##代码##And here's how you might iterate through the uppercase alphabet this way:
以下是您可以通过这种方式遍历大写字母的方法：
##代码##回答by amaurea
Well... If you really want them all, and you want it to be something script-like, you could do this, I guess:
嗯......如果你真的想要它们，并且你想要它像脚本一样，你可以这样做，我想：
##代码##But the list is pretty huge, and not very useful. Awk may not be the most elegant way of generating the binary integers from 0 to 0x110000 - substitute something more elegant if you find it.
但是这个列表非常庞大，而且不是很有用。awk 可能不是生成从 0 到 0x110000 的二进制整数的最优雅的方法——如果你找到它，用更优雅的方法代替它。
Edit: Oh, I see you only wanted ascii. Well, I will let this answer stay here in case somebody else actually wants all the UTF printable characters.
编辑：哦，我看到你只想要 ascii。好吧，我会让这个答案留在这里，以防其他人真的想要所有 UTF 可打印字符。
