bash 如何在固定长度的块中打印可变长度的行？
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How to printf a variable length line in fixed length chunks?
提问by user1187987
I need to to analyze (with grep) and print (with some formatting) the content of an
app's log.
我需要分析（使用grep）并打印（使用某种格式）应用程序日志的内容。
This log contains text data in variable length lines. What I need is, after some grepping, loop each line of this output and print it with a maximum fixed length of 50 characters. If a line is longer than 50 chars, it should print a newline and then continue with the rest in the following line and so on until the line is completed.
此日志包含可变长度行中的文本数据。我需要的是，在一些 grepping 之后，循环此输出的每一行并以最大固定长度 50 个字符打印它。如果一行超过 50 个字符，它应该打印一个换行符，然后在下一行继续其余的行，依此类推，直到该行完成。
I tried to use printfto do this, but it's not working and I don't know why. It just outputs the lines in same fashion of echo, without any consideration about printfformatting, though the \tcharacter (tab) works.
我试图用来printf这样做，但它不起作用，我不知道为什么。它只是以与 相同的方式输出行echo，而不考虑printf格式，尽管\t字符（制表符）有效。
function printContext
{
str=""
log=""
tmp="/tmp/deluge/$$"
rm -f $tmp
echo ""
echo -e "\tLog entries for $str :"
ln=$(grep -F "$str" "$log" &> "$tmp" ; cat "$tmp" | wc -l)
if [ $ln -gt 0 ];
then
while read line
do
printf "\t%50s\n" "$line"
done < $tmp
fi
}
What's wrong? I Know that I can make a substring routine to accomplish this task, but printfshould be handy for stuff like this.
怎么了？我知道我可以制作一个子字符串例程来完成这个任务，但是printf对于这样的事情应该很方便。
回答by anubhava
Instead of:
代替：
printf "\t%50s\n" "$line"
use
用
printf "\t%.50s\n" "$line"
to truncate your line to 50 characters only.
将您的行截断为仅 50 个字符。
回答by frankc
I'm not sure about printf but seeing as how perl is installed everywhere, how about a simple 1 liner?
我不确定 printf 但看到 perl 是如何安装的，一个简单的 1 liner 怎么样？
echo $ln | perl -ne ' while( m/.{1,50}/g ){ print "$&\n" } '
回答by glenn Hymanman
Here's a clunky bash-only way to break the string into 50-character chunks
这是将字符串分成 50 个字符的块的笨拙方法
i=0
chars=50
while [[ -n "${y:$((chars*i)):$chars}" ]]; do
printf "\t%s\n" "${y:$((chars*i)):$chars}"
((i++))
done
