在 Bash 循环中递增变量

Incrementing a variable inside a Bash loop

提问by maephisto

I'm trying to write a small script that will count entries in a log file, and I'm incrementing a variable (USCOUNTER) which I'm trying to use after the loop is done.

我正在尝试编写一个小脚本来计算日志文件中的条目，并且我正在增加一个变量 ( USCOUNTER)，我在循环完成后尝试使用该变量。

But at that moment USCOUNTERlooks to be 0 instead of the actual value. Any idea what I'm doing wrong? Thanks!

但在那一刻USCOUNTER看起来是 0 而不是实际值。知道我做错了什么吗？谢谢！

FILE=

tail -n10 mylog > $FILE

USCOUNTER=0

cat $FILE | while read line; do
  country=$(echo "$line" | cut -d' ' -f1)
  if [ "US" = "$country" ]; then
        USCOUNTER=`expr $USCOUNTER + 1`
        echo "US counter $USCOUNTER"
  fi
done
echo "final $USCOUNTER"

It outputs:

它输出：

US counter 1
US counter 2
US counter 3
..
final 0

回答by fedorqui 'SO stop harming'

You are using USCOUNTERin a subshell, that's why the variable is not showing in the main shell.

您USCOUNTER在子 shell中使用，这就是变量未显示在主 shell 中的原因。

Instead of cat FILE | while ..., do just a while ... done < $FILE. This way, you avoid the common problem of I set variables in a loop that's in a pipeline. Why do they disappear after the loop terminates? Or, why can't I pipe data to read?:

而不是cat FILE | while ...，只做一个while ... done < $FILE. 这样，您就可以避免我在管道中的循环中设置变量的常见问题。为什么它们在循环终止后消失？或者，为什么我不能通过管道读取数据？：

while read country _; do
  if [ "US" = "$country" ]; then
        USCOUNTER=$(expr $USCOUNTER + 1)
        echo "US counter $USCOUNTER"
  fi
done < "$FILE"

Note I also replaced the `` expression with a $().

注意我还用 $() 替换了 `` 表达式。

I also replaced while read line; do country=$(echo "$line" | cut -d' ' -f1)with while read country _. This allows you to say while read var1 var2 ... varNwhere var1contains the first word in the line, $var2and so on, until $varNcontaining the remaining content.

我也换成while read line; do country=$(echo "$line" | cut -d' ' -f1)了while read country _。这使您可以说while read var1 var2 ... varN在那里var1包含在该行的第一个字，$var2等等，直到$varN包含剩余的内容。

回答by Aleks-Daniel Jakimenko-A.

• Always use -rwith read.
• There is no need to use cut, you can stick with pure bash solutions.
  • In this case passing reada 2nd var (_) to catch the additional "fields"
• Prefer [[ ]]over [ ].
• Use arithmetic expressions.
• Do not forget to quote variables! Link includes other pitfalls as well

• 始终-r与 read 一起使用。
• 无需使用cut，您可以坚持使用纯 bash 解决方案。
  • 在这种情况下，传递read第二个 var ( _) 以捕获额外的“字段”
• 身高[[ ]]超过[ ]。
• 使用算术表达式。
• 不要忘记引用变量！链接还包括其他陷阱

while read -r country _; do
  if [[ $country = 'US' ]]; then
    ((USCOUNTER++))
    echo "US counter $USCOUNTER"
  fi
done < "$FILE"

回答by geekzspot

minimalist

极简主义者

counter=0
((counter++))
echo $counter

回答by anubhava

You're getting final 0because your while loopis being executed in a sub (shell) process and any changes made there are not reflected in the current (parent) shell.

你得到final 0是因为你while loop正在一个子（shell）进程中执行，并且在那里所做的任何更改都没有反映在当前（父）shell中。

Correct script:

正确的脚本：

while read -r country _; do
  if [ "US" = "$country" ]; then
        ((USCOUNTER++))
        echo "US counter $USCOUNTER"
  fi
done < "$FILE"

回答by jobwat

I had the same $count variable in a while loop getting lost issue.

我在 while 循环丢失问题中有相同​​的 $count 变量。

@fedorqui's answer(and a few others) are accurate answers to the actual question: the sub-shell is indeed the problem.

@fedorqui 的答案（以及其他一些答案）是对实际问题的准确答案：子外壳确实是问题所在。

But it lead me to another issue: I wasn't piping a file content... but the output of a series of pipes & greps...

但这让我想到了另一个问题：我不是在通过管道传输文件内容......而是一系列管道和 grep 的输出......

my erroring sample code:

我的错误示​​例代码：

count=0
cat /etc/hosts | head | while read line; do
  ((count++))
  echo $count $line
done
echo $count

and my fix thanks to the help of this thread and the process substitution:

多亏了这个线程和进程替换的帮助，我的修复：

count=0
while IFS= read -r line; do
  ((count++))
  echo "$count $line"
done < <(cat /etc/hosts | head)
echo "$count"

回答by Walter A

USCOUNTER=$(grep -c "^US " "$FILE")

回答by thanh

Using the following 1 line command for changing many files name in linux using phrase specificity:

使用以下 1 行命令使用短语特异性更改 linux 中的许多文件名：

find -type f -name '*.jpg' | rename 's/holiday/honeymoon/'

For all files with the extension ".jpg", if they contain the string "holiday", replace it with "honeymoon". For instance, this command would rename the file "ourholiday001.jpg" to "ourhoneymoon001.jpg".

对于所有扩展名为“.jpg”的文件，如果它们包含字符串“holiday”，请将其替换为“honeymoon”。例如，此命令会将文件“ourholiday001.jpg”重命名为“ourhoneymoon001.jpg”。

This example also illustrates how to use the find command to send a list of files (-type f) with the extension .jpg (-name '*.jpg') to rename via a pipe (|). rename then reads its file list from standard input.

此示例还说明了如何使用 find 命令发送扩展名为 .jpg (-name '*.jpg') 的文件列表 (-type f) 以通过管道 (|) 重命名。rename 然后从标准输入读取它的文件列表。

回答by trax

Incrementing a variable can be done like that:

增加一个变量可以这样完成：

  _my_counter=$[$_my_counter + 1]

Counting the number of occurrence of a pattern in a column can be done with grep

可以使用 grep 计算列中模式出现的次数

 grep -cE "^([^ ]* ){2}US"

-ccount

-c数数

([^ ]* )To detect a colonne

([^ ]* )检测冒号

{2}the colonne number

{2}冒号

USyour pattern

US你的模式