blocks|key|579715|text|更好的是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|579716|for+i+in+xa*;+do
++++sed+-i+'s/asd/dfg/g'+$i
done|code-block|syntax|javascript|579717|因为没有人知道那里有多少文件，而且很容易打破命令行限制。|579718|下面是当文件太多时会发生的情况：|579719|#+grep+-c+aaa+*
-bash:+/bin/grep:+Argument+list+too+long
#+for+i+in+*;+do+grep+-c+aaa+$i;+done
0
...+(output+skipped)
#|579720|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Better yet:

<pre><code>for i in xa*; do
 sed -i 's/asd/dfg/g' $i
done
</code></pre>

because nobody knows how many files are there, and it's easy to break command line limits.

Here's what happens when there are too many files:

<pre><code># grep -c aaa *
-bash: /bin/grep: Argument list too long
# for i in *; do grep -c aaa $i; done
0
... (output skipped)
#
</code></pre>

blocks|key|579945|text|我很惊讶没有人提到要查找的-exec参数，虽然它会为每个匹配的文件名启动一个进程，但它是针对这种用例的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|579946|find+.+-type+f+-name+'xa*'+-exec+sed+-i+'s/asd/dsg/g'+{}+\;|code-block|syntax|javascript|579947|或者，可以使用xargs，这将调用更少的进程：|579948|find+.+-type+f+-name+'xa*'+%7C+xargs+sed+-i+'s/asd/dsg/g'|579949|或者更简单地在find中使用%2B+exec+variant而不是;，以允许find为每个子流程调用提供多个文件：|offset|length|style|CODE|579950|find+.+-type+f+-name+'xa*'+-exec+sed+-i+'s/asd/dsg/g'+{}+%2B|579951|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/6085272/1941755^0|0|0|0|0|E|1|V|1|G|C|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|14|8|@$M|15|N|16|O|P]|$M|17|N|18|O|P]]|9|@$M|19|N|1A|1|1B]]|A|$]]|$1|Q|3|R|5|D|7|1C|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

I'm surprised nobody has mentioned the -exec argument to find, which is intended for this type of use-case, although it will start a process for each matching file name:

<pre><code>find . -type f -name 'xa*' -exec sed -i 's/asd/dsg/g' {} \;
</code></pre>

Alternatively, one could use xargs, which will invoke fewer processes:

<pre><code>find . -type f -name 'xa*' | xargs sed -i 's/asd/dsg/g'
</code></pre>

Or more simply use the <code>+</code> <a href="https://stackoverflow.com/a/6085272/1941755">exec variant</a> instead of <code>;</code> in find to allow find to provide more than one file per subprocess call:

<pre><code>find . -type f -name 'xa*' -exec sed -i 's/asd/dsg/g' {} +
</code></pre>

blocks|key|579822|text|您可以同时使用grep和sed。这允许您以递归方式搜索子目录。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|579823|Linux:+grep+-r+-l+<old>+*+%7C+xargs+sed+-i+'s/<old>/<new>/g'
OS+X:+grep+-r+-l+<old>+*+%7C+xargs+sed+-i+''+'s/<old>/<new>/g'

For+grep:
++++-r+recursively+searches+subdirectories+
++++-l+prints+file+names+that+contain+matches
For+sed:
++++-i+extension+(Note:+An+argument+needs+to+be+provided+on+OS+X)|code-block|syntax|javascript|579824|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You could use grep and sed together. This allows you to search subdirectories recursively.

<pre><code>Linux: grep -r -l &lt;old&gt; * | xargs sed -i 's/&lt;old&gt;/&lt;new&gt;/g'
OS X: grep -r -l &lt;old&gt; * | xargs sed -i '' 's/&lt;old&gt;/&lt;new&gt;/g'

For grep:
 -r recursively searches subdirectories 
 -l prints file names that contain matches
For sed:
 -i extension (Note: An argument needs to be provided on OS X)
</code></pre>

blocks|key|579772|text|这些命令在Mac自带的默认sed中不起作用。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|579773|来自man+1+sed|579774|-i+extension
+++++++++++++Edit+files+in-place,+saving+backups+with+the+specified
+++++++++++++extension.++If+a+zero-length+extension+is+given,+no+backup+
+++++++++++++will+be+saved.++It+is+not+recommended+to+give+a+zero-length
+++++++++++++extension+when+in-place+editing+files,+as+you+risk+corruption
+++++++++++++or+partial+content+in+situations+where+disk+space+is+exhausted,+etc.|code-block|syntax|javascript|579775|已尝试|579776|sed+-i+'.bak'+'s/old/new/g'+logfile*|579777|和|579778|for+i+in+logfile*;+do+sed+-i+'.bak'+'s/old/new/g'+$i;+done|579779|两者都工作得很好。|579780|entityMap^0|D|3|0|2|9|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|15|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|16|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|17|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|18|8|@]|D|@]|E|$K|L]]|$1|U|3|V|5|6|7|19|8|@]|D|@]|E|$]]|$1|W|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|X|$]]

Those commands won't work in the default <code>sed</code> that comes with Mac OS X. 

From <code>man 1 sed</code>:

<pre><code>-i extension
 Edit files in-place, saving backups with the specified
 extension. If a zero-length extension is given, no backup 
 will be saved. It is not recommended to give a zero-length
 extension when in-place editing files, as you risk corruption
 or partial content in situations where disk space is exhausted, etc.
</code></pre>

Tried

<pre><code>sed -i '.bak' 's/old/new/g' logfile*
</code></pre>

and

<pre><code>for i in logfile*; do sed -i '.bak' 's/old/new/g' $i; done
</code></pre>

Both work fine.

blocks|key|364007|text|@PaulR发布了这篇评论，但人们应该将其视为答案(这个答案最适合我的需求)：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|364008|sed+-i+'s/abc/xyz/g'+xa*|code-block|syntax|javascript|364009|这将适用于中等数量的文件，可能在几十个数量级，但probably+not+on+the+order+of+millions。|364010|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/q/10445934/393280#comment13486175_10445934|1|https://superuser.com/q/257250/45163^0|0|6|0|0|0|O|11|1|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]|$1|K|3|-4|5|6|7|13|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

<a href="https://stackoverflow.com/q/10445934/393280#comment13486175_10445934">@PaulR</a> posted this as a comment, but people should view it as an answer (and this answer works best for my needs):

<pre><code>sed -i 's/abc/xyz/g' xa*
</code></pre>

This will work for a moderate amount of files, probably on the order of tens, but <a href="https://superuser.com/q/257250/45163">probably not on the order of millions</a>.

blocks|key|363989|text|另一种更通用的方法是使用find|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|363990|sed+-i+'s/asd/dsg/g'+$(find+.+-type+f+-name+'xa*')|code-block|syntax|javascript|363991|entityMap^0|C|4|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Another more versatile way is to use <code>find</code>:

<pre><code>sed -i 's/asd/dsg/g' $(find . -type f -name 'xa*')
</code></pre>

blocks|key|580015|text|我正在使用find来完成类似的任务。它非常简单:您必须将其作为参数传递给sed，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|580016|sed+-i+'s/EXPRESSION/REPLACEMENT/g'+`find+-name+"FILE.REGEX"`|580017|通过这种方式，您不必编写复杂的循环，并且很容易看到您要更改哪些文件，只需在运行sed之前运行find即可。|580018|entityMap^0|5|4|10|3|0|0|1P|0|13|3|1A|4|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|K|$]]

I'm using <code>find</code> for similar task. It is quite simple: you have to pass it as an argument for <code>sed</code> like this:

<code>sed -i 's/EXPRESSION/REPLACEMENT/g' `find -name "FILE.REGEX"`</code>

This way you don't have to write complex loops, and it is simple to see, which files you are going to change, just run <code>find</code> before you run <code>sed</code>.

blocks|key|580077|text|如果您能够运行脚本，下面是我针对类似情况所做的工作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|580078|使用字典/hashMap(关联数组)和sed命令的变量，我们可以遍历数组来替换几个字符串。在name_pattern中包含通配符将允许在文件中使用特定目录(source_dir)中的模式(这可能类似于name_pattern='File*.txt'+)进行就地替换。所有更改都写入destin_dir中的logfile中|offset|length|style|CODE|580079|#!/bin/bash
source_dir=source_path
destin_dir=destin_path
logfile='sedOutput.txt'
name_pattern='File.txt'

echo+"--Begin+$(date)--"+%7C+tee+-a+$destin_dir/$logfile
echo+"Source_DIR=$source_dir+destin_DIR=$destin_dir+"

declare+-A+pairs=(+
++++['WHAT1']='FOR1'
++++['OTHER_string_to+replace']='string+replaced'
)

for+i+in+"${!pairs[@]}";+do
++++j=${pairs[$i]}
++++echo+"[$i]=$j"
++++replace_what=$i
++++replace_for=$j
++++echo+"+"
++++echo+"Replace:+$replace_what+for:+$replace_for"
++++find+$source_dir+-name+$name_pattern+%7C+xargs+sed+-i+"s/$replace_what/$replace_for/g"+
++++find+$source_dir+-name+$name_pattern+%7C+xargs+-I{}+grep+-n+"$replace_for"+{}+/dev/null+%7C+tee+-a+$destin_dir/$logfile
done

echo+"+"
echo+"----End+$(date)---"+%7C+tee+-a+$destin_dir/$logfile|code-block|syntax|javascript|580080|首先，声明成对数组，每对都是一个替换字符串，然后在文件File.txt中将WHAT1替换为FOR1，将OTHER_string_to+replace替换为string+replaced。在循环中读取数组，该对的第一个成员被检索为replace_what=$i，第二个成员被检索为replace_for=$j。find命令在目录中搜索文件名(可能包含通配符)，而sed+-i命令在相同的文件中替换先前定义的内容。最后，我添加了一个重定向到日志文件的grep，以记录在文件中所做的更改。|580081|这在我的GNU+Bash+4.3+sed+4.2.2中很有效，并且基于VasyaNovikov对Loop+over+tuples+in+bash的回答。|580082|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/9713104/loop-over-tuples-in-bash^0|0|J|3|1A|C|26|A|2S|O|3W|A|48|7|0|0|R|8|11|5|19|4|1F|N|25|F|36|F|3V|E|4A|4|50|6|67|4|0|4|C|H|9|1C|O|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@$D|10|E|11|F|G]|$D|12|E|13|F|G]|$D|14|E|15|F|G]|$D|16|E|17|F|G]|$D|18|E|19|F|G]|$D|1A|E|1B|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|1C|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|1D|8|@$D|1E|E|1F|F|G]|$D|1G|E|1H|F|G]|$D|1I|E|1J|F|G]|$D|1K|E|1L|F|G]|$D|1M|E|1N|F|G]|$D|1O|E|1P|F|G]|$D|1Q|E|1R|F|G]|$D|1S|E|1T|F|G]|$D|1U|E|1V|F|G]|$D|1W|E|1X|F|G]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1Y|8|@$D|1Z|E|20|F|G]|$D|21|E|22|F|G]]|9|@$D|23|E|24|1|25]]|A|$]]|$1|Q|3|-4|5|6|7|26|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

If you are able to run a script, here is what I did for a similar situation:

Using a dictionary/hashMap (associative array) and variables for the <code>sed</code> command, we can loop through the array to replace several strings. Including a wildcard in the <code>name_pattern</code> will allow to replace in-place in files with a pattern (this could be something like <code>name_pattern='File*.txt'</code> ) in a specific directory (<code>source_dir</code>).
All the changes are written in the <code>logfile</code> in the <code>destin_dir</code>

<pre><code>#!/bin/bash
source_dir=source_path
destin_dir=destin_path
logfile='sedOutput.txt'
name_pattern='File.txt'

echo "--Begin $(date)--" | tee -a $destin_dir/$logfile
echo "Source_DIR=$source_dir destin_DIR=$destin_dir "

declare -A pairs=( 
 ['WHAT1']='FOR1'
 ['OTHER_string_to replace']='string replaced'
)

for i in "${!pairs[@]}"; do
 j=${pairs[$i]}
 echo "[$i]=$j"
 replace_what=$i
 replace_for=$j
 echo " "
 echo "Replace: $replace_what for: $replace_for"
 find $source_dir -name $name_pattern | xargs sed -i "s/$replace_what/$replace_for/g" 
 find $source_dir -name $name_pattern | xargs -I{} grep -n "$replace_for" {} /dev/null | tee -a $destin_dir/$logfile
done

echo " "
echo "----End $(date)---" | tee -a $destin_dir/$logfile
</code></pre>

First, the pairs array is declared, each pair is a replacement string, then <code>WHAT1</code> will be replaced for <code>FOR1</code> and <code>OTHER_string_to replace</code> will be replaced for <code>string replaced</code> in the file <code>File.txt</code>. In the loop the array is read, the first member of the pair is retrieved as <code>replace_what=$i</code> and the second as <code>replace_for=$j</code>. The <code>find</code> command searches in the directory the filename (that may contain a wildcard) and the <code>sed -i</code> command replaces in the same file(s) what was previously defined. Finally I added a <code>grep</code> redirected to the logfile to log the changes made in the file(s). 

This worked for me in <code>GNU Bash 4.3</code> <code>sed 4.2.2</code> and based upon VasyaNovikov's answer for <a href="https://stackoverflow.com/questions/9713104/loop-over-tuples-in-bash">Loop over tuples in bash</a>.

The following command is correctly changing the contents of 2 files.

<pre><code>sed -i 's/abc/xyz/g' xaa1 xab1 
</code></pre>

But what I need to do is to change several such files dynamically and I do not know the file names. I want to write a command that will read all the files from current directory starting with <code>xa*</code> and <code>sed</code> should change the file contents.

Change multiple files

日志文件

Linux

以下命令正在正确更改%2个文件的内容。sed -i 's/abc/xyz/g' xaa1 xab1 但我需要做的是动态更改几个这样的文件，而我不知道文件名。我想写一个命令，它将读取当前目录中的所有文件，从xa*开始，sed将更改文件内容。

问更改多个文件
EN

回答 8

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