将反斜杠转义的字符保存到 bash 中的变量
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Saving backslash-escaped characters to variable in bash
提问by mingos
I've just written a bash script that takes some info from the mysql database and reads it line by line, extracting tab-separated columns into separate variables, something like this:
我刚刚编写了一个 bash 脚本，它从 mysql 数据库中获取一些信息并逐行读取它，将制表符分隔的列提取到单独的变量中，如下所示：
oldifs=$IFS
result=result.txt
$mysql -e "SELECT id,foo,bar,baz FROM $db.$table" -u $user --password=$pass -h $server > $result
cat $result | grep -e ^[0-9].*$ | while IFS=$'\t' read id foo bar baz
do
# some code
done
IFS=$oldifs
Now, while this works OK and I'm satisfied with the result (especially since I'm going to move the query t oanother script and let cron regenerate the result.txt file contents once a week or so, since I'm dealing with a table that changes maybe once or twice a year), I'm curious about the possibility of putting the query's result in a variable instead of a file.
现在，虽然这工作正常并且我对结果感到满意（特别是因为我要将查询移动到另一个脚本并让 cron 每周重新生成一次 result.txt 文件内容，因为我正在处理一个每年可能更改一次或两次的表），我很好奇将查询结果放在变量而不是文件中的可能性。
I have noticed that in order to echo out backslash-excaped characters, I need to tell the command explicitly to interpret such characters as special chars:
我注意到为了回显出反斜杠转义字符，我需要明确地告诉命令将这些字符解释为特殊字符：
echo -e "some\tstring\n"
But, being a bash noob that I am, I have no idea how to place the backslash escaped characters (the tabs and newlines from the query) inside a variable and just work with it the same way I'm working with the external file (just changing the catwith echo -e). I tried this:
但是，作为一个 bash 菜鸟，我不知道如何将反斜杠转义字符（查询中的制表符和换行符）放置在变量中，并以与处理外部文件相同的方式使用它（只是改变了cat用echo -e）。我试过这个：
result=`$mysql -e "SELECT id,foo,bar,baz FROM $db.$table" -u $user --password=$pass -h $server`
but the backslash escaped characters are converted into spaces this way :(. How can I make it work?
但是反斜杠转义字符以这种方式转换为空格:(。我怎样才能让它工作？
回答by bobbogo
To get the output of a command, use $(...). To avoid wordsplitting and other bash processing you will need to quote. Single quotes ('$(...)') will not work as the quoting is too strong.
要获取命令的输出，请使用$(...). 为避免分词和其他 bash 处理，您需要引用。单引号 ( '$(...)') 将不起作用，因为引用太强了。
Note that once the output is in your variable, you will probably need to (double) quote it wherever you use it if you need to preserve anything that's in $IFS.
请注意，一旦输出在您的变量中，如果您需要保留$IFS.
$ listing="$(ls -l)"
$ echo "$listing"
回答by bmk
Could you try to set double quotes around $result- thus echo -e "$result"?
您可以尝试在周围设置双引号$result- 因此echo -e "$result"？
回答by mikeserv
% awk '/^[0-9]/ { print , , , }' <<SQL | set -- -
> $("${mysql}" -e "SELECT id,foo,bar,baz FROM $db.$table" -u $user --password=$pass -h $server)
> SQL
% printf '%s\t' "${@}"
<id> <foo> <bar> <baz>
You might get some use out of this. The heredocshould obviate any escaping issues, awkwill separate on tabs by default, and setaccepts the input as a builtin argvarray. printfisn't necessary, but it's better than echo- especially when working with escape characters.
你可能会从中得到一些用处。在heredoc应该避免任何逸出的问题，awk将默认上分离的标签，并且set接受输入作为内置argv阵列。printf不是必需的，但它比echo- 特别是在使用转义字符时更好。
You could also use readas you did above - but to better handle backslashes use the -rargument if you go that route. The above method would work best as a function and you could then iterate over your variables with shiftand similar.
您也可以read像上面那样使用- 但是-r如果您走那条路，请使用参数更好地处理反斜杠。上述方法最适合作为函数使用，然后您可以使用shift和 类似的方法迭代变量。
-Mike
-麦克风
