在 bash 中格式化标准输入
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Format stdin in bash
提问by User1
I have a multi-line string coming from another program that I want to convert to a SQL command. I was hoping that printf could help me, but it doesn't seem to work:
我有一个来自另一个程序的多行字符串，我想将其转换为 SQL 命令。我希望 printf 可以帮助我，但它似乎不起作用：
echo -e '1\n2\n3'|printf 'SELECT %s INTO MyTable'
I was hoping to see:
我希望看到：
SELECT '1 2 3' INTO MyTable
But I got:
但我得到了：
SELECT INTO MyTable
How can I get the %s to read stdin?
我怎样才能让 %s 读取标准输入？
回答by Jürgen H?tzel
Use xargsto transform stdinto program arguments:
使用xargs将stdin转换为程序参数：
echo -n -e '1\n2\n3' |xargs -0 printf 'SELECT %s INTO MyTable'
回答by Paused until further notice.
Give this a try:
试试这个：
printf_stdin() { local stdin; read -d '' -u 0 stdin; printf "$@" "$stdin"; }
echo -e '1\n2\n3' | printf_stdin 'SELECT %s INTO MyTable'
回答by D.Shawley
You can't. The printfshell command formats its arguments not standard input so what you can do is provide the output of a command as a single argument:
你不能。在printfshell命令格式化它的参数不是标准输入，所以你可以做的是提供了一个命令，一个参数的输出：
bash$ printf "SELECT '%s' INTO MyTable" "`echo -e '1\n2\n3'`"
SELECT '1
2
3' INTO MyTable
bash$
Edit: a solution in Awk
编辑：Awk 中的解决方案
bash$ echo -e '1\n2\n3' | awk -v 'ORS=' '
BEGIN { print "SELECT \"" }
{ print ##代码##, "\n" }
END { print "\" INTO MyTable" }'
SELECT "1
2
3
" INTO MyTable
bash$
I'll leave stripping the final newline as an exercise to the reader. If you want to do anything more complex in the printf, then you will have to come up with some more creative awk script.
我将剥离最后的换行符作为练习留给读者。如果您想在printf.
