我如何间接在bash变量赋值从两个标准中，一个文件需要多行数据，和执行的输出 [英] How do I indirectly assign a variable in bash to take multi-line data from both Standard In, a File, and the output of execution

问题描述

我发现有很多片段在这里和在其他地方是回答这个问题的一部分。我甚至设法在一个低效率的方式来做到这一点在许多步骤。如果有可能，我真的想找到执行单行，将执行此任务，而不是分配给一个变​​量，并复制了几次来执行任务。

例如

  executeToVar（）
{
    ＃注意到的Arg1：以存储变量名
    ＃所有剩余的参数被执行
    当地STORE_INvar =$ {1};转移
    的eval $ {STORE_INvar} = \\$（$ @2  - ;＆放大器; 1）\\
}
 

总体不工作，即 $ executeToVar SOME_VAR的ls -l *＃将真正填补 SOME_VAR 与输出在执行的ls -l * 命令是从参数的其余采取。但是，如果命令是为了输出空行末，（对于如 - 回声-e -n的'\\ n \\ N123 \\ n456 \\ n789 \\ n \\ n 里面应该有在开始2个新的生产线和年底），这些是由庆典的副执行过程中剥离。我已经在类似于该其他职位看出，这已经通过将令牌的x到流，例如结束解决打开子执行到是这样的：

 的eval $ {STORE_INvar} = \\$（$ @2＆GT;＆放大器; 1;回声-nx）\\＃＆LT;  - （添加回声-nx ）
＃，然后，如果它不是一个间接引用一个变种：
STORE_INvar = $ {STORE_INvar％X}
＃但是无论我多么玩：
的eval$ {STORE_INvar}= \\$ {STORE_INvar％×}\\
＃我无法间接地从末尾移除的X.
 

反正，我还需要在这2个其他的变种，一个分配 STDIN流来var和一个文件的内容分配给VAR哪些我认为将是涉及 $的变化（猫$ {1}），或者 $（猫$ {1： - }）来给我一个' - '如果没有文件名。但是，没有将工作，​​直到我能理清去除，以保证多行变量的准确分配时所需要的x的。

我也尝试（但无济于事）：

  IFS =''读-d''$ {} STORE_INvar＆LT;＆LT;＆LT;$（$ @;回声-n X）
EVAL \\'$ {} STORE_INvar = $ {！STORE_INvar％X}'\\
 

解决方案

这是接近最优的 - 但降评估

。

  executeToVar（）{本地的varName = $ 1;转移; printf的-v$ 1％S$（$ @）; }
 

在一个问题这个提法仍是尾随换行符的 $（）带。如果你想为$，你需要添加自己的尾随字符子shell里面，剥离其关闭自己p $ pvent。

  executeToVar（）{
  当地的varName = $ 1;转移;
  当地VAL =$（的printf％S X;$ @;的printf％S X）; VAL = $ {VAL＃X}
  printf的-v$ varname的％S$ {VAL％X}
}
 

---

如果您想从标准输入到一个变量中读取的所有内容，这是特别容易：

 ＃这需要自动分配的fd 4.1的bash
readToVar（）{
  如果[[$ 2及和放大器; $ 2 = - ]！;然后
    EXEC {read_in_fd}＆LT;$ 2＃从指定的文件复制
  其他
    EXEC {read_in_fd}＆LT;＆安培; 0＃从标准输入复制
  科幻
  IFS =读-r -d''$ 1＆LT;＆安培; $＃read_in_fd从FD读
  EXEC {read_in_fd}＆LT;＆安培;  - ＃接近，FD
}
 

...作为：

  readToVar VAR＆LT; ≤（：跑的东西在这里读它的输出字节为字节）
 

......或者......

  readToVar变种文件名
 

---

测试这些：

  bash3-3.2 $ executeToVar VAR的printf的'\\ n \\ N123 \\ n456 \\ n789 \\ n \\ n'
bash3-3.2 $申报-p VAR
声明 -  VAR =123
456
789
 

......然后......

  bash4-4.3 $ readToVar VAR2＆LT; ≤（的printf的'\\ n \\ N123 \\ n456 \\ n789 \\ n \\ n）
bash4-4.3 $申报-p VAR2
声明 -  VAR2 =123
456
789
 

I have found many snippets here and in other places that answer parts of this question. I have even managed to do this in many steps in an inefficient manner. If it is possible, I would really like to find single lines of execution that will perform this task, rather than having to assign to a variable and copy it a few times to perform the task.

e.g.

executeToVar ()
{
    # Takes Arg1: NAME OF VARIABLE TO STORE IN
    # All Remaining Arguments Are Executed
    local STORE_INvar="${1}" ; shift
    eval ${STORE_INvar}=\""$( "$@" 2>&1 )"\"
}

Overall does work, i.e. $ executeToVar SOME_VAR ls -l * # will actually fill SOME_VAR with the output of the execution of the ls -l * command that is taken from the rest of the arguments. However, if the command was to output empty lines at the end, (for e.g. - echo -e -n '\n\n123\n456\n789\n\n' which should have 2 x new lines at the start and the end ) these are stripped by bash's sub-execution process. I have seen in other posts similar to this that this has been solved by adding a token 'x' to the end of the stream, e.g. turning the sub-execution into something like:

eval ${STORE_INvar}=\""$( "$@" 2>&1 ; echo -n x )"\" # <-- ( Add echo -n x )
# and then if it wasn't an indirect reference to a var:
STORE_INvar=${STORE_INvar%x} 
# However no matter how much I play with:
eval "${STORE_INvar}"=\""${STORE_INvar%x}"\"  
# I am unable to indirectly remove the x from the end.

Anyway, I also need 2 x other variants on this, one that assigns the STDIN stream to the var and one that assigns the contents of a file to the var which I assume will be variations of this involving $( cat ${1} ), or maybe $( cat ${1:--} ) to give me a '-' if no filename. But, none of that will work until I can sort out the removal of the x that is needed to ensure accurate assignment of multi line variables.

I have also tried (but to no avail):

IFS='' read -d '' "${STORE_INvar}" <<<"$( $@ ; echo -n x )"
eval \"'${STORE_INvar}=${!STORE_INvar%x}'\"

解决方案

This is close to optimal -- but drop the eval.

executeToVar() { local varName=$1; shift; printf -v "$1" %s "$("$@")"; }

The one problem this formulation still has is that $() strips trailing newlines. If you want to prevent that, you need to add your own trailing character inside the subshell, and strip it off yourself.

executeToVar() {
  local varName=$1; shift;
  local val="$(printf %s x; "$@"; printf %s x)"; val=${val#x}
  printf -v "$varName" %s "${val%x}"
}

---

If you want to read all content from stdin into a variable, this is particularly easy:

# This requires bash 4.1 for automatic fd allocation
readToVar() {
  if [[ $2 && $2 != "-" ]]; then
    exec {read_in_fd}<"$2"              # copy from named file
  else
    exec {read_in_fd}<&0                # copy from stdin
  fi
  IFS= read -r -d '' "$1" <&$read_in_fd # read from the FD
  exec {read_in_fd}<&-                  # close that FD
}

...used as:

readToVar var < <( : "run something here to read its output byte-for-byte" )

...or...

readToVar var filename

---

Testing these:

bash3-3.2$ executeToVar var printf '\n\n123\n456\n789\n\n'
bash3-3.2$ declare -p var
declare -- var="

123
456
789

"

...and...

bash4-4.3$ readToVar var2 < <(printf '\n\n123\n456\n789\n\n')
bash4-4.3$ declare -p var2
declare -- var2="

123
456
789

"

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