如何拦截和删除 bash 中的命令行参数
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How to intercept and remove a command line argument in bash
提问by inspector-g
After "upgrading" to Mavericks and Xcode 5, I have a variety of minor problems to deal with to make Xcode compile some of my older projects.
在“升级”到 Mavericks 和 Xcode 5 之后，我有各种小问题需要处理，以使 Xcode 编译我的一些旧项目。
It appears that Xcode is passing a new argument to the ldlinker, and there's really no stopping Xcode from doing so. An older version of ld, which I need for a variety of reasons, gives an error when seeing an argument it doesn't know (so my projects cannot compile).
看起来 Xcode 正在向ld链接器传递一个新参数，而且实际上并没有阻止 Xcode 这样做。ld由于各种原因我需要一个旧版本的，当看到一个它不知道的参数时会出错（所以我的项目无法编译）。
What I need is a thin wrapper over my older version of ldto remove the "bad" arguments under certain circumstances. I thought that a bash shell script would be perfect, but bash is not my forte.
我需要的是在我的旧版本上的一个薄包装器，ld以在某些情况下删除“坏”参数。我认为 bash shell 脚本会很完美，但 bash 不是我的强项。
Here's what I've got:
这是我所拥有的：
# Look for conditions necessary to use older ld
... # (placeholder, obviously)
# Run older ld (pseudo condition)
if [ <old_ld_condition> ]; then
ARGS=''
for var in "$@"; do
# Ignore known bad arguments
if [ "$var" = '-dependency_info' ]; then
continue
fi
ARGS="$ARGS $var"
done
/path/to/old/ld "$ARGS"
else
/path/to/new/ld "$@"
fi
However, running /path/to/old/ld "$ARGS"results in ldinterpreting the entire $ARGSstring as one argument. Running /path/to/old/ld $ARGSresults in ldreceiving unescaped versions of previously escaped strings.
但是，运行会/path/to/old/ld "$ARGS"导致ld将整个$ARGS字符串解释为一个参数。运行/path/to/old/ld $ARGS结果会ld收到以前转义字符串的未转义版本。
Clearly, I'm misunderstanding something about the nature of $@, how to manipulate it, and how to pass that manipulation to the older ld. Thanks everyone.
很显然，我误解一些有关的性质$@，如何处理它，以及如何操纵这传递给上了年纪ld。谢谢大家。
回答by anubhava
This should work:
这应该有效：
# Run older ld (pseudo condition)
if [ <old_ld_condition> ]; then
ARGS=()
for var in "$@"; do
# Ignore known bad arguments
[ "$var" != '-dependency_info' ] && ARGS+=("$var")
done
/path/to/old/ld "${ARGS[@]}"
else
/path/to/new/ld "$@"
fi
回答by user3159253
You should use Bash Arrays if you really wantto stay with bash:
如果你真的想继续使用 bash，你应该使用 Bash Arrays ：
declare -a ARGS
for var in "$@"; do
# Ignore known bad arguments
if [ "$var" = '-dependency_info' ]; then
continue
fi
ARGS[${#ARGS[@]}]="$var"
done
now "${ARGS[@]}"can be used just as "$@". man bashfor more information.
现在"${ARGS[@]}"可以像"$@". man bash想要查询更多的信息。
