Question

我正在尝试编写一个bash脚本，以自动将我的构建和源目录（具有共享对象二进制文件）中的文件分别复制到我的系统/usr/lib和/usr/include目录。

当我运行以下内容时：

#!/bin/bash

RELEASE_ARG = "--release"
DEBUG_ARG   = "--debug"

DEST_NAME = "libEngine.so"

SRC_NAME = "libEngineSO.so"

ENGINE_DBG_DEST = "/usr/lib/Stitches/Debug/$DEST_NAME"
ENGINE_REL_DEST = "/usr/lib/Stitches/Release/$DEST_NAME"

ENGINE_REL_SRC = "$STITCH_ROOT/Engine__Release/$SRC_NAME"
ENGINE_DBG_SRC = "$STITCH_ROOT/Engine__Debug/$SRC_NAME"

if [ ! -z $1 ]; then 
    #foo
    echo "No argument specified; please specify either ${RELEASE_ARG} or $DEBUG_ARG. K BAI\n"
    exit
elif [ -z $1 -eq RELEASE_ARG ]; then
    #wat
    echo "If you actually chose this option, you better be damn ready to distribute...\n"
    echo "Updating Release directory in $ENGINE_REL_DEST\n"

    cp $ENGINE_REL_SRC $ENGINE_REL_DEST

elif [ -z $1 -eq DEBUG_ARG ]; then
    #reasonable and realistic
    echo "Good choice...\n"
    echo "Updating Debug directory in $ENGINE_DBG_DEST\n"
    cp $ENGINE_DBG_SRC $ENGINE_DBG_DEST
fi

echo "Sex ;)"

我得到了这个输出：

x@x ~/Programming/Stitches-Qt/scripts $ sh postbuild.sh --debug
postbuild.sh: 3: postbuild.sh: DEBUG: not found
postbuild.sh: 5: postbuild.sh: RELEASE_ARG: not found
postbuild.sh: 6: postbuild.sh: DEBUG_ARG: not found
postbuild.sh: 8: postbuild.sh: DEST_NAME: not found
postbuild.sh: 10: postbuild.sh: SRC_NAME: not found
postbuild.sh: 12: postbuild.sh: ENGINE_DBG_DEST: not found
postbuild.sh: 13: postbuild.sh: ENGINE_REL_DEST: not found
postbuild.sh: 15: postbuild.sh: ENGINE_REL_SRC: not found
postbuild.sh: 16: postbuild.sh: ENGINE_DBG_SRC: not found
No argument specified; please specify either  or . K BAI

正如您所看到的，我必须使用不正确的变量或其他东西。我如何完成我在这里要做的事情？

P.S。，我知道除非我是root用户，否则我不能这样做，所以我在/usr/lib |中创建了一个子目录。 /usr/include所以我可以将其添加到我的用户名中。所以，这应该有用......

Answer 1

你不能在bash中使用任意空格。

RELEASE_ARG="--release"
DEBUG_ARG="--debug"

使用test时应引用。

if [ ! -z "$1" ]; then

学习Bash - 如何正确声明和打印/嵌入变量？

1 个答案: