ubuntu - ifeq 不比较两个变量

Question

ifeq 不比较两个变量

我正在制作一个程序来将文件移动到桌面，与操作系统中选择的语言无关。在葡萄牙语中，桌面文件夹称为“Área de trabalho”。由于“Área de Trabalho”是三个分开的单词，所以在写目录的时候需要加上单引号('Área de Trabalho')，否则命令会出错。

    mv Minesweeper $HOME/Área de Trabalho/  #command

    mv: target 'Trabalho/' is not a directory  #Error

我正在使用此命令来获取用户的桌面文件夹的目录:

    xdg-user-dir DESKTOP

如果目录等于主目录加上“/Área de Trabalho”，我想改变

这是代码:

    desktop=$(shell xdg-user-dir DESKTOP) # I'm using this command to get the directory of the desktop folder of the user
    desktopVar="$$HOME/Área de Trabalho" # And if the directory is equals to the home directory plus "/Área de Trabalho"

    move:
    ifeq (${desktop},${desktopVar})
        desktop=${HOME}/'Área de Trabalho' # I want to change the directory to ${HOME} plus "/'Área de Trabalho'", so the command doesn't results in error
        @echo yes
    endif
        @echo ${desktopVar}
        @echo ${desktop}

但是 ifeq 说两个变量是相等的并且不会改变任何东西。

这是输出:

    /home/daniel/Área de Trabalho
    /home/daniel/Área de Trabalho

Answer 1

这是一个应该可以工作的makefile(不，这次是真的:-)我实际上测试了这个:)

desktop:=$(shell xdg-user-dir DESKTOP)
desktopVar:=${HOME}/Área de Trabalho  #note: using _MAKE_ variable ${HOME} here

$(info desktop:     [${desktop}])
$(info desktopVar:  [${desktopVar}])

    # note: using double quotes around ifeq arguments -- not required
    #       here, but it's good practice:
    # also note: this is not inside of a recipe, so it's done by the
    #            makefile at parse time

ifneq ("${desktop}","${desktopVar}")
$(info changing desktop)
desktop:=${HOME}/Área de Trabalho
endif


$(info desktop:     [${desktop}])
$(info desktopVar:  [${desktopVar}])

all:
        @echo running $@

输出:

scripts> make -f test.mk
desktop:     [/users/foo/Desktop]
desktopVar:  [/users/foo/Área de Trabalho  ]
changing desktop
desktop:     [/users/foo/Área de Trabalho]
desktopVar:  [/users/foo/Área de Trabalho  ]
running all