[英]problems with checking for a directory in bash shell script
I am writing a batch-processing script bash that needs to first check to see if a folder exists to know whether or not to run a certain python script that will create and populate the folder. 我正在编写一个批处理脚本bash,需要首先检查是否存在一个文件夹,以了解是否运行某个将创建和填充该文件夹的python脚本。 I have done similar things before that do fine with changing the directories and finding directories from a stored variable, but for some reason I am just missing something here.
我之前做过类似的事情,可以更改目录并从存储的变量中查找目录,但出于某种原因,我只是在这里遗漏了一些东西。
Here is roughly how the script is working. 以下是脚本的工作原理。
if [ -d "$net_output" ]
then
echo "directory exists"
else
echo "directory does not exist"
fi
when I run this script, I usually echo $net_output in the line before to see what it will evaluate to. 当我运行这个脚本时,我通常在该行中回显$ net_output以查看它将评估的内容。 When the script runs I get my else block of code saying "Directory does not exist", but when I then copy and paste the $net_output directory path that is echoed before into the shell terminal, it changes directories just fine, proving that the directory does in fact exist.
当脚本运行时,我得到我的其他代码块说“目录不存在”,但是当我将之前回显的$ net_output目录路径复制并粘贴到shell终端时,它会更改目录就好了,证明该目录确实存在。 I am using Ubuntu 12.04 on a Dell machine.
我在戴尔机器上使用Ubuntu 12.04。
Thank you in advance for any help that someone can offer. 提前感谢您提供任何帮助。 Let me know what additional information I can provide.
让我知道我可以提供哪些其他信息。
The most common cases I've encountered when someone posts a problem like this are the following: 当有人发布这样的问题时我遇到的最常见的情况如下:
1. The variable contains literal quotes. 1.变量包含文字引号。 Bash does not recursively parse quotes, it only parses the "outer" quotes given on the command line.
Bash不会递归地解析引号,它只解析命令行中给出的“外部”引号。
$ mkdir "/tmp/dir with spaces"
$ var='"/tmp/dir with spaces"'
$ echo "$var"
"/tmp/dir with spaces"
$ [ -d "/tmp/dir with spaces" ]; echo $?
0
$ [ -d "$var" ]; echo $? # equivalent to [ -d '"/tmp/dir with spaces"' ]
1
2. The variable contains a relative path, and the current directory is not what you expected. 2.变量包含相对路径,当前目录不是您所期望的。 Check that the value of
echo "$PWD"
outputs what you expected. 检查
echo "$PWD"
的值是否输出您的预期。
3. The variable was read from a file with dos line endings, CRLF (\\r\\n). 3.从具有dos行结尾的文件CRLF(\\ r \\ n)读取变量。 Unix and unix-like systems use just LF (\\n) for line endings.
Unix和unix类系统只使用LF(\\ n)作为行结尾。 If that's the case, the path will contain a CR (\\r) at the end.
如果是这种情况,路径将在末尾包含CR(\\ r)。 A CR at the end of a line will be "invisible" in a terminal.
一行末尾的CR将在终端中“不可见”。 Check with
printf '%q\\n' "$var"
while debugging the script. 在调试脚本时检查
printf '%q\\n' "$var"
。 See BashFAQ 52 on how to get rid of them. 请参阅BashFAQ 52 ,了解如何摆脱它们。
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