我也有同样的问题。确保你的脚本没有

#!/bin/sh 

在你的脚本顶部。相反，你应该补充

#!/bin/bash

您的脚本语法是有效的bash和良好的。

故障可能原因:

Your bash is not really bash but ksh or some other shell which doesn't understand bash's parameter substitution. Because your script looks fine and works with bash. Do ls -l /bin/bash and check it's really bash and not sym-linked to some other shell. If you do have bash on your system, then you may be executing your script the wrong way like: ksh script.sh or sh script.sh (and your default shell is not bash). Since you have proper shebang, if you have bash ./script.sh or bash ./script.sh should be fine.

我也有同样的问题。确保你的脚本没有

#!/bin/sh 

在你的脚本顶部。相反，你应该补充

#!/bin/bash

我发现这个问题要么是由标记的答案引起的，要么是在bash声明之前有一行或空格

尝试使用bash命令显式地运行脚本，而不是将其作为可执行文件执行。

看起来“+x”会导致问题:

root@raspi1:~# cat > /tmp/btest
#!/bin/bash

jobname="job_201312161447_0003"
jobname_pre=${jobname:0:16}
jobname_post=${jobname:17}
root@raspi1:~# chmod +x /tmp/btest
root@raspi1:~# /tmp/btest
root@raspi1:~# sh -x /tmp/btest
+ jobname=job_201312161447_0003
/tmp/btest: 4: /tmp/btest: Bad substitution