简单的bash脚本；这种语法有什么问题? [英] Simple bash script; what's wrong with this syntax?

问题描述

我正在阅读bash脚本教程，这似乎应该可以工作，但是我显然缺少了一些东西:

I'm reading bash script tutorials and this seems like it should work, but I clearly am missing something:

isframecount=0
framecount=0
while read p; do
  if ["$isframecount" -eq 0]
    then 
    echo $p
    $isframecount = 1
    else
    echo "frame count"
    echo $p
    $isframecount = 0
  fi
done < $filenam

当我尝试运行以上命令时，出现找不到命令错误"消息；有指针吗?

I get "command not found errors" when I try to run the above; any pointers?

推荐答案

当心哪些地方重要，哪些地方不重要.在这种情况下，if ["$isframecount" -eq 0]应该为if [ "$isframecount" -eq 0 ](请参见[之后和]之前的空格).

Watch out for spaces where they matters and where there should be not. In this case, if ["$isframecount" -eq 0] should be if [ "$isframecount" -eq 0 ] (see the spaces after [ and before ]).

这样做的原因是[实际上是程序的名称...自己查看...键入ls /bin/[ ...现在，如果没有空间，那么bash会寻找程序名为["0"或类似名称的东西，您的路径中肯定不存在

The reason for this is that [ is actually the name of a program... See by yourself... Type ls /bin/[... Now, if there is no space, then bash will look for a program named ["0" or something similar, which most certainly does not exist in your path.

然后，相反的情况...在变量分配中，=周围必须没有空格.所以$isframecount = 1应该是isframecount=1.请注意，我也删除了美元符号，这是要走的路.

Then, the opposite situation... There must be no space around the = in variable assignment. So $isframecount = 1 should be isframecount=1. Note that I also removed the dollar sign, that's the way to go.

这篇关于简单的bash脚本；这种语法有什么问题?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！