Bash - 比较两个命令的输出
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Bash - comparing output of two commands
提问by antoninkriz
I have this code:
我有这个代码：
#!/bin/bash
CMDA=$(curl -sI website.com/example.txt | grep Content-Length)
CMDB=$(curl -sI website.com/example.txt | grep Content-Length)
if [ "CMDA" == "CMDB" ];then
echo "equal";
else
echo "not equal";
fi
with this output
有了这个输出
root@abcd:/var/www/html# bash ayy.sh
not equal
which should be "equal" instead of "not equal". What did I do wrong?
这应该是“相等”而不是“不相等”。我做错了什么？
Thnaks
纳克斯
回答by janos
You forgot the $for the variables CMDAand CMDBthere. This is what you need:
你忘记了$变量CMDA和CMDB那里。这是你需要的：
if [ "$CMDA" = "$CMDB" ]; then
I also changed the ==operator to =,
because man testonly mentions =,
and not ==.
我也将==运算符更改为=，因为man test只提到了=，而不是==。
Also, you have some redundant semicolons. The whole thing a bit cleaner:
此外，您还有一些多余的分号。整个事情更干净一点：
if [ "$CMDA" = "$CMDB" ]; then
echo "equal"
else
echo "not equal"
fi
回答by Aus
You are comparing string "CMDA" with "CMDB", you should instead compare the variables using $ like ${CMDA}
您正在将字符串“CMDA”与“CMDB”进行比较，您应该使用 $ like ${CMDA} 来比较变量
