检查 bash 中 if 语句中的两个条件
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check two conditions in if statement in bash
提问by MLSC
I have problem in writing if statement
我在写 if 语句时遇到问题
var1=`COMMAND | grep <Something>`
if [ -n "$var1" ] && if [[ "$var1" != string ]]; then
.
.
.
fi
I want to write if statement that check:
我想写 if 语句来检查：
If var1is not null ANDif string(could be helloword) is not in var1 then do stuff.
如果VAR1不为空和是否串（可以是hello文字）是不是在VAR1然后做的东西。
How can I do that?
我怎样才能做到这一点？
回答by fedorqui 'SO stop harming'
Just use something like this:
只需使用这样的东西：
if [ -n "$var1" ] && [[ ! $var1 == *string* ]]; then
...
fi
See an example:
看一个例子：
$ v="hello"
$ if [ -n "$v" ] && [[ $v == *el* ]] ; then echo "yes"; fi
yes
$ if [ -n "$v" ] && [[ ! $v == *ba* ]] ; then echo "yes"; fi
yes
The second condition is a variation of what is indicated in String contains in bash.
第二个条件是String contains in bash 中指示的内容的变体。
回答by ooga
Other possibilities:
其他可能性：
if [ -n "$var1" -a "$var1" != string ]; then ...
if [ "${var1:-xxx}" != "string" ]; then ...
回答by opalenzuela
You should rephrase the condition like this:
你应该像这样改写条件：
var1=`COMMAND | grep <Something>`
if [ -n "$var1" ] && [[ "$var1" != "string" ]]; then
.
.
.
fi
or the equivalent:
或等价物：
if test -n "$var1" && test "$var1" != "string"
then
...
fi
