我想检查一个字符串是否以“节点”开始，例如。“node001”。类似的
if [ $HOST == user* ]
then
echo yes
fi
我怎样才能正确地做呢?
我进一步需要组合表达式来检查HOST是否为“user1”或以“node”开头
if [ [[ $HOST == user1 ]] -o [[ $HOST == node* ]] ];
then
echo yes
fi
> > > -bash: [: too many arguments
我怎样才能正确地做呢?
当前回答
保持简单
word="appel"
if [[ $word = a* ]]
then
echo "Starts with a"
else
echo "No match"
fi
其他回答
保持简单
word="appel"
if [[ $word = a* ]]
then
echo "Starts with a"
else
echo "No match"
fi
if [ [[ $HOST == user1 ]] -o [[ $HOST == node* ]] ];
then
echo yes
fi
不起作用，因为所有的[、[[和test都识别相同的非递归语法。请参阅Bash手册页上的条件表达式部分。
说句题外话，SUSv3说
The KornShell-derived conditional command (double bracket [[]]) was removed from the shell command language description in an early proposal. Objections were raised that the real problem is misuse of the test command ([), and putting it into the shell is the wrong way to fix the problem. Instead, proper documentation and a new shell reserved word (!) are sufficient. Tests that require multiple test operations can be done at the shell level using individual invocations of the test command and shell logicals, rather than using the error-prone -o flag of test.
你需要这样写，但是test不支持:
if [ $HOST == user1 -o $HOST == node* ];
then
echo yes
fi
Test使用=表示字符串相等，更重要的是它不支持模式匹配。
Case / esac对模式匹配有很好的支持:
case $HOST in
user1|node*) echo yes ;;
esac
它还有一个额外的好处，那就是它不依赖于Bash，而且语法是可移植的。从单一Unix规范，Shell命令语言:
case word in
[(]pattern1) compound-list;;
[[(]pattern[ | pattern] ... ) compound-list;;] ...
[[(]pattern[ | pattern] ... ) compound-list]
esac
你可以选择你想要检查的字符串部分:
if [ "${HOST:0:4}" = user ]
对于你接下来的问题，你可以使用OR:
if [[ "$HOST" == user1 || "$HOST" == node* ]]
我调整了@markrushakoff的答案，使其成为一个可调用的函数:
function yesNo {
# Prompts user with $1, returns true if response starts with y or Y or is empty string
read -e -p "
$1 [Y/n] " YN
[[ "$YN" == y* || "$YN" == Y* || "$YN" == "" ]]
}
像这样使用它:
$ if yesNo "asfd"; then echo "true"; else echo "false"; fi
asfd [Y/n] y
true
$ if yesNo "asfd"; then echo "true"; else echo "false"; fi
asfd [Y/n] Y
true
$ if yesNo "asfd"; then echo "true"; else echo "false"; fi
asfd [Y/n] yes
true
$ if yesNo "asfd"; then echo "true"; else echo "false"; fi
asfd [Y/n]
true
$ if yesNo "asfd"; then echo "true"; else echo "false"; fi
asfd [Y/n] n
false
$ if yesNo "asfd"; then echo "true"; else echo "false"; fi
asfd [Y/n] ddddd
false
下面是一个更复杂的版本，提供了一个指定的默认值:
function toLowerCase {
echo "$1" | tr '[:upper:]' '[:lower:]'
}
function yesNo {
# $1: user prompt
# $2: default value (assumed to be Y if not specified)
# Prompts user with $1, using default value of $2, returns true if response starts with y or Y or is empty string
local DEFAULT=yes
if [ "$2" ]; then local DEFAULT="$( toLowerCase "$2" )"; fi
if [[ "$DEFAULT" == y* ]]; then
local PROMPT="[Y/n]"
else
local PROMPT="[y/N]"
fi
read -e -p "
$1 $PROMPT " YN
YN="$( toLowerCase "$YN" )"
{ [ "$YN" == "" ] && [[ "$PROMPT" = *Y* ]]; } || [[ "$YN" = y* ]]
}
像这样使用它:
$ if yesNo "asfd" n; then echo "true"; else echo "false"; fi
asfd [y/N]
false
$ if yesNo "asfd" n; then echo "true"; else echo "false"; fi
asfd [y/N] y
true
$ if yesNo "asfd" y; then echo "true"; else echo "false"; fi
asfd [Y/n] n
false
虽然我发现这里的大多数答案都是正确的，但其中许多都包含不必要的bashism。POSIX参数展开为您提供所需的一切:
[ "${host#user}" != "${host}" ]
and
[ "${host#node}" != "${host}" ]
${var#expr}从${var}中剥离匹配expr的最小前缀并返回。因此，如果${host}不以用户(节点)开头，${host#user} (${host#node})与${host}相同。
Expr允许fnmatch()通配符，因此${host#node??朋友也能工作。
