/ usr / bin / perl -e“statement”不接受变量

Question

我想将以下perl脚本作为oneliner执行：

perl -e 'print "my $var = "Hello "; print $var."World\n";'

以下面的错误消息结束：

Bareword found where operator expected at -e line 1, near ""$var = "Hello"
(Missing operator before Hello?)
String found where operator expected at -e line 1, near "Hello "; print $var.""
(Do you need to predeclare Hello?)
Backslash found where operator expected at -e line 1, near "World\"
String found where operator expected at -e line 1, at end of line
(Missing semicolon on previous line?)
syntax error at -e line 1, near ""$var = "Hello "
Can't find string terminator '"' anywhere before EOF at -e line 1.

然而，以下声明完美无缺：

perl -e 'print "Hello "; print "World\n";'

似乎perl -e不接受变量。这是真的，如果是，为什么？

Answer 1

您还有一个"：

perl -e 'print "my $var = "Hello "; print $var."World\n";'   
#      here  __^

删除它。

但它并不清楚你想要什么。