如何通过引用将方法返回的散列元素插入到字符串中？ [英] How to interpolate hash element via reference returned by a method into a string?

问题描述

我想插入对字符串的散列引用，但此方法不起作用。
如何插入 $ self> Test-> {text} ？

 ＃$ self> Test-> {text}包含test 123 ok
 printValue is：$ self-> Test-> {text} \ N; ＃但不工作
  

输出：

  Test = HASH（0x2948498） - > Test-> {text} 
  

解决方案

方法调用不会在双引号内部进行插值，所以您最终得到字符串化引用，然后是 - > Test-> {text } 。

简单的做法是利用 print 获取参数列表的事实：

  printValue is：，$ self-> Test-> {text}，\\\
; 
  

您也可以使用连接：

  printValue is：。 $ self> Test-> {text}。 \\\
; 
  

您也可以使用 printf
$ p $ printfValue is％s \\\
，$ self> Test-> {text} ;


或者您可以使用这个愚蠢的技巧：

  printValue is：@ {[$ self-> Test-> {text}]} \\\
; 
  




I want to interpolate hash reference to string, but this method is not working. How does one interpolate $self->Test->{text} ?

# $self->Test->{text} contains "test 123 ok"
print "Value is: $self->Test->{text} \n";   # but not working

output:

Test=HASH(0x2948498)->Test->{text} 

解决方案

Method calls won't get interpolated inside double quotes, so you end up with the stringified reference followed by ->Test->{text}.

The simple way to do it is to take advantage of the fact that print takes a list of arguments:

print "Value is: ", $self->Test->{text}, "\n";

You could also use concatenation:

print "Value is: " . $self->Test->{text} . "\n";

You could also use the tried-and-true printf

printf "Value is %s\n", $self->Test->{text};

Or you can use this silly trick:

print "Value is: @{ [ $self->Test->{text} ] }\n";

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