Perl 中的字符串与“eq"比较与“==" [英] String compare in Perl with "eq" vs "=="
问题描述
我(一个完整的 Perl 新手)在 if
语句中进行字符串比较:
I am (a complete Perl newbie) doing string compare in an if
statement:
如果我执行以下操作:
if ($str1 == "taste" && $str2 == "waste") { }
我看到了正确的结果(即如果条件匹配,它会评估then"块).但我看到了这些警告:
I see the correct result (i.e. if the condition matches, it evaluates the "then" block). But I see these warnings:
在第 x 行的数字 eq (==) 中,参数taste"不是数字.
在第 x 行的数字 eq (==) 中,参数waste"不是数字.
Argument "taste" isn't numeric in numeric eq (==) at line number x.
Argument "waste" isn't numeric in numeric eq (==) at line number x.
但如果我这样做:
if ($str1 eq "taste" && $str2 eq "waste") { }
即使满足 if 条件,它也不会评估then"块.
Even if the if condition is satisfied, it doesn't evaluate the "then" block.
这里,$str1
是 taste
,$str2
是 waste
.
Here, $str1
is taste
and $str2
is waste
.
我应该如何解决这个问题?
How should I fix this?
推荐答案
首先,eq是用来比较字符串的;== 用于比较数字.
First, eq is for comparing strings; == is for comparing numbers.
即使满足if"条件,它也不会评估then"块.
Even if the "if" condition is satisfied, it doesn't evaluate the "then" block.
我认为您的问题是您的变量不包含您认为的内容.我认为您的 $str1
或 $str2
包含类似taste
"之类的东西.通过在 if 之前打印来检查它们:print "str1='$str1'
";
.
I think your problem is that your variables don't contain what you think they do. I think your $str1
or $str2
contains something like "taste
" or so. Check them by printing before your if: print "str1='$str1'
";
.
可以使用 chomp($str1);
函数删除尾随的换行符.
The trailing newline can be removed with the chomp($str1);
function.
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