perl - 拆分空白字符并去除空字段

Question

($red, $tapinfo) = split(/:/, $line);
@fields = split(/\s+/, $tapinfo);

在数组字段中，我看到甚至添加了空间。我想消除空格，以便字段只包含非空格字符。请评论可能出了什么问题。

Answer 1

我假设您正在谈论剩余的前导空格，因此@fields看起来像：

$VAR1 = [
          '',    # empty field
          'foo',
          'bar'
        ];

这是因为当/\s+/您split应该使用默认值' '（单个空格字符）时，您正在使用。此默认行为将在拆分字符串之前去除前导空格。换句话说，你应该这样做：

@fields = split(' ', $tapinfo);

这记录在perldoc -f split中：

As another special case, "split" emulates the default behavior
of the command line tool awk when the PATTERN is either omitted
or a *literal string* composed of a single space character (such
as ' ' or "\x20", but not e.g. "/ /"). In this case, any leading
whitespace in EXPR is removed before splitting occurs, and the
PATTERN is instead treated as if it were "/\s+/"; in particular,
this means that *any* contiguous whitespace (not just a single
space character) is used as a separator. However, this special
treatment can be avoided by specifying the pattern "/ /" instead
of the string " ", thereby allowing only a single space
character to be a separator.

Answer 2

默认split情况下与

my @list = $string =~ /\S+/g;

即它找到所有非空白字符的连续子串。

您可以使用正则表达式，但要从中获取默认行为split，请将单个文字空格字符作为第一个参数传递。不是正则表达式。文档说这个

作为另一种特殊情况，当 PATTERN 被省略或由单个空格字符组成的文字字符串（例如 ' ' 或 "\x20" ，但不是例如 // ）时， split 模拟命令行工具 awk 的默认行为。在这种情况下，EXPR 中的任何前导空格在拆分发生之前都会被删除，而 PATTERN 则被视为 /\s+/ ；特别是，这意味着任何连续的空格（不仅仅是单个空格字符）都用作分隔符。