如何从PHP文本中删除空白行？

Question

我需要在PHP中删除空白行（带空格或绝对空白）。我使用此正则表达式，但不起作用：

$str = ereg_replace('^[ \t]*$\r?\n', '', $str);
$str = preg_replace('^[ \t]*$\r?\n', '', $str);

我想要一个结果：

blahblah

blahblah

   adsa


sad asdasd

将：

blahblah
blahblah
   adsa 
sad asdasd

Answer 1

// New line is required to split non-blank lines
preg_replace("/(^[\r\n]*|[\r\n]+)[\s\t]*[\r\n]+/", "\n", $string);

上面的正则表达式说：

/(^[\r\n]*|[\r\n]+)[\s\t]*[\r\n]+/
    1st Capturing group (^[\r\n]*|[\r\n]+)
        1st Alternative: ^[\r\n]*
        ^ assert position at start of the string
            [\r\n]* match a single character present in the list below
                Quantifier: Between zero and unlimited times, as many times as possible, giving back as needed [greedy]
                \r matches a carriage return (ASCII 13)
                \n matches a fine-feed (newline) character (ASCII 10)
        2nd Alternative: [\r\n]+
            [\r\n]+ match a single character present in the list below
            Quantifier: Between one and unlimited times, as many times as possible, giving back as needed [greedy]
            \r matches a carriage return (ASCII 13)
            \n matches a fine-feed (newline) character (ASCII 10)
    [\s\t]* match a single character present in the list below
        Quantifier: Between zero and unlimited times, as many times as possible, giving back as needed [greedy]
        \s match any white space character [\r\n\t\f ]
        \tTab (ASCII 9)
    [\r\n]+ match a single character present in the list below
        Quantifier: Between one and unlimited times, as many times as possible, giving back as needed [greedy]
        \r matches a carriage return (ASCII 13)
        \n matches a fine-feed (newline) character (ASCII 10)