在PHP中只用一个空格替换多个空格和换行符

Question

我有一个包含多个换行符的字符串.

字符串:

This is         a dummy text.               I need              




to                                      format
this.

期望的输出:

This is a dummy text. I need to format this.

我正在使用这个:

$replacer  = array("\r\n", "\n", "\r", "\t", "  ");
$string = str_replace($replacer, "", $string);

但它没有按要求/要求工作.有些单词之间没有空格.

实际上我需要用单个空格分隔的所有单词转换字符串.

Answer 1

我鼓励你使用preg_replace:

# string(45) "This is a dummy text . I need to format this."
$str = preg_replace( "/\s+/", " ", $str );

演示:http://codepad.org/no6zs3oo

您可能已在" . "第一个示例的部分中注意到.紧跟着标点符号的空格应该可以完全删除.快速修改允许这样:

$patterns = array("/\s+/", "/\s([?.!])/");
$replacer = array(" ","$1");

# string(44) "This is a dummy text. I need to format this."
$str = preg_replace( $patterns, $replacer, $str );

演示:http://codepad.org/ZTX0CAGD