Question

如何剥离第一个（在任何可见文本之前）从变量中提取的文本中输入/返回空格（从textarea提交）？

Answer 1

像这样：

$str = str_replace("\n\r", '', $content);

注意：如果您想将新功能转换为<br />，请改用nl2br。

另请注意，"\n\r"适用于Windows系统，在{iix / linux>上为\n。

<强>更新

试试这个正则表达式：

$str = preg_replace(/[\n\r{2,}]+/, "\n", $content);

Answer 2

$userinput = trim($userinput)

trim() works for followings, though you can strip some of those giving a char list
    * " " (ASCII 32 (0x20)), an ordinary space.
    * "\t" (ASCII 9 (0x09)), a tab.
    * "\n" (ASCII 10 (0x0A)), a new line (line feed).
    * "\r" (ASCII 13 (0x0D)), a carriage return.
    * "\0" (ASCII 0 (0x00)), the NUL-byte.
    * "\x0B" (ASCII 11 (0x0B)), a vertical tab.

Answer 3

ltrim在文本之前只删除空格。你可以在第二个参数中添加一些你不需要的额外字符。

$userinput = ltrim($userinput, $charsYouDontNeed);

它不起作用的事实可能意味着你试图删除的字符不是真正的新行字符

Answer 4

尝试：

$str = preg_replace('/[\n\r{2,}]+/', "\n", $content);

PHP String Parsing Strip * First *从表单textarea输入/返回空格

4 个答案: