展开反向正则表达式以允许 1 个空格字符
如何更改下面的模式以允许 1 个空格字符？
$name = 'too long name';
$pattern_name = '/[^a-zA-Z]/';
if (preg_match($pattern_name,$name)) { // remove any non-letter characters
$name = preg_replace($pattern_name,'',$name);
$errors['name'] = 'Invalid characters found and removed in name';
}
使用这些模式中的任何一个都不起作用：
$pattern_name = '/[^a-zA-Z ?]/';
$pattern_name = '/[^a-zA-Z] ?/';
预期结果是匹配，因为 $name 中存在 2 个空格字符，因此 if 语句应该为 true，并且替换函数将更新 $name，因此其值将变为“too longname”。
How can I alter the pattern below to allow 1 space character ?
$name = 'too long name';
$pattern_name = '/[^a-zA-Z]/';
if (preg_match($pattern_name,$name)) { // remove any non-letter characters
$name = preg_replace($pattern_name,'',$name);
$errors['name'] = 'Invalid characters found and removed in name';
}
Using either of these patterns does not work:
$pattern_name = '/[^a-zA-Z ?]/';
$pattern_name = '/[^a-zA-Z] ?/';
Expected result is a match, since 2 space characters exists in $name, thus the if-statement should be true and the replace function will update $name so its value will become "too longname".
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评论（3）
您必须使您的模式更加明确。如果最多可以有一个空格，并且它必须被字母包围，那么：
You'll have to make your pattern more explicit. If you can have one space at maximum, and it must be surrounded by letters, then:
它应该像在括号中添加空格一样简单。
It should be as simple as adding a space in the brackets.
我会反转正则表达式，而不是尝试查找无效字符，而是匹配有效名称（这是您正在做的事情吗？）。这给了我们这个正则表达式：
/[a-zA-Z]+ [a-zA-Z]+/。匹配有效字符、一个空格，然后匹配更多有效字符。I'd invert the regex, and instead of trying to find invalid characters, match valid names (is that what you are doing?). That gives us this regex:
/[a-zA-Z]+ [a-zA-Z]+/. Match valid characters, one space and then more valid characters.