简单的问题,但我很头疼解决这个游戏。示例正则表达式。
[a-zA-Z0-9\s]
[whitespace]Stack[whitespace]Overflow - not allow
Stack[whitespace]Overflow - allow
Stack[whitespace]Overflow[whitespace] - not allow
让我知道
更新 regex from JG并且它正在运行。
function regex($str)
{
$check = preg_replace('/^[a-zA-Z0-9][a-zA-Z0-9\s]+[a-zA-Z0-9]$|^[a-zA-Z0-9]*$/', "", $str);
if (empty($check)) {
return true;
} else {
return false;
}
}
$str = 'Stack Overflow ';
$validator = regex($str);
if ($validator) {
echo "OK » " .$str;
} else {
echo "ERROR » " . $str;
}
答案 0 :(得分:5)
尝试:
/^\S.*\S$|^\S$/
如果您只想要字母,数字和下划线,以及两个单词,不能少,不再需要:
/^\w+\s+\w+$/
没有下划线,
/^\p{Alnum}+\s+\p{Alnum}+$/
虽然,在一些正则表达式样式(特别是PHP,我现在看到的是目标)中,你使用它:
/^[[:alnum:]]+\s+[[:alnum:]]+$/
如果可以接受任何数量的此类单词和数字:
/^\w[\w\s]*\w$|^\w$/
答案 1 :(得分:2)
为什么你真的要使用正则表达式?
trim(默认情况下)删除:
* " " (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.
所以你需要的只是:
function no_whitespace($string)
{
return trim($string) === $string;
}
就是这样!
$tests = array
(
' Stack Overflow',
'Stack Overflow',
'Stack Overflow '
);
foreach ($tests as $test)
{
echo $test."\t:\t".(no_whitespace($test) ? 'allowed' : 'not allowed').PHP_EOL;
}
答案 2 :(得分:1)
根据我的理解,你想要一个正则表达式,它不允许你的字符串在开头或结尾都有空格。这些方面应该有用:
/^[a-zA-Z0-9][a-zA-Z0-9\s]+[a-zA-Z0-9]$|^[a-zA-Z0-9]*$/
Python中的一个例子:
import re
test = ["Stack Overflow",
"Stack&!Overflow",
" Stack Overflow",
"Stack Overflow ",
"x",
"",
" "]
regex = re.compile(r'^[a-zA-Z0-9][a-zA-Z0-9\s]+[a-zA-Z0-9]$|^[a-zA-Z0-9]*$')
for s in test:
print "'"+s+"'", "=>", "match" if regex.match(s) != None else "non-match"
输出:
'Stack Overflow' => match
'Stack&!Overflow' => non-match
' Stack Overflow' => non-match
'Stack Overflow ' => non-match
'x' => match
'' => match
' ' => non-match
答案 3 :(得分:0)
if ($string =~ /^\S.*\S$/){
print "allowed\n";
} else {
print "not allowed\n";
}