[英]Regex to find multiple words that don't match (Negative words matching)
I need to allow only a specific word per time per string, but exists multiple valid words. 每个字符串每次只允许一个特定的单词,但是存在多个有效单词。
Using the below regex i am not getting the expected result: 使用下面的正则表达式我没有得到预期的结果:
preg_replace('/\b^(true|false|TRUE|FALSE)\b/', '', 'false'); // Returns an empty string, but i expect 'false'
preg_replace('/\b^(true|false|TRUE|FALSE)\b/', '', 'test') // Returns 'test', but i expect an empty string
Someone knows what's wrong? 有人知道怎么了吗?
EDIT 1 编辑1
An example for a long input: 长输入的示例:
preg_replace('/\b^(true|false|TRUE|FALSE)\b/', '', 'This regex allow only boolean, such as true, false, TRUE and FALSE');
It prints: 它打印:
// This regex allow only boolean, such true, false, TRUE and FALSE
But i expect an empty string, because only a sigle word should be a valid match 但是我期望一个空字符串,因为只有一个单词是有效的匹配
Your regex is wrong for what you're trying to do. 您的正则表达式对于您要执行的操作是错误的。 You will need negative lookahead.
您将需要负前瞻。
Try this code: 试试这个代码:
$re = '/\b(?!(?:true|false|TRUE|FALSE))\w+\b/';
$str = 'I need true to allow only false specific FALSE words in a TRUE string';
$repl = preg_replace($re, "", $str);
//=> true false FALSE TRUE
Not sure I well understand your needs, but is that OK for you: 不确定我是否完全了解您的需求,但是您是否可以:
if (preg_match('/^(true|false|TRUE|FALSE)$/', $string, $match)) {
echo "Found : ",$m[1],"\n";
} else {
echo "Not found\n";
}
In this case you don't need a regex, you can use in_array
: 在这种情况下,您不需要正则表达式,可以使用
in_array
:
$arr = array('true', 'false', 'TRUE', 'FALSE');
$result = (in_array($str, $arr, true)) ? $str : '';
Is this what you want? 这是你想要的吗?
(?=.)(true|false|TRUE|FALSE|\\s*)(?!.) (=?)(TRUE | FALSE | TRUE | FALSE | \\ S *)(?!)。
$regex = '/(?=.)(true|false|TRUE|FALSE|\\s{0,})(?!.)/';
$testString = ''; // Fill this in
preg_match($regex, $testString, $matches);
// the $matches variable contains the list of matches
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