PHP - 正则表达式只允许字母和数字
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PHP - regex to allow letters and numbers only
提问by Jason94
I have tried:
我试过了：
preg_match("/^[a-zA-Z0-9]", $value)
but im doing something wrong i guess.
但我猜我做错了什么。
回答by SW4
1. Use PHP's inbuilt ctype_alnum
1.使用PHP内置的ctype_alnum
You dont need to use a regex for this, PHP has an inbuilt function ctype_alnumwhich will do this for you, and execute faster:
您不需要为此使用正则表达式，PHP 有一个内置函数ctype_alnum可以为您执行此操作，并且执行速度更快：
<?php
$strings = array('AbCd1zyZ9', 'foo!#$bar');
foreach ($strings as $testcase) {
if (ctype_alnum($testcase)) {
echo "The string $testcase consists of all letters or digits.\n";
} else {
echo "The string $testcase does not consist of all letters or digits.\n";
}
}
?>
2. Alternatively, use a regex
2. 或者，使用正则表达式
If you desperately want to use a regex, you have a few options.
如果您非常想使用正则表达式，您有几个选择。
Firstly:
首先：
preg_match('/^[\w]+$/', $string);
\wincludes more than alphanumeric (it includes underscore), but includes all
of \d.
\w包括多个字母数字（包括下划线），但包括所有\d.
Alternatively:
或者：
/^[a-zA-Z\d]+$/
Or even just:
或者甚至只是：
/^[^\W_]+$/
回答by Kendall Hopkins
You left off the /(pattern delimiter) and $(match end string).
您省略了/(模式分隔符) 和$(匹配结束字符串)。
preg_match("/^[a-zA-Z0-9]+$/", $value)
回答by KingCrunch
- Missing end anchor $
- Missing multiplier
- Missing end delimiter
- 缺少结束锚 $
- 缺少乘数
- 缺少结束分隔符
So it should fail anyway, but if it may work, it matches against just one digit at the beginning of the string.
所以无论如何它都应该失败，但如果它可以工作，它只会匹配字符串开头的一位数字。
/^[a-z0-9]+$/i
回答by Andrew
As the OP said that he wants letters and numbers ONLY (no underscore!), one more way to have this in phpregex is to use posix expressions:
正如 OP 所说，他只想要字母和数字（没有下划线！），在phpregex 中实现这一点的另一种方法是使用 posix 表达式：
/^[[:alnum:]]+$/
Note: This will not work in Java, JavaScript, Python, Ruby, .NET
注意：这不适用于 Java、JavaScript、Python、Ruby、.NET
回答by pooja
try this way .eregi("[^A-Za-z0-9.]", $value)
试试这种方式 .eregi("[^A-Za-z0-9.]", $value)
