java 如何检查一个字符串是否至少包含另一个字符串中的一个字符？
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How do I check if a string contains at least one character from another string?
提问by Dave
I'm using Java 6. I have a string that contains "special" characters -- "!@#$%^&*()_". How do I write a Java expression to check if another string, "password", contains at least one of the characters defined in the first string? I have
我使用的是 Java 6。我有一个包含“特殊”字符的字符串——“!@#$%^&*()_”。如何编写 Java 表达式来检查另一个字符串“密码”是否至少包含第一个字符串中定义的一个字符？我有
regForm.getPassword().matches(".*[\~\!\@\#\$\%\^\&\*\(\)\_\+].*")
but I don't want to hard-code the special characters, rather load them into a string from a properties file. So I'm having trouble figuring out how to escape everything properly after I load it from the file.
但我不想对特殊字符进行硬编码，而是将它们从属性文件加载到字符串中。所以我在从文件中加载它后无法弄清楚如何正确地转义所有内容。
回答by Mikita Belahlazau
You can try creating regex from string that contains special characters and escape symbols using Pattern.quote. Try this:
您可以尝试使用Pattern.quote从包含特殊字符和转义符号的字符串创建正则表达式。试试这个：
String special = "!@#$%^&*()_";
String pattern = ".*[" + Pattern.quote(special) + "].*";
regFrom.getPassword().matches(pattern);
回答by anubhava
I think simple looping the regex to check each character might work better and will work for all the cases:
我认为简单的循环正则表达式来检查每个字符可能会更好，并且适用于所有情况：
String special = "!@#$%^&*()_";
boolean found = false;
for (int i=0; i<special.length(); i++) {
if (regFrom.getPassword().indexOf(special.charAt(i)) > 0) {
found = true;
break;
}
}
if (found) { // password has one of the allowed characters
//...
//...
}
回答by SJuan76
One option is to use StringTokenizer, and see if it returns more than 1 substring. It has a constructor that allows specifying the characters to split by.
一种选择是使用StringTokenizer, 并查看它是否返回超过 1 个子字符串。它有一个构造函数，允许指定要拆分的字符。
Anyway, my favourite option would be just iterating the characters and using String.indexOf.
无论如何，我最喜欢的选择就是迭代字符并使用String.indexOf.
