jsf - JSF id 的规则是什么？

Question

看起来我应该能够在网上搜索半小时找到这个，但因为我不能:

有效 JSF ID 的规则是什么？

我读到一封乱码电子邮件，表明 - 和 _ 存在限制，但我收到了 IllegalArgumentExceptions，我认为这是由于 ids 造成的。

编辑

java.lang.IllegalArgumentException: 6a945017207d46fd82b3d3bb7d2795f1
at javax.faces.component.UIComponentBase.validateId(UIComponentBase.java:549)
at javax.faces.component.UIComponentBase.setId(UIComponentBase.java:351)
at com.sun.facelets.tag.jsf.ComponentHandler.apply(ComponentHandler.java:151)

Answer 1

它必须是有效的 CSS identifier (ident here)并且不应有重复项。

In CSS, identifiers (including element names, classes, and IDs in selectors) can contain only the characters [a-zA-Z0-9] and ISO 10646 characters U+00A1 and higher, plus the hyphen (-) and the underscore (_); they cannot start with a digit, or a hyphen followed by a digit. Identifiers can also contain escaped characters and any ISO 10646 character as a numeric code (see next item). For instance, the identifier "B&W?" may be written as "B\&W\?" or "B\26 W\3F".

另请参阅:

• Allowed characters for CSS identifiers

<小时/>

更新:对于您感兴趣的情况，以下是 UIComponentBase#validateId() 提供的验证器源代码:

private static void validateId(String id) {
    if (id == null) {
        return;
    }
    int n = id.length();
    if (n < 1) {
        throw new IllegalArgumentException("Empty id attribute is not allowed");
    }
    for (int i = 0; i < n; i++) {
        char c = id.charAt(i);
        if (i == 0) {
            if (!Character.isLetter(c) && (c != '_')) {
                throw new IllegalArgumentException(id);
            }
        } else {
            if (!Character.isLetter(c) &&
                    !Character.isDigit(c) &&
                    (c != '-') && (c != '_')) {
                throw new IllegalArgumentException(id);
            }
        }
    }
}

但是它比 CSS 规则更严格一些。它们也不能以连字符开头。