illegal character in authority at indx 7:

Sorry, as an AI language model, I cannot provide a specific answer without additional context.

However, this error message typically indicates that there is a problem with the format of a URL. The "authority" refers to the part of the URL after "://" and before the next "/" character.

An "illegal character" in this context means that there is a character that is not allowed or recognized in the authority section of a URL. This can happen if there is a typo or if the URL is not properly encoded.

If you provide more information about the specific URL or context in which you encountered this error, I may be able to provide more specific guidance.

相关问题

java.lang.IllegalArgumentException: Illegal character in query at index 235:

这个错误表示在查询字符串的索引 235 处存在非法字符。在 Java 中，URL 查询字符串应该是经过编码的，不允许包含一些特殊字符。你需要检查查询字符串中是否包含了这样的非法字符，并进行相应的处理。

你可以尝试使用 `URLEncoder` 类来对查询字符串进行编码，例如：

String encodedQuery = URLEncoder.encode(query, "UTF-8");
String url = "https://example.com/search?query=" + encodedQuery;

这将使用 UTF-8 编码对查询字符串进行编码，然后将其追加到 URL 中。这样可以确保查询字符串中的特殊字符被正确地编码，避免引发异常。

java.lang.IllegalArgumentException: Illegal character in query at index 167:

java.lang.IllegalArgumentException是Java中的一个异常类，它表示方法接收到了一个不合法的参数。在你提供的引用中，异常信息是"Illegal character in query at index 167:?"。这个异常通常在处理URL或查询字符串时出现，表示在查询字符串中存在非法字符。

以下是一个示例代码，演示了如何处理这个异常：

try {
    // 假设这里有一段代码处理URL或查询字符串
    // ...
} catch (IllegalArgumentException e) {
    System.out.println("Illegal character in query: " + e.getMessage());
    // 其他异常处理逻辑
}

在这个示例中，我们使用try-catch语句来捕获IllegalArgumentException异常。如果异常被抛出，我们将打印异常信息并执行其他异常处理逻辑。