bash - 在 URL 中使用 Bash 变量会导致 "curl: (3) Illegal characters found in URL"

Question

问题

我尝试在调用 cURL 命令时使用变量，但它包含文字 $line 而不是实际值。

while read line; do
  curl "https://x.com/v1/PhoneNumbers/$line?Type=carrier" -u "x:x" 
done < "${1:-/dev/stdin}"

上下文

我将一个数字列表传递给脚本，尝试逐行读取它们。

Answer 1

从评论中我们知道您收到以下错误:

curl: (3) Illegal characters found in URL

如果这样格式化:

while IFS="$IFS"$'\r' read line; do
  curl "https://x.com/v1/PhoneNumbers/$line?Type=carrier" -u "x:x" 
done < "${1:-/dev/stdin}"

你的命令应该可以工作。

问题在于您在输入行末尾附加了 \r (以便输入的每一行都以 \r\n 序列结尾)。默认情况下，read 不会删除尾随的r。如果我们想要 read 修剪这些字符，我们必须将此字符添加到 IFS 环境变量中进行 read，如下所示: IFS= “$IFS”$'\r')”读取...。

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这是来自 Charles Duffy 的精彩评论:

Personally, I'd suggest IFS=$' \t\n\r', not referring to the old $IFS value -- why make your code's behavior contextually dependent?

又一条宝贵意见；这次来自chepner :

Granted, a valid line probably isn't going to contain a \r, but conceptually, you don't want to treat a carriage return as whitespace; you just want to strip the \r that is part of the \r\n line ending. Instead of modifying IFS, read the line normally, then strip it with line=${line%$'\r'} before calling curl.

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