Question

我有一个脚本，我想找出HTTP请求的状态代码。但是，if语句从未评估为真，我不明白为什么。

#!/bin/sh

set -e

CURL='/usr/bin/curl'
CURL_ARGS='-o - -I -s'
GREP='/usr/bin/grep'

url="https://stackoverflow.com"

res=$($CURL $CURL_ARGS $url | $GREP "HTTP/1.1")

echo $res # This outputs 'HTTP/1.1 200 OK'
echo ${#res} # This outputs 16, even though it should be 15

if [ "$res" == "HTTP/1.1 200 OK" ]; then # This never evaluates to true
  echo "It worked"
  exit 1
fi

echo "It did not work"

我检查了res的长度，它是16，我在浏览器的控制台中检查了它，它是15，所以我通过删除两端的空格来修剪它，但它仍然没有评估为真。

res_trimmed="$(echo "${res}" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')"

它仍然没有＆＃ 39;工作。

可能出现什么问题？任何帮助表示赞赏。感谢。

Answer 1

更好的实践实施可能如下：

#!/usr/bin/env bash
#              ^^^^- ensure that you have bash extensions available, rather than being
#                    only able to safely use POSIX sh syntax. Similarly, be sure to run
#                    "bash yourscript", not "sh yourscript".

set -o pipefail  # cause a pipeline to fail if any component of it fails

url="https://stackoverflow.com"

# curl -f == --fail => tell curl to fail if the server returns a bad (4xx, 5xx) response
res=$(curl -fsSI "$url" | grep "HTTP/1.1") || exit
res=${res%$'\r'}  # remove a trailing carriage return if present on the end of the line

if [ "$res" = "HTTP/1.1 200 OK" ]; then
  echo "It worked" >&2
  exit 0            # default is the exit status of "echo". Might just pass that through?
fi

echo "It did not work" >&2
exit 1

Answer 2

您的问题是您从命令替换返回时遇到了流浪字符。要消除，只匹配有效字符，例如

GREP='/usr/bin/grep -o'
...
res=$($CURL $CURL_ARGS $url | $GREP 'HTTP/1.1[A-Za-z0-9 ]*')

其他改变

echo "'$res'" # This outputs 'HTTP/1.1 200 OK'

示例使用/输出

$ sh curltest.sh
'HTTP/1.1 200 OK'
15
It worked

Bash字符串比较不eval为true

2 个答案: