Shell Curl返回http状态代码和响应
我使用curl获取http头以查找http状态代码并返回响应。我使用命令获取http头Shell Curl返回http状态代码和响应,shell,curl,Shell,Curl,我使用curl获取http头以查找http状态代码并返回响应。我使用命令获取http头 curl -I http://localhost 为了得到响应，我使用命令 curl http://localhost 只要使用-I标志，我就只得到标题，响应就不再存在了。有没有一种方法可以在一个命令中同时获得http响应和头/http状态代码？我可以通过查看curl文档获得一个解决方案，该文档指定使用-作为输出以将输出获取到stdout curl -o - http://localhost 为了只使用
curl -I http://localhost
curl http://localhost
只要使用-I标志，我就只得到标题，响应就不再存在了。有没有一种方法可以在一个命令中同时获得http响应和头/http状态代码？我可以通过查看curl文档获得一个解决方案，该文档指定使用
-curl -o - http://localhost
curl -o /dev/null -s -w "%{http_code}\n" http://localhost
request_cmd="$(curl -i -o - --silent -X GET --header 'Accept: application/json' --header 'Authorization: _your_auth_code==' 'https://example.com')"
http_status=$(echo "$request_cmd" | grep HTTP | awk '{print $2}')
echo $http_status
output_response=$(echo "$request_cmd" | grep body)
echo $output_response
我使用此命令打印状态代码，无需任何其他输出。此外，它将仅执行HEAD请求并遵循重定向（分别是
-I-Lsh -c '[ $(curl -o -I -L -s -w "%{http_code}" http://localhost) -eq 200 ]'
详细模式将告诉您一切
curl -v http://localhost
curl http://localhost -w ", %{http_code}"
您可以随意更改分隔符。您需要的是
-icurl -i http://localhost
curl -v http://localhost
对于编程使用，我使用以下内容：
curlwithcode() {
code=0
# Run curl in a separate command, capturing output of -w "%{http_code}" into statuscode
# and sending the content to a file with -o >(cat >/tmp/curl_body)
statuscode=$(curl -w "%{http_code}" \
-o >(cat >/tmp/curl_body) \
"$@"
) || code="$?"
body="$(cat /tmp/curl_body)"
echo "statuscode : $statuscode"
echo "exitcode : $code"
echo "body : $body"
}
curlwithcode https://api.github.com/users/tj
statuscode : 200
exitcode : 0
body : {
"login": "tj",
"id": 25254,
...
}
我发现这个问题是因为我想独立访问响应和内容，以便为用户添加一些错误处理 Curl允许您自定义输出您可以将HTTP状态代码打印出来，并将内容写入另一个文件
curl -s -o response.txt -w "%{http_code}" http://example.com
http_response=$(curl -s -o response.txt -w "%{http_code}" http://example.com)
if [ $http_response != "200" ]; then
# handle error
else
echo "Server returned:"
cat response.txt
fi
%{http_code}--write out这是一种检索正文“和”状态代码并将其格式化为正确的json或任何适合您的格式的方法。有些人可能会争辩说，这是因为写入格式选项的使用不正确，但当我需要脚本中的主体代码和状态代码来检查状态代码并从服务器转发响应时，这对我来说是有效的
curl-xget-w“{stderr}{\“status\”：\“{http\u code}\”，“body\”：\“{stdout}\””）-s-o-https://github.com” 2>&1
{
"status" : <status code>,
"body" : <body of response>
}
{
“地位”：，
“正文”：
}
使用-w write format选项，由于首先打印stderr，因此可以使用var http_代码格式化输出，并将响应主体放在一个值（body）中，然后使用var stdout进行封装。然后将您的stderr输出重定向到stdout，您将能够将http_代码和响应正文合并成一个整洁的输出
while : ; do curl -sL -w "%{http_code} %{url_effective}\\n" http://host -o /dev/null; done
curl-D-$curl-D-http://localhost:1234/foo
HTTP/1.1200ok
连接：保持活力
传输编码：分块
内容类型：application/json
日期：2020年7月29日星期三20:59:21 GMT
{“数据”：[“out.csv”]}
-D---dump header-D/dev/stderr-sS$curl-sSD/dev/stderrhttp://localhost:1231/foo |jq。
HTTP/1.1200ok
连接：保持活力
传输编码：分块
内容类型：application/json
日期：2020年7月29日星期三格林尼治标准时间21:08:22
{
“数据”：[
“out.csv”
]
}
$curl-sSD/dev/stderrhttp://localhost:1231 >/dev/null
HTTP/1.1200ok
连接：保持活力
传输编码：分块
内容类型：application/json
日期：2020年7月29日星期三格林威治标准时间21:20:02
-IHEADGET-HEADHEAD附加一行“HTTP\u code:200”最后，再grep关键字“http_code:”并提取响应代码
result=$(curl -w "\nhttp_code:%{http_code}" http://localhost)
echo "result: ${result}" #the curl result with "http_code:" at the end
http_code=$(echo "${result}" | grep 'http_code:' | sed 's/http_code://g')
echo "HTTP_CODE: ${http_code}" #the http response code
result=$(curl -w "\nhttp_code:%{http_code}" http://localhost)
echo "result: ${result}" #the curl result with "http_code:" at the end
http_code=$(echo "${result}" | grep 'http_code:' | sed 's/http_code://g')
echo "HTTP_CODE: ${http_code}" #the http response code
{
IFS= read -rd '' out
IFS= read -rd '' http_code
IFS= read -rd '' status
} < <({ out=$(curl -sSL -o /dev/stderr -w "%{http_code}" 'https://httpbin.org/json'); } 2>&1; printf '\0%s' "$out" "$?")
echo out $out
echo http_code $http_code
echo status $status
out { "slideshow": { "author": "Yours Truly", "date": "date of publication", "slides": [ { "title": "Wake up to WonderWidgets!", "type": "all" }, { "items": [ "Why <em>WonderWidgets</em> are great", "Who <em>buys</em> WonderWidgets" ], "title": "Overview", "type": "all" } ], "title": "Sample Slide Show" } }
http_code 200
status 0
#!/bin/bash
output=$(curl -w "\n%{http_code}" "$@")
res=$?
if [[ "$res" != "0" ]]; then
echo -e "$output"
exit $res
fi
if [[ $output =~ [^0-9]([0-9]+)$ ]]; then
httpCode=${BASH_REMATCH[1]}
body=${output:0:-${#httpCode}}
echo -e "$body"
if (($httpCode < 200 || $httpCode >= 300)); then
# Remove this is you want to have pure output even in
# case of failure:
echo
echo "Failure HTTP response code: ${httpCode}"
exit 1
fi
else
echo -e "$output"
echo
echo "Cannot get the HTTP return code"
exit 1
fi
./curl-wrapper.sh www.google.com
# ...the same output as pure curl would return...
echo $?
# 0
./curl-wrapper.sh www.google.com/no-such-page
# ...the same output as pure curl would return - plus the line
# below with the failed HTTP code, this line can be removed if needed:
#
# Failure HTTP response code: 404
echo $?
# 1
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:38 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:41 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:43 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:45 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:47 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:49 GMT
$ curl -kv https://www.example.org
$ curl -kv https://www.example.org 2>&1 | grep -i 'HTTP/1.1 ' | awk '{print $3}'| sed -e 's/^[ \t]*//'
curl -s -i -k --location --request POST ''${HOST}':${PORT}/api/14/project/'${PROJECT_NAME}'/jobs/import' \
--header 'Content-Type: application/yaml' \
--header 'X-Rundeck-Auth-Token: '${JOB_IMPORT_TOKEN}'' \
--data "$(cat $yaml_file)" &>/dev/stdout | tee -a $response_file
return_code=$(cat $response_file | head -3 | tail -1 | awk {'print $2'})
if [ "$return_code" != "200" ]; then
echo -e "\Job import api call failed with rc: $return_code, please rerun or change pipeline script."
exit $return_code
else
echo "Job import api call completed successfully with rc: $return_code"
fi