检查状态代码和内容的 wget 响应

Question

我正在使用 bash 为 docker 容器编写健康检查脚本。它应该检查 URL 返回状态 200 及其内容。

有没有一种方法可以使用 wget 来检查 URL returns 200 并且 URL 的内容包含某个词（在我的例子中这个词是 "DB_OK") 而最好只调用一次？

StatusString=$(wget -qO- localhost:1337/status)

#does not work
function available_check(){
  if [[ $StatusString != *"HTTP/1.1 200"* ]];
      then AvailableCheck=1
      echo "availability check failed"
  fi
}

#checks that statusString contains "DB_OK", works
function db_check(){
  if [[ $StatusString != *"DB_OK"* ]];
      then DBCheck=1
      echo "db check failed"
  fi
}

Answer 1

#!/bin/bash

# -q removed, 2>&1 added to get header and content
StatusString=$(wget -O- 'localhost:1337/status' 2>&1)

function available_check{
  if [[ $StatusString == *"HTTP request sent, awaiting response... 200 OK"* ]]; then
    echo "availability check okay"
  else
    echo "availability check failed"
    return 1
  fi
}

function db_check{
  if [[ $StatusString == *"DB_OK"* ]]; then
    echo "content check okay"
  else
    echo "content check failed"
    return 1
  fi
}

if available_check && db_check; then
  echo "everything okay"
else
  echo "something failed"
fi

输出：

availability check okay
content check okay
everything okay

检查状态代码和内容的 wget 响应

Check wget response for both status code and content

bash

wget