[英]Check wget response for both status code and content
我正在使用 bash 為 docker 容器編寫運行狀況檢查腳本。 它應該檢查 URL 返回狀態 200 及其內容。
有沒有辦法使用 wget 來檢查 URL 返回 200 並且 URL 內容包含某個單詞(在我的情況下該單詞是“DB_OK”),同時最好只調用一次?
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 [[ $StatusString != *"DB_OK"* ]];
then DBCheck=1
echo "db check failed"
fi
}
#!/bin/bash
# -q removed, 2>&1 added to get header and content
StatusString=$(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
輸出:
availability check okay
content check okay
everything okay
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