文件 'sample.json'(JSON):桑達:如何使用文件的contemt替換模式
{
"jsonrpc": "2.0",
"method": "configuration.import",
"params": {
"format": "xml",
"source": "REPLACE_ME_WITH_XMLSOURCE"
},
"auth": "91ea4764dcab42e8317b399c42985792",
"id": 1
}
文件 'source.xml'(XML,1個和雙引號長線):
<?xml version="1.0" encoding="UTF-8"?><zabbix_export>...</zabbix_export>
所需的結果(JSON和XML作爲PARAM的值):
...
"source": "<?xml version=\"1.0\" encoding=\"UTF-8\"?><zabbix_export>...</zabbix_export>"
...
我試圖把 'source.xml' 的內容,變量和SED使用它,但withou牛逼的成功:
# x=$(cat source.xml)
# sed "s/REPLACE_ME_WITH_XMLSOURCE/$x/" sample.json
sed: -e expression #1, char 80: unknown option to `s'
# sed "s/XMLSOURCE/"$x"/" sample.json
sed: -e expression #1, char 17: unterminated `s' command
# sed "s/XMLSOURCE/"$x"/" sample.json
...
我試着使用變化的sed「/ REPLACE_ME_WITH_XMLSOURCE/R source.xml」,但我的sed福是不夠的,解決這個問題..
我想谷歌和搜索結果,但..你看這個問題..
感謝您的任何意見
這個問題涉及到[API的zabbix configuration.import(https://www.zabbix.com/documentation/2.2/manual/api/reference/configuration) – user2583148