我想让cURL / Elasticsearch理解以普通字符串形式传递的HTTP查询参数,同时通过命令对其进行url编码。
如果我通过cURL
运行此HTTP GET,以将查询提交给Elasticsearch:
curl \
-H 'Content-Type: application/json' \
-XGET '127.0.0.1:9200/movies/movie/_search?q=%2Byear%3A%3E1980+%2Btitle%3Astar%20wars&pretty'
然后我就可以检索所需的文档。
但是,如果我运行此cURL
查询:
curl \
-H 'Content-Type: application/json' \
--data-urlencode "pretty" \
--data-urlencode "q=+year:>1980 +title:star wars&pretty" \
-XGET '127.0.0.1:9200/movies/movie/_search'
然后我得到这个错误:
{
"error": {
"root_cause": [{
"type": "json_parse_exception",
"reason": "Unrecognized token 'pretty': was expecting ('true', 'false' or 'null')\n at [Source: org.elasticsearch.transport.netty4.ByteBufStreamInput@7856627; line: 1, column: 8]"
}],
"type": "json_parse_exception",
"reason": "Unrecognized token 'pretty': was expecting ('true', 'false' or 'null')\n at [Source: org.elasticsearch.transport.netty4.ByteBufStreamInput@7856627; line: 1, column: 8]"
},
"status": 500
}
我正在使用:
cURL
版本7.47.0,该版本应了解命令参数--data-urlencode
答案 0 :(得分:1)
--data-urlencode
将发送POST和URL编码正文。您必须使用-G
or --get
来发送GET请求并在URL中附加用--data-urlencode
指定的数据:
curl -G -v \
-H 'Content-Type: application/json' \
--data-urlencode "pretty=true" \
--data-urlencode "q=+year:>1980 +title:star wars" \
'127.0.0.1:9200/movies/movie/_search'