如何解析cassandra中文本列的JSON值

Question

我有一列文本类型包含 JSON 值。

{
    "customer": [
        {
            "details": {
                "customer1": {
                    "name": "john",
                    "addresses": {
                        "address1": {
                            "line1": "xyz",
                            "line2": "pqr"
                        },
                        "address2": {
                            "line1": "abc",
                            "line2": "efg"
                        }
                    }
                }
                "customer2": {
                    "name": "robin",
                    "addresses": {
                        "address1": null
                    }
                }
            }
        }
    ]
}

如何使用查询提取列的 'address1' JSON 字段？

首先我尝试获取 JSON 值，然后我将继续解析。

SELECT JSON customer from text_column;

我的查询出现以下错误。

com.datastax.driver.core.exceptions.SyntaxError: line 1:12 no viable alternative at input 'customer' (SELECT [JSON] customer...)
com.datastax.driver.core.exceptions.SyntaxError: line 1:12 no viable alternative at input 'customer' (SELECT [JSON] customer...)

Cassandra 版本 2.1.13

Answer 1

您不能在 Cassandra v2.1.x CQL v3.2.x[=11 中使用 SELECT JSON =]

对于 Cassandra v2.1.x CQL v3.2.x：

SELECT 之后唯一支持的操作是：

不同
计数 (*)
计数 (1)
column_name AS new_name
写入时间 (column_name)
TTL (column_name)
dateOf()、now()、minTimeuuid()、maxTimeuuid()、unixTimestampOf()、typeAsBlob() 和 blobAsType()

在Cassandra v2.2.x CQL v3.3.x中引入：SELECTJSON

With SELECT statements, the new JSON keyword can be used to return each row as a single JSON encoded map. The remainder of the SELECT statment behavior is the same.

The result map keys are the same as the column names in a normal result set. For example, a statement like “SELECT JSON a, ttl(b) FROM ...” would result in a map with keys "a" and "ttl(b)". However, this is one notable exception: for symmetry with INSERT JSON behavior, case-sensitive column names with upper-case letters will be surrounded with double quotes. For example, “SELECT JSON myColumn FROM ...” would result in a map key "\"myColumn\"" (note the escaped quotes).

The map values will JSON-encoded representations (as described below) of the result set values.

Answer 2

如果您的 Cassandra 版本是 2.1x 及以下，您可以使用基于 Python 的方法。使用 Cassandra-Python API

编写 python 脚本

这里你必须先获取你的行，然后使用python json的loads方法，这会将你的json文本列值转换为JSON对象这将是 Python 中的字典。然后您可以使用 Python 字典并提取所需的嵌套键。请参阅下面的代码片段。

from cassandra.cluster import Cluster
from cassandra.auth import PlainTextAuthProvider
import json

if __name__ == '__main__':
   
    auth_provider = PlainTextAuthProvider(username='xxxx', password='xxxx')
    cluster = Cluster(['0.0.0.0'],
                      port=9042, auth_provider=auth_provider)
    session = cluster.connect("keyspace_name")
    print("session created successfully")
    rows = session.execute('select * from user limit 10')

    for user_row in rows:
        customer_dict = json.loads(user_row.customer)
        print(customer_dict().keys()

如何解析cassandra中文本列的JSON值

How to parse JSON value of a text column in cassandra

json

cassandra

cassandra-2.1