I have a column of text type be contain JSON value.
{
"customer": [
{
"details": {
"customer1": {
"name": "john",
"addresses": {
"address1": {
"line1": "xyz",
"line2": "pqr"
},
"address2": {
"line1": "abc",
"line2": "efg"
}
}
}
"customer2": {
"name": "robin",
"addresses": {
"address1": null
}
}
}
}
]
}
How can I extract 'address1' JSON field of column with query?
First I am trying to fetch JSON value then I will go with parsing.
SELECT JSON customer from text_column;
With my query, I get following error.
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 version 2.1.13
You can't use SELECT JSON in Cassandra v2.1.x CQL v3.2.x
For Cassandra v2.1.x CQL v3.2.x :
The only supported operation after SELECT are :
In Cassandra v2.2.x CQL v3.3.x Introduce : SELECT JSON
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.