On 02/28/2018 08:34 PM, tj5527 wrote:
I create a table with the command `CREATE TABLE test1 (key text primary key,
obj json);` with two records inserted
# select * from test1;
key1|{"a":1,"b":2,"c":{"x":99},"status":"done"} # record 1
key2|{ "key": key2 } # record 2
The second record is not valid json because "key2" is not quoted.
Causing the error.
Dan.
Now I want to retrieve the record that is not marked with status "done" (so basically it is
expected to return the second record i.e. record 2 with key2). The command I use is `select * from test1,
json_tree(test1.obj) where json_tree.value <> "done";` But it returns
key1|{"a":1,"b":2,"c":{"x":99},"status":"done"}|a|1|integer|1|2|0|$.a|$
key1|{"a":1,"b":2,"c":{"x":99},"status":"done"}|b|2|integer|2|4|0|$.b|$
key1|{"a":1,"b":2,"c":{"x":99},"status":"done"}|c|{"x":99}|object||6|0|$.c|$
key1|{"a":1,"b":2,"c":{"x":99},"status":"done"}|x|99|integer|99|8|6|$.c.x|$.c
Error: malformed JSON
What is the correct SQL command to achieve such effect?
SQLite version I use is 3.11.0 2016-02-15 17:29:24
Thanks
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