Return more than one elements in the final list

[ghost]

I am using the books (json) that you are using in the tutorial and I would like to retrieve the result of the following query:

$..[@.author is 'Evelyn Waugh' and @.title is 'Sword of Honour'] and $..[@.author is 'Nigel Rees' and @.title is 'Sayings of the Century']

this returns only 1 object: [{ "category": "reference", "price": 8.95, "title": "Sayings of the Century", "author": "Nigel Rees" }]

I would expect to get two entities back, one for Sword of Honour and one for Sayings of the Century. Is this feasible?

[adriank]

Hi,

I wonder if it's not a bug (it should return true probably), but generally, if you want to merge results from two expressions, use + instead of and between $..[...'Evelyn and $..[...'Nigel.

Greetings, Adrian Kalbarczyk

http://kalbarczyk.co

On Tue, Jan 9, 2018 at 9:19 AM, std::lef notifications@github.com wrote:

I am using the books (json) that you are using in the tutorial and I would like to retrieve the result of the following query:

$..[@.author is 'Evelyn Waugh' and @.title is 'Sword of Honour'] and $..[@.author is 'Nigel Rees' and @.title is 'Sayings of the Century']

this returns only 1 object: [{ "category": "reference", "price": 8.95, "title": "Sayings of the Century", "author": "Nigel Rees" }]

I would expect to get two entities back, one for Sword of Honour and one for Sayings of the Century. Is this feasible?

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[ghost]

Hello Adrian,

Thanks for your prompt reply. I believe there is a bug in the code. Here is an example

[
  {
    "a": 10,
    "b": 11
  },
  {
    "a": 20,
    "b": 21
  }
]

 $..*[@.a is 10 and @.b is 11]
[{
  "a": 10,
  "b": 11
}]
 $..*[@.a is 10 and @.b is 11] and  $..*[@.a is 20 and @.b is 21]
[{
  "a": 20,
  "b": 21
}]
$..*[@.a is 20 and @.b is 11] and  $..*[@.a is 20 and @.b is 21]
[{
  "a": 20,
  "b": 21
}]

**$..*[@.a is 10 and @.b is 11]** returns the first element, however 
**$..*[@.a is 20 and @.b is 11] and  $..*[@.a is 20 and @.b is 21]** returns only the second. As you said above I would expect to return true since the first expression had list.size>0 and the second subexpression also. 

If you take a look closely at the third expression **$..*[@.a is 20 and @.b is 11] and  $..*[@.a is 20 and @.b is 21]** returns the second object, however the first element should not even match (a is 10 in the JSON and not 20). 

Any chance to "offer' this functionality?

Thanks,
Lef

[adriank]

Hi,

I'm not coding in Python since 2013, so there is little chance that I would be able to fix it in foreseeable future. You can try with len():

len($..[@.a is 20 and @.b is 11]) and len($..[@.a is 20 and @.b is 21])

Greetings, Adrian Kalbarczyk

http://kalbarczyk.co

On Tue, Jan 9, 2018 at 11:21 AM, std::lef notifications@github.com wrote:

Hello Adrian,

Thanks for your prompt reply. I believe there is a bug in the code. Here is an example

[ { "a": 10, "b": 11 }, { "a": 20, "b": 21 } ]

$..[@.a is 10 and @.b is 11] [{ "a": 10, "b": 11 }] $..[@.a is 10 and @.b is 11] and $..[@.a is 20 and @.b is 21] [{ "a": 20, "b": 21 }] $..[@.a is 20 and @.b is 11] and $..*[@.a is 20 and @.b is 21] [{ "a": 20, "b": 21 }]

*$..[@.a is 10 and @.b is 11] returns the first element, however $..[@.a is 20 and @.b is 11] and $..[@.a is 20 and @.b is 21]** returns only the second. As you said above I would expect to return true since the first expression had list.size>0 and the second subexpression also.

If you take a look closely at the third expression $..[@.a is 20 and @.b is 11] and $..[@.a is 20 and @.b is 21] returns the second object, however the first element should not even match (a is 10 in the JSON and not 20).

Any chance to "offer' this functionality?

Thanks, std::lef

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[ghost]

Thanks it worked.

[adriank]

OK, I have a project where I'm using the ObjectPath myself and I know a little bit more now.

The is operator is taken directly from Python and behaves in the exact same way. Refer to Python documentation to see how it works.

>>> 3 and 4 => 4

So that's why your query returns the last object.

Should it work like that in ObjectPath? I need to think about that.