Question

我有以下内容：

Mat flooded=new Mat();
Point flood=new Point(1,1);
// floodedmat = Mat.zeros(myMat2.size(), CvType.CV_8UC1);
Imgproc.floodFill(myMat2, flooded, flood, new Scalar(255, 255, 255));
Utils.matToBitmap(flooded, copy);

我正在过滤对象'{“name”：“Bob”，“Occup”：“janitor”}'

我如何归还鲍勃的职业（“看门人”）？

SELECT * 
FROM (
   SELECT '{"people": [{"name": "Bob", "occupation": "janitor"}, {"name": "Susan", "occupation": "CEO"}]}'::jsonb as data
) as b 
WHERE data->'people' @> '[{"name":"Bob"}]'::jsonb;

返回

SELECT data->'people'->>'occupation'  
FROM (
   SELECT '{"people": [{"name": "Bob", "occupation": "janitor"}, {"name": "Susan", "occupation": "CEO"}]}'::jsonb as data
) as b 
WHERE data->'people' @> '[{"name":"Bob"}]'::jsonb;

寻找：

?column?
--------
NULL

Answer 1

如果你不关心jsonb所在的行上的任何其他内容，你可以从jsonb中取出所有元素，然后将它们作为单独的元素来选择

SELECT data->>'occupation' as occupation
FROM (
  SELECT jsonb_array_elements(
    '{"people": 
       [
         {"name": "Bob", "occupation": "janitor"}, 
         {"name": "Susan", "occupation": "CEO"}
       ]
     }'::jsonb->'people') as data) as b
WHERE data @> '{"name":"Bob"}';

结果

职业
  -----------
   看门
  （1排）

Answer 2

您的“people”元素是一个数组。您可以使用jsonb_array_elements函数获取数组的元素。之后，您只需过滤person->>'name'：

SELECT  person->>'occupation' as occupation
FROM    (
        SELECT  person.value as person
        FROM    (
                SELECT     
                  '{"people": 
                     [
                       {"name": "Bob", "occupation": "janitor"}, 
                       {"name": "Susan", "occupation": "CEO"}
                     ]
                   }'::jsonb as data
                ) a
        CROSS JOIN
                jsonb_array_elements(data->'people') as person
        ) b
WHERE   person->>'name' = 'Bob';

请注意，->>会返回文字，而->会返回jsonb。

我如何选择“职业”？

2 个答案: