Question

我需要一些有关SQL查询的信息。这是我的SQL表；

CustomData是表中的列名。
CustomData有更多列。
它表示列中的列。
如果我使用此查询；

SELECT CustomData FROM Entities

结果；

[
    {
        "Name": "Telefon",
        "Value": "5417416039"
    },
    {
        "Name": "Adres",
        "Value": "AYDOĞAN MAH. 7.SOK. HACIHÜSEYİN APT. KAT:3 NO:6"
    }
]

我想在CustomData WHERE Name = Telefon中获取价值

我该怎么做？

已编辑

<?php

  header('Access-Control-Allow-Origin: *');
  header('Access-Control-Allow-Headers: X-Requested-With');
  header('Access-Control-Allow-Headers: Content-Type');
  header('Access-Control-Allow-Methods:POST,GET,OPTIONS,DELETE,PUT');

  $serverName = "x.x.x.x"; //serverName\instanceName
  $connectionInfo = array( "Database"=>"hidden","UID"=>"sa","PWD"=>"hidden", "CharacterSet" => "UTF-8");
  $consamba = sqlsrv_connect( $serverName, $connectionInfo);

  $sql3 = "SELECT CustomData FROM Entities WHERE EntityTypeId='1'";
  $result3 = sqlsrv_query($consamba,$sql3);
  $json = array();
  while($row3 = sqlsrv_fetch_array($result3,SQLSRV_FETCH_ASSOC)){
      $json[] = $row3;
  }
  echo json_encode($json);


 ?>

Answer 1

类似这样的东西：

declare @entities table (id int, customdata nvarchar(max))
insert  into @entities(id,customdata)
values (1,'
[
    {
        "Name": "Telefon",
        "Value": "5417416039"
    },
    {
        "Name": "Adres",
        "Value": "AYDOĞAN MAH. 7.SOK. HACIHÜSEYİN APT. KAT:3 NO:6"
    }
]')

select e.id, c.Value Telefon
from @entities e
cross apply openjson(e.customdata) 
            with( 
                   Name nvarchar(200) '$.Name',
                   value nvarchar(200) '$.Value'
                 ) c
where c.name = N'Telefon'

Answer 2

Select CustomData

From Entities

Where CustomData LIKE '%Telefon%'

会让你很遥远

Answer 3

原始答案（对于SQL Server 2016 +）：

CustomData列中的数据似乎是有效的JSON数组，因此一种可能的方法是使用OPENJSON()和JSON_VALUE()（{{1} }支持是在SQL Server 2016中引入的。当您将JSON与默认架构一起使用时，结果是一个包含列OPENJSON()，key和value的表，因此您需要包括其他type运算符解析CROSS APPLY数组中的每个项目。

表格：

JSON

PHP：

CREATE TABLE Entities (
   EntityTypeId varchar(1),
   CustomData nvarchar(max)
)
INSERT INTO Entities
   (EntityTypeId, CustomData)
VALUES ('1', N'[
    {
        "Name": "Telefon",
        "Value": "5417416039"
    },
    {
        "Name": "Adres",
        "Value": "AYDOĞAN MAH. 7.SOK. HACIHÜSEYİN APT. KAT:3 NO:6"
    }
]')

更新：

SQL Server 2014不支持<?php header('Access-Control-Allow-Origin: *'); header('Access-Control-Allow-Headers: X-Requested-With'); header('Access-Control-Allow-Headers: Content-Type'); header('Access-Control-Allow-Methods:POST,GET,OPTIONS,DELETE,PUT'); $serverName = "x.x.x.x"; $connectionInfo = array("Database"=>"hidden", "UID"=>"sa", "PWD"=>"hidden", "CharacterSet" => "UTF-8"); $consamba = sqlsrv_connect($serverName, $connectionInfo); if ($consamba === false) { echo "Error (sqlsrv_connect): ".print_r(sqlsrv_errors(), true); exit; } $sql3 = " SELECT e.CustomData, JSON_VALUE(j.[value], '$.Value') AS Telefon FROM Entities e CROSS APPLY OPENJSON(e.CustomData) j WHERE EntityTypeId = '1' AND JSON_VALUE(j.[value], '$.Name') = N'Telefon' "; $result3 = sqlsrv_query($consamba, $sql3); if ($result3 === false) { echo "Error (sqlsrv_query): ".print_r(sqlsrv_errors(), true); exit; } $json = array(); while ($row3 = sqlsrv_fetch_array($result3, SQLSRV_FETCH_ASSOC)){ $json[] = $row3["Telefon"]; } echo json_encode($json); sqlsrv_free_stmt($result3); sqlsrv_close($consamba); ?>，因此一种可能的方法是使用JSON来解析JSON部分中的PHP数据：

json_decode()

SQL查询列中的更多信息

已编辑

3 个答案: