SQL查询列中的更多信息

Question

我需要一些关于 SQL 查询的信息。这是我的 SQL 表；

CustomData 是我表中的列名。
CustomData 有更多列。
表示列中的列。
如果我使用这个查询；

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

原始答案（适用于 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() 解析此数据（SQL Server 2016 中引入了JSON 支持）。当您将OPENJSON() 与默认架构一起使用时，结果是一个包含key、value 和type 列的表，因此您需要包含额外的CROSS APPLY 运算符来解析JSON 数组中的每个项目。

表：

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"
    }
]')

PHP：

<?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);
?>

更新：

SQL Server 2014 不支持JSON，因此一种可能的方法是使用json_decode() 解析PHP 部分中的JSON 数据：

<?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 CustomData FROM Entities";
$result3 = sqlsrv_query($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)){
    $decoded = json_decode($row3["CustomData"], true);
    foreach($decoded as $item) {
        if ($item["Name"] == "Telefon") {
            $json[] = $item["Value"];
        }   
    }   
}
echo json_encode($json);

sqlsrv_free_stmt($result3);
sqlsrv_close($consamba);
?>

Answer 2

类似这样的：

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 3

Select CustomData

From Entities

Where CustomData LIKE '%Telefon%'

会让你走得很远