使用php从数据库中获取多个json

Question

我有两个表 ERRORS 和 PRODUCTS，它们由两列相关联（error_code 和 issue）我想使用 php 获取响应 json。这些是我的文件：

表格：

错误：

id_error	error_code	issue	cause	solution
1	A01	issue1	cause1.1	solution1.1
2	A01	issue1	cause1.2	solution1.2
3	A01	issue1	cause1.3	solution1.3
4	A02	issue2	cause2.1	solution2.1

产品：

id_error	product	issue
A01	P1	issue1
A01	P2	issue1
A02	P1	issue2

这是我的 .php 文件

<?php
include('functions.php');
//$tipo = $_GET['txtTi'];
$array = array();   
if ($resultset = getSQLResultSet("
    SELECT errors.error_code,
           errors.issue,
           errors.cause,
           errors.solution,
           products.error_code,
           products.product,
           products.issue 
    FROM errors 
    JOIN products
        ON errors .issue = products.issue
            AND errors.error_code = products.error_code
    WHERE errors.error_code = 'A01'")
){
    $p = array();
    $c = array ();
    $s = array();
    while ($row = $resultset->fetch_array(MYSQLI_NUM)){
        $e = array();
        $e['error_code'] = $row[0];
        $e['issue'] = $row[1];
        $c['cause'] = $row[2];
        $e['cause'] = $c;
        $s['solution'] = $row[3];
        $e['solution'] = $s;
        $p['product'] = $row[5];
        $e['products'] = $p;
        array_push($array,$e);
    }
    echo json_encode($array);
}
?>

我有这个 json：

[
   {
      "error_code":"A01",
      "issue":"issue1",
      "cause":{
         "cause":"cause1.1"
      },
      "solution":{
         "solution":"solution1.1"
      },
      "products":{
         "product":"P1"
      }
   },
{
      "error_code":"A01",
      "issue":"issue1",
      "cause":{
         "cause":"cause1.2"
      },
      "solution":{
         "solution":"solution1.2"
      },
      "products":{
         "product":"P1"
      }
   },{
      "error_code":"A01",
      "issue":"issue1",
      "cause":{
         "cause":"cause1.3"
      },
      "solution":{
         "solution":"solution1.3"
      },
      "products":{
         "product":"P1"
      }
   },
(...)

]

但这是我想要的 json

[
{
 "error_code":"A01",
 "issue":"issue1",
 "cause":{
         "cause":"cause1.1",
         "cause":"cause1.2",
         "cause":"cause1.3"
          },
 "solution":{
          "solution":"solution1.1",
          "solution":"solution1.2",
          "solution":"solution1.3"
          },
 "products":{
          "product":"P1",
          "product":"P2"
          }
       }
]

Answer 1

我想你可以对你的数据库行进行分组并用逗号分隔它们，以便它们是平坦的，然后在你迭代时解包/分解它们。

SELECT errors.error_code,
       errors.issue,
       GROUP_CONCAT(DISTINCT errors.cause) AS causes,
       GROUP_CONCAT(DISTINCT errors.solution) AS solutions,
       GROUP_CONCAT(DISTINCT products.product) AS products
FROM errors 
JOIN products
    ON errors.issue = products.issue
       AND errors.error_code = products.error_code
WHERE errors.error_code = 'A01'
GROUP BY errors.error_code,
         errors.issue;

error_code	issue	causes	solutions	products
A01	issue1	cause1.1,cause1.2,cause1.3	solution1.1,solution1.2,solution1.3	P1,P2
A01	issue2	cause1.4	solution1.4	P4

---

View on DB Fiddle

$result = [];
while ($row = $resultset->fetch_array(MYSQLI_ASSOC)) {
    $result[] = [
        'error_code' => $row['error_code'],
        'issue' => $row['issue'],
        'causes' => explode(',', $row['causes']),
        'solutions' => explode(',', $row['solutions']),
        'products' => explode(',', $row['products'])
    ];
}
echo json_encode($result);

有几种方法可以做到这一点。以上只是一种方式。

Answer 2

查询下一种形式的数据：

SELECT error_code, issue, cause, NULL solution, NULL product
FROM ERRORS
UNION ALL
SELECT error_code, issue, NULL, solution, NULL
FROM ERRORS
UNION ALL
SELECT id_error, issue, NULL, NULL, product
FROM PRODUCTS
ORDER BY error_code, issue, cause IS NULL, solution IS NULL

（请参阅fiddle），然后在您的 PHP 代码中迭代此行集。

PS。由于键名重复，您无法将此数据聚合到 MySQL 上所需的 JSON 中。