Php 一个对象json中的两个查询mysql

我想将两个表转换为json，如下所示：

[
   {
      "date":"2013-07-20",
      "id":"123456",
      "year":"2013",
      "people":[
         {
            "name":"First",
            "age":"60",
            "city":"1"
         },
         {
            "name":"second",
            "age":"40",
            "city":"2"
         },
         {
            "name":"third",
            "age":"36",
            "city":"1"
         }
      ]
   }
]

但我的代码的结果是：

[
   {
      "date":"2013-07-20",
      "id":"123456",
      "year":"2013",}
      ,{
      "people":[
         {
            "name":"First",
            "age":"60",
            "city":"1"
         },
         {
            "name":"second",
            "age":"40",
            "city":"2"
         },
         {
            "name":"third",
            "age":"36",
            "city":"1"
         }
      ]
   }
]

代码为数组“people”创建了一个新对象，我希望它们位于同一个对象中

$result = mysql_query("SELECT * FROM data where id='123456'");
$fetch = mysql_query("SELECT name,age,city FROM people where id='123456'"); 

$json = array();
$json2['people'] = array();

  while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
    $json[] = $row;
  }

  while ($row = mysql_fetch_assoc($fetch)){
    $row_temp["name"]=$row["name"];
    $row_temp["age"] = $row["age"];
    $row_temp["city"] = $row["city"];

   array_push($json2['people'],$row_temp);
   }

    array_push($json, $json2);

echo Json_encode($json);

如何使数组与表“data”位于同一对象中

---

非常感谢

您非常接近，但是您希望People数组是外部数组的直接值，并将其包装在一个额外的数组中

另外，请注意，您正在使用的MySQL库已被弃用。这意味着它将在未来的版本中从PHP中删除。您应该将MySQL_*系列函数的调用替换为或

---

我想你可以试试这个

$result = mysql_query("SELECT * FROM data where id='123456'");
$fetch = mysql_query("SELECT name,age,city FROM people where id='123456'"); 

// I think, you'll get a single row, so no need to loop
$json = mysql_fetch_array($result, MYSQL_ASSOC);

$json2 = array();
while ($row = mysql_fetch_assoc($fetch)){
    $json2[] = array( 
        'name' => $row["name"],
        'age' => $row["age"],
        'city' => $row["city"]
    );
}
$json['people'] = $json2;
echo json_encode($json);

print\r（$json）

的结果应该是这样的

Array
(
    [date] => 2013-07-20
    [year] => 2013
    [id] => 123456
    [people] => Array
        (
            [0] => Array
                (
                    [name] => First
                    [age] => 60
                    [city] => 1
                )

            [1] => Array
                (
                    [name] => second
                    [age] => 40
                    [city] => 2
                )

        )

)

[
    {
        "date" : "2013-07-20",
        "year":"2013",
        "id":"123456",
        "people":
        [
            {
                "name" : "First",
                "age" : "60",
                "city" : "1"
            },
            {
                "name" : "second",
                "age" : "40",
                "city" : "2"
            }
        ]
    }
]

echo json\u encode（$json）

的结果应该是

{
    "date" : "2013-07-20",
    "year":"2013",
    "id":"123456",
    "people":
    [
        {
            "name" : "First",
            "age" : "60",
            "city" : "1"
        },
        {
            "name" : "second",
            "age" : "40",
            "city" : "2"
        }
    ]
}

如果您执行

echo-json\u-encode（array（$json））

，那么您将把整个

json

包装在一个数组中，类似这样

Array
(
    [date] => 2013-07-20
    [year] => 2013
    [id] => 123456
    [people] => Array
        (
            [0] => Array
                (
                    [name] => First
                    [age] => 60
                    [city] => 1
                )

            [1] => Array
                (
                    [name] => second
                    [age] => 40
                    [city] => 2
                )

        )

)

[
    {
        "date" : "2013-07-20",
        "year":"2013",
        "id":"123456",
        "people":
        [
            {
                "name" : "First",
                "age" : "60",
                "city" : "1"
            },
            {
                "name" : "second",
                "age" : "40",
                "city" : "2"
            }
        ]
    }
]

---

您可以等待使用键

people

，直到您加入两个数组的最后一刻，使其工作。在此之前，只需将数据加载到

$json

和

$json2

中即可

$json = array('date' => '2013', 'id' => '123456', 'year' => '2013');

$result = mysql_query("SELECT * FROM data where id='123456'");
$fetch = mysql_query("SELECT name,age,city FROM people where id='123456'"); 

$json = array();
$json2 = array();

while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
    $json[] = $row;
}

while ($row = mysql_fetch_assoc($fetch)){
    $row_temp["name"]=$row["name"];
    $row_temp["age"] = $row["age"];
    $row_temp["city"] = $row["city"];
    array_push($json2, $row_temp);
}

$json['people'] = $json2;

echo Json_encode($json);

---

谢谢你的回答。我尝试了三个示例，它们都响应并返回了相同的代码。{“0”：{“日期”：“2013-07-20”，“id”：“123456”，“年”：“2013”}，“人”：[…]}我希望它是这样的[{“日期”：“2013-07-20”，“id”：“123456”，“年”：“2013”，“人”：[…]}]你知道如何纠正它吗？非常感谢。也许，您应该发布查询结果。您是指数据库中的查询结果或json编码的？两个查询的结果，而不是json。

SELECT*FROM data where id='123456'date year id 2013-07-20 2013 123456 SELECT name，age，城市来源于人，其中id='123456'姓名年龄城市id第一个60 1 123456第二个40 2 123456城市36 1 123456