Question

我需要JSON以下面的格式显示。对象是“用户”

  {
    "users": [{
        "ver": "1.5",
        "name": "Cupcake",
        "api": "API level 3"
    }, {
        "ver": "7.0",
        "name": "Nougat",
        "api": "API level 24"
    }]
}

当我使用这个PHP时

<?php
$link = mysqli_connect("localhost", "***", "***", "***");
if (mysqli_connect_error()) {
        die("Could not connect to database");
}

$query = "SELECT * FROM users";
if ($result=mysqli_query($link, $query)) {

 {
   while($row =mysqli_fetch_assoc($result))

  $emparray[] = $row;

    echo json_encode($emparray);
 }

} else {
echo "It failed";
}
?>

我回来了这个JSON

   [{
    "ver": "1.5",
        "name": "Cupcake",
        "api": "API level 3"
}, {
    "ver": "7.0",
        "name": "Nougat",
        "api": "API level 24"
}
}]

当我使用这个PHP时

  <?php

$link = mysqli_connect("localhost", "***", "***", "***");
if (mysqli_connect_error()) {
    die("Could not connect to database");
}

$query = "SELECT * FROM users";
if ($result=mysqli_query($link, $query)) {

 {
   while($row =mysqli_fetch_assoc($result))
  $emparray[] = $row;

    echo json_encode(array('users' => $emparray),JSON_FORCE_OBJECT);
  }

} else {
echo "It failed";
}
?>

我明白了

 {
    "users": {
        "0": {
            "ver": "1.5",
        "name": "Cupcake",
        "api": "API level 3"
        },
        "1": {
            "ver": "7.0",
        "name": "Nougat",
        "api": "API level 24"
        }
    }
}

后者几乎是我需要的，但显示{“0”：而不是所需的[和]

任何人都可以帮忙纠正这个问题吗？

Answer 1

尝试

if ($result = mysqli_query($link, $query))
{
  while ($row = mysqli_fetch_assoc($result))
  {
    $emparray[] = $row;
  }

  echo json_encode(array('users' => $emparray));
}
else
{
  echo "It failed";
}

删除JSON_FORCE_OBJECT常量。

将mySQL查询显示为JSON对象

1 个答案: