Question

我正在创建一个从SQL数据库中提取数据的应用程序。我的PHP代码echos json数据，但它没有很好地形成。我当前的输出没有结果数组的名称，我希望它 - 我已经提供了一个链接，指向我希望输出看起来如何。任何帮助将不胜感激。

<?php
// database settings 
$db_username = "****";
$db_password = "****";
$db_name = "****";
$db_host = "****";


   //open connection to mysql db
    $connection = mysqli_connect("$db_host","$db_username","$db_password","$db_name") or die("Error " . mysqli_error($connection));

    //fetch table rows from mysql db
    $sql = "select * from markers";
    $result = mysqli_query($connection, $sql) or die("Error in Selecting " . mysqli_error($connection));

    //create an array
    $emparray = array();
    while($row =mysqli_fetch_assoc($result))
    {
        $emparray[] = $row;
    }


$data = array($emparray);

 echo json_encode($data);
    //close the db connection
    mysqli_close($connection);
?>

Final JSON Current Output

Answer 1

如果替换

，该怎么办？

$data = array($emparray);

与

$data = array('markers' => $emparray);

应输出类似

的内容

JSON : {"markers":[{"id":1,"lorem":"ipsum"},{"id":2,"lorem":"dolor"}]}

php > var_dump($test);
array(1) {
  ["markers"]=>
  array(2) {
    [0]=>
    array(2) {
      ["id"]=>
      int(1)
      ["lorem"]=>
      string(5) "ipsum"
    }
    [1]=>
    array(2) {
      ["id"]=>
      int(2)
      ["lorem"]=>
      string(5) "dolor"
    }
  }
}

尝试使用PHP

1 个答案: