Php : json_encode 在字符串的开头返回 NULL 数组

Question

这是我的 Php 代码。我是 php 的新人，我想得到 JSON.

的适当回应

 $db_handle = mysql_connect($dbhost, $dbuser, $dbpass);

   if(! $db_handle  )
   {
     die('Could not connect: ' . mysql_error());
   }
   else 
    $response["message"] = "Connected Succsessfully";
     //echo json_encode($response);


$db_found = mysql_select_db($db, $db_handle);

// check for required fields
   if($db_found)
{
    // mysql inserting a new row
    mysql_query('SET CHARACTER SET utf8');
$query = "SELECT * FROM users";
$result = mysql_query($query);
//iterate over all the rows
$emparray[] = array();
while($row = mysql_fetch_assoc($result)){
    //iterate over all the fields
       $emparray[] = $row;
}

   echo json_encode($emparray);

这是我的回复：

[[],{"id":"1","username":"Liyari","password":"Karachi"},{"id":"12","username":"ssssss","password":"111111"},{"id":"121","username":"135","password":"Qasim"},{"id":"17","username":"nafees","password":"99999"},{"id":"134","username":"","password":""},{"id":"133","username":"","password":""},{"id":"135","username":"mustafa","password":"45612"},{"id":"131","username":"mustafa","password":"45612"},{"id":"132","username":"","password":""}]

问）为什么我在 Response 开始时得到空数组。

Answer 1

移除

$emparray[] = array();

或设为

$emparray = array();

那就可以正常工作了。

Answer 2

当你写类似

的东西时

$foo[] = $something;

在 php 中，您正在向数组末尾添加一个元素。

与

$emparray[] = array();

您正在添加和数组作为第一个元素，因为 $emparray 之前没有初始化，那是因为您在响应开始时得到一个空数组。

你应该改变

$emparray[] = array();

来自

$emparray = array();

Php : json_encode 在字符串的开头返回 NULL 数组

Php : json_encode is returning NULL array at the start of the string

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