Question

我尝试从我的服务器获取用希腊语编写的数据。它们存储得很好，但是当我读它们时，我遇到了一个问题。

[{"id":"22","game":"??? 3 - 0 ?????","date":"27th August, 2016"}]

我正在添加

JSON_UNESCAPED_UNICODE

在我的json_encode()函数中但知道我明白了。

 Notice: Use of undefined constant JSON_UNESCAPED_UNICODE - assumed 'JSON_UNESCAPED_UNICODE' in /var/www/vhosts/theo-android.co.uk/httpdocs/ael/android/last_game_json.php on line 25

 Warning: json_encode() expects parameter 2 to be long, string given in /var/www/vhosts/theo-android.co.uk/httpdocs/ael/android/last_game_json.php on line 25

这是我的PHP代码，它读取JSON。

<?php 
include("../init.php");

$string="";
$newString="";
$get_posts = "select * from last_game";

error_reporting(E_ALL); 
ini_set("display_errors", 1);

$run_posts = mysqli_query($con, $get_posts); 

$posts_array = array();

while ($posts_row = mysqli_fetch_array($run_posts)) {
    $row_array['id'] = $posts_row['id'];
    $row_array['game'] = $posts_row['game'];
    $row_array['date'] = $posts_row['date'];

    array_push($posts_array,$row_array);            
 }

 $string = json_encode($posts_array, JSON_UNESCAPED_UNICODE);      
 echo $string;

任何想法如何解决？

感谢。

西奥。

修改

我这样做但是我收到了错误。

<?php 
include("init.php");

$get_posts = "select * from last_game";
error_reporting(E_ALL); 
ini_set("display_errors", 1);
$run_posts = mysqli_query($con,$get_posts); 

 $posts_array = array();

 while ($posts_row = mysqli_fetch_array($run_posts)){


               $row_array['id'] = $posts_row['id'];
               $row_array['game'] =$posts_row['game'];
               $row_array['date'] = $posts_row['date'];


               array_push($posts_array,$row_array);

 }
    $str = '\u0391\u0395\u039b 3 - 0 \u039e\u03b1\u03bd\u03b8\u03b7';
    $str = preg_replace_callback('/\\\\u([0-9a-fA-F]{4})/', function ($match) {
  (return mb_convert_encoding(pack('H*', $match[1]), 'UTF-8', 'UCS-2BE');
   ( }, $str);
  //$string = json_encode(utf8_encode($posts_array));
  //$response = utf8_encode($string,true);
  //echo $response;
   print($str);

?>

在这一行：

return mb_convert_encoding(pack('H*', $match[1]), 'UTF-8', 'UCS-2BE');

Answer 1

在访问数据库之前，最好在建立连接后立即执行以下操作：

mysqli_query($con, "SET CHARACTER SET utf8");
mysqli_query($con, "SET NAMES 'utf8'");
mysqli_query($con, "SET SESSION collation_connection = 'utf8_general_ci'");

还要确保您的表是使用UTF-8编码为文本字段创建的。要检查这一点，请使用：

SHOW CREATE TABLE last_game;

如果您看到latin1，我们想要将其更改为utf8，utf8_general_ci或utf8_unicode_ci。

在输入和存储数据时，还要确保数据为UTF-8。

编辑：您似乎正在从JSON解析中获取转义的unicode。可能是因为你有一个旧版本的PHP。你可以这样做来转换它：

php > $str = '\u0391\u0395\u039b 3 - 0 \u039e\u03b1\u03bd\u03b8\u03b7';
php > $str = preg_replace_callback('/\\\\u([0-9a-fA-F]{4})/', function ($match) {
php (     return mb_convert_encoding(pack('H*', $match[1]), 'UTF-8', 'UCS-2BE');
php ( }, $str);
php >
php > print($str);
ΑΕΛ 3 - 0 Ξανθη
php >

有关如何在较旧的PHP版本中模拟JSON_UNESCAPED_UNICODE的其他示例，请参阅here。

修改：将修改后的代码更改为：

<?php 
include("../init.php");

// Probably just put this in init.php.
mysqli_query($con, "SET CHARACTER SET utf8");
mysqli_query($con, "SET NAMES 'utf8'");
mysqli_query($con, "SET SESSION collation_connection = 'utf8_general_ci'");

$string="";
$newString="";
$get_posts = "select * from last_game";

error_reporting(E_ALL); 
ini_set("display_errors", 1);

$run_posts = mysqli_query($con, $get_posts); 

$posts_array = array();

while ($posts_row = mysqli_fetch_array($run_posts)) {
    $row_array['id'] = $posts_row['id'];
    $row_array['game'] = $posts_row['game'];
    $row_array['date'] = $posts_row['date'];

    array_push($posts_array, $row_array);            
}

$string = json_encode($posts_array);
$string = preg_replace_callback('/\\\\u([0-9a-fA-F]{4})/', function ($match) {
    return mb_convert_encoding(pack('H*', $match[1]), 'UTF-8', 'UCS-2BE');
}, $string);

echo $string;

希腊字母和JSON响应

1 个答案: