How to Json Decode in PHP using Services_JSON package of pear(PHP Catchable fatal error: Object of class stdClass could not be converted to string )

Question

我想 encode/decode 使用 JSON 的数组。因为我有 php 5.1.6 ，所以我正在使用 pear 的 (http://pear.php.net/pepr/pepr-proposal-show.php?id=198) 包。 使用它我可以编码但是，我无法解码我试图阅读文档，但不明白 anything.Here 是我的代码：

<?php  include("/home/gpreeti/php/JSON.php");
$json = new Services_JSON();
$marks = array(
            "mohammad" => array (
               "physics" => 35,
               "maths" => 30,
               "chemistry" => 39
            ),

            "qadir" => array (
               "physics" => 30,
               "maths" => 32,
               "chemistry" => 29
            ),

            "zara" => array (
               "physics" => 31,
               "maths" => 22,
               "chemistry" => 39
            )
         );
$marks=$json->encode($marks);
print"$marks\n";
$marks = $json->decode($marks);
#var_dump($marks);
print"$marks";
?>

在 运行 上，我得到了这个

{"mohammad":{"physics":35,"maths":30,"chemistry":39},"qadir":{"physics":30,"maths":32,"chemistry":29},"zara":{"physics":31,"maths":22,"chemistry":39}}
PHP Catchable fatal error:  Object of class stdClass could not be converted to string in /servers/scratch05/gpreeti/php_pgms/test_json.php on line 26

请帮助， 谢谢

Answer 1

当您有 array/object 时，您需要使用 print_r / var_dump 显示它。打印用于字符串

$marks=$json->encode($marks);
print"$marks\n";
$marks = $json->decode($marks);
#var_dump($marks);
print_r($marks); //change this line

编辑 要获取数组，您需要进行以下更改：

$json = new Services_JSON(SERVICES_JSON_LOOSE_TYPE);
$marks = $json->decode($marks);