如何将json解析为php

Question

i have a json data retrieved from data base ie

$result=array();
$query="SELECT * FROM fish";
$result1 = mysql_query($query, $conn);
while ($table = mysql_fetch_array($result1, MYSQL_ASSOC)){
   $result[]=$table;
}
   echo json_encode($result);

and this give me the result

 [{"fish_id":"1","name":"first fish update","info":"this is my first fish update","image":"http:\\/\\/www.localhost\\/cafe\\/pics\\/logout (1).gif"}]  

but from another page when i call this json data ie

 $input = file_get_contents("http://localhost/fish/fish-json.php"); \n$json=json_decode($input); \necho $json->fish_id;  

it give me the error

 Notice: Trying to get property of non-object in /var/www/fish/json-to-php.php on line 13 Call Stack: 0.0005 318764 1. {main}() /var/www/fish/json-to-php.php:0  

Answer 1

Is an array of object, so

echo $json[0]->fish_id;

To loop

if (!is_array($json)) die('...');
foreach ($json as $key=>$fish)
{
  echo $fish->fish_id;
}

Answer 2

Try

$input = file_get_contents("http://localhost/fish/fish-json.php");
$json=json_decode($input);
echo $json['fish_id'];

Answer 3

try echo $json['fish_id']; i might suspect it convert it to an array

Answer 4

try this first:

echo $json;

to check if you got valid json. Eventually, the php in fish-json.php was not executed.

Answer 5

By default json_decode returns an stdClass object. You have to give a second parameter for this function TRUE.

$json=json_decode($input, TRUE);
echo $json[0]['fish_id'];