Json解析问题，并带有错误尝试获取非对象的属性

Question

I want to parse following json for value of

JSON (saved in my.json):

{
"mackoniv":{"entry":"","lastExit":"","userId":"OPENGOVTJOBS","userNick":"mack"},
"johanna":{"entry":"","lastExit":"","userId":"FREEJOBALERT","userNick":"jone"}
}

Code used :

$json = file_get_contents('my.json');
$json_data = json_decode($json,true);
$userToCheck='johanna';
echo 'userId'.$json_data->$userToCheck->userId;

The above code gives error "Trying to get property of non-object", which i understand as $userToCheck isnt an object of $json_data but how do i access the data of "mackoniv" or even "johanna" when userToCheck part is not to be hardcoded.

If i try following way, it gives same error.

echo 'userId'.$json_data[$userToCheck]['userId'];

Answer 1

first of all json_decode($jsondata, true); would return an array. try to var_dump your $json to make sure you've got the file and then var_dump $json_data.

Answer 2

If you write it like below it should work.

        $json = file_get_contents('my.json');
        $json_data = json_decode($json, true);
        $userToCheck = 'johanna';
        echo 'userId = ' . $json_data[$userToCheck]['userId']; 

Basically change the line:

echo 'userId' . $json_data->$userToCheck->userId; 

with:

echo 'userId = ' . $json_data[$userToCheck]['userId'];

Answer 3

if json_decode second paramether is true, it returns an array, if it is false (default), returns an object. So, if you want to use it as an object, just decode it like: $json_data = json_decode($json); . If this doesn't work, check what echo json_last_error_msg(); returns. If it returns "No error", then the problem is that your file_get_contents() is not refering to the correct file.