duanlian1978
2018-09-18 15:48 阅读 306
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php - 如何从数组对象中获取值

I use php code and I have a problem with getting value of an array object. When I use var_dump($arr), I get this (the code below is formatted for better readability):

object(League\OAuth2\Client\Provider\GoogleUser)#5 (1) {
    ["response":protected]=> array(5) {
        ["emails"]=> array(1) {
            [0]=> array(1) {
                ["value"]=> string(21) "thienlam129@gmail.com"
            }
        }
        ["id"]=> string(21) "115281634466837725533"
        ["displayName"]=> string(18) "thiên lâm trần"
        ["name"]=> array(2) {
            ["familyName"]=> string(6) "trần"
            ["givenName"]=> string(11) "thiên lâm"
        }
        ["image"]=> array(1) {
            ["url"]=> string(98) "https://lh3.googleusercontent.com/-SpWfKGTcQt8/AAAAAAAAAAI/AAAAAAAAAB8/IlGQQnvy7so/photo.jpg?sz=50"
        }
    }
}

Please, tell me how to get emails value.

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2条回答 默认 最新

  • 已采纳
    dongmu3187 dongmu3187 2018-09-18 18:56

    You could cast the object as an array as such:

    var_dump((array) $object);
    

    Or

    $array = (array) $object;
    

    And access all its properties but you might have to array walk and filter the name of the class out.

    But this is a hack and an example of bad practise. Accessing protected properties by its methods is the way to go or else request json object instead, if possible.

    Naturally, you would say

    $object = (object) $inputObject;
    $value = (string) $object->getValue();
    

    Or

    $value = (array) $object->getArray();
    
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  • doushi1960 doushi1960 2018-09-18 18:46

    The main reason you're facing this problem is probably because the object is not an array at all.

    As mentioned in the comments, the GoogleUser object has a public getEmail method which you can use to retrieve the email address.

    You can use it like this:

    $email = $googleUser->getEmail();
    

    (In the above code sample I took the liberty of calling it a $googleUser rather than an $arr because it's not an arr but a google user)

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