[英]Php and JSON : object(stdClass)
I have decoded a JSON file into a variable ( $tmp
). 我已将JSON文件解码为变量( $tmp
)。 var_dump($tmp)
gives: var_dump($tmp)
给出:
object(stdClass)#31 (3) {
["c"]=> int(2)
["r"]=> int(2)
["d"]=> object(stdClass)#32 (4) {
["1"]=> string(2) "un"
["2"]=> string(4) "deux"
["3"]=> string(5) "trois"
["4"]=> string(6) "quatre"
}
}
I want to retrieve for example "un" so I do $tmp->d["1"]
but it doesn't work. 我想检索例如“un”所以我做$tmp->d["1"]
但它不起作用。 I've got the following error: 我有以下错误:
Fatal error: Cannot use object of type stdClass as array in File.php on line 17
json_decode takes an additional paramater that will turn your json string into an array instead of an object json_decode采用额外的参数,将您的json字符串转换为数组而不是对象
json_decode($json_str, true)
As comment noted, your d
property of your json object is an object not an array, so you can't access it with array notation (as you see there is an error) 正如注释所指出的,你的json对象的d
属性是一个对象而不是一个数组,所以你不能用数组表示法访问它(你看到有一个错误)
I believe 我相信
$tmp->d->{'1'}
// "un"
should work in accessing it 应该努力访问它
php has a default function json_encode and json_decode php有一个默认函数json_encode和json_decode
$arrayOfValues = array();
$jsonString = json_encode($arrayOfValues);
and 和
$arrayOfValues = json_decode($jsonString);
with this function you can use variabled with json. 使用此函数,您可以使用带有json的variabled。
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