在 PHP 中访问 JSON 对象名称
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Access JSON object name in PHP
提问by xadoc
I have the following JSON:
我有以下 JSON：
{
"nickname": "xadoc",
"level": 4,
"loc": "Tulsa, OK, USA",
"score": 122597,
"money": 29412.5,
"streetNum": 8,
"streets": {
"-91607259\/387798111": {
"name": "Alam\u00e9da Ant\u00f3nio S\u00e9rgio",
"value": 243,
"type": 1
},
"-91016823\/388182402": {
"name": "Autoestrada do Norte",
"value": 18304,
"type": 1
},
"-86897820\/399032795": {
"name": "Autoestrada do Norte",
"value": 12673,
"type": 1
},
"-973092846\/479475465": {
"name": "19th Ave",
"value": 7794,
"type": 1
},
"-974473223\/480054888": {
"name": "23rd Ave NE",
"value": 33977,
"type": 1
}
}
}
I'm desperately trying to access the dynamic object names like "-91607259\/387798111", how can I do it?
我拼命地尝试访问动态对象名称，例如"-91607259\/387798111"，我该怎么做？
Right now I have:
现在我有：
$jsonurl = "http://www.monopolycitystreets.com/player/stats?nickname=$username&page=1";
$json = file_get_contents($jsonurl,0,null, $obj2 = json_decode($json);
foreach ( $obj2->streets as $street )
{
//Here I want to print the "-91607259\/387798111" for each street, please help
//echo $street[0]; gives "Fatal error: Cannot use object of type stdClass as array"
//echo $street gives "Catchable fatal error: Object of class stdClass could not be converted to string"
echo '<th>'.$street->name.'</th><td>'."M ".number_format($street->value, 3, ',', ',').'</td>';
}
回答by Peter Bailey
I would imagine that the simplest thing to do is to decode into associative arrays instead of stdClass objects
我想最简单的事情是解码成关联数组而不是 stdClass 对象
$obj2 = json_decode( $json, true );
foreach ( $obj2['streets'] as $coords => $street )
{
echo $coords;
}
回答by Pascal MARTIN
Considering this piece of code :
考虑这段代码：
$json = '{"nickname":"xadoc","level":4,"loc":"Tulsa, OK, USA","score":122597,"money":29412.5,"streetNum":8,"streets":{"-91607259\/387798111":{"name":"Alam\u00e9da Ant\u00f3nio S\u00e9rgio","value":243,"type":1},"-91016823\/388182402":{"name":"Autoestrada do Norte","value":18304,"type":1},"-86897820\/399032795":{"name":"Autoestrada do Norte","value":12673,"type":1},"-973092846\/479475465":{"name":"19th Ave","value":7794,"type":1},"-974473223\/480054888":{"name":"23rd Ave NE","value":33977,"type":1}}}';
$obj2 = json_decode($json);
var_dump($obj2);
You'll get :
你会得到 ：
object(stdClass)[1]
public 'nickname' => string 'xadoc' (length=5)
public 'level' => int 4
public 'loc' => string 'Tulsa, OK, USA' (length=14)
public 'score' => int 122597
public 'money' => float 29412.5
public 'streetNum' => int 8
public 'streets' =>
object(stdClass)[2]
public '-91607259/387798111' =>
object(stdClass)[3]
public 'name' => string 'Alaméda António Sérgio' (length=25)
public 'value' => int 243
public 'type' => int 1
public '-91016823/388182402' =>
object(stdClass)[4]
public 'name' => string 'Autoestrada do Norte' (length=20)
public 'value' => int 18304
public 'type' => int 1
public '-86897820/399032795' =>
object(stdClass)[5]
public 'name' => string 'Autoestrada do Norte' (length=20)
public 'value' => int 12673
public 'type' => int 1
public '-973092846/479475465' =>
object(stdClass)[6]
public 'name' => string '19th Ave' (length=8)
public 'value' => int 7794
public 'type' => int 1
public '-974473223/480054888' =>
object(stdClass)[7]
public 'name' => string '23rd Ave NE' (length=11)
public 'value' => int 33977
public 'type' => int 1
Which means you can loop over the streets like this :
这意味着你可以像这样在街道上循环：
foreach ( $obj2->streets as $id => $street ) {
echo $id;
var_dump($street);
echo '<hr />';
}
With this, for each $street, you'll get the corresponding key into $id-- and the data into $street.
这样，对于每个$street，您将获得相应的键$id- 并将数据放入$street.
Or you can directly access one this way :
或者您可以通过这种方式直接访问一个：
$street = $obj2->streets->{'-86897820/399032795'};
var_dump($street);
Which will get you :
这会让你：
object(stdClass)[5]
public 'name' => string 'Autoestrada do Norte' (length=20)
public 'value' => int 12673
public 'type' => int 1
Your $obj2->streetis an object, which means you cannot use array-syntax access ; this explains the "Fatal error: Cannot use object of type stdClass as array" if you try to use this :
你$obj2->street是一个对象，这意味着你不能使用数组语法访问；这解释了“ Fatal error: Cannot use object of type stdClass as array”，如果您尝试使用它：
$obj2->streets['-86897820/399032795'];
But the properties of your object have quite "strange" names ; which means you cannot do this :
但是你的对象的属性有很“奇怪”的名字；这意味着你不能这样做：
$obj2->streets->-86897820/399032795;
Which gives Parse error: syntax error, unexpected '-', expecting T_STRING or T_VARIABLE or '{' or '$'
这使 Parse error: syntax error, unexpected '-', expecting T_STRING or T_VARIABLE or '{' or '$'
Nor that :
也不是：
$obj2->streets->'-86897820/399032795';
Which also gives Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING, expecting T_STRING or T_VARIABLE or '{' or '$'
这也给 Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING, expecting T_STRING or T_VARIABLE or '{' or '$'
Happily, you can use {}to kind of "protect" the name of your keys, and get everything working ;-)
(I cannot find the page in the manual that explains this syntax nor give its name... If anyone know... )
令人高兴的是，您可以使用{}来“保护”您的密钥名称，并使一切正常；-)
（我在手册中找不到解释此语法的页面，也找不到其名称...如果有人知道... )
回答by Hamid Araghi
A simple way to extract data from Json stringis to use jsone_decode()function. For example:
从中提取数据的一种简单方法Json string是使用jsone_decode()函数。例如：
$json_data = '{"nickname":"xadoc","level":4,"loc":"Tulsa}';
$decoded_data = json_decode($json_data);
Then you can simply access the members of $decoded_datausing ->operator:
然后您可以简单地访问$decoded_datausing->运算符的成员：
echo "$decoded_data->nickname \n";
echo "$decoded_data->level \n";
And if your extracted members from $decoded_dataare json strings again, then you can use json_decode()again!
如果您从$decoded_data中提取的成员json string再次是s，那么您可以json_decode()再次使用！
回答by jeroen
I can′t try it right now, but if you do a:
我现在不能尝试，但如果你做：
var_dump($obj2);
you should be able to see exactly how to access your information.
您应该能够准确了解如何访问您的信息。
