Question

这是json

[{"location":"USA","email":"test@test.com","sex":"male","age":"Unkown","other":null,"profile":{"net":["55","56"],"networks":[{"site_url":"http://site.com","network":"test","username":"mike"},{"site_url":"http://site.com/2","network":"test2","username":"mike2"}]},"name":"Mike Jones","id":111}]

我想知道如何能够回应所有网络，以便它为2中的每一个回应site_url，网络和用户。

我怎么会在那里得到“名字”呢？

坦克！

Answer 1

使用json_decode（） http://php.net/manual/en/function.json-decode.php

  $data = json_decode(...your sstring ...);
  echo $data[0]->name;

Answer 2

使用json_decode解码JSON数据。然后，您可以使用foreach迭代数组，并使用另一个foreach访问每个数组项的 site_url ：

$arr = json_decode($json);
foreach ($arr as $obj) {
    foreach ($obj->profile->networks as $network) {
        echo $network->site_url;
    }
}

Answer 3

http://lt2.php.net/json_decode

Answer 4

以aviv的答案为基础......

$data = json_decode(...your sstring ...);
echo $data[0]->location; // USA
...
echo $data[0]->profile->net[0]; // 55
echo $data[0]->profile->net[1]; // 56
echo $data[0]->profile->networks[0]->site_url; // http://site.com
echo $data[0]->profile->networks[0]->network;  // test
echo $data[0]->profile->networks[0]->username; // mike
echo $data[0]->profile->networks[1]->site_url; // http://site.com/2
echo $data[0]->profile->networks[1]->network;  // test2
echo $data[0]->profile->networks[1]->username; // mike2
echo $data[0]->name; // Mike Jones

Answer 5

这是一个直截了当的例子来做你要求的两件事（参见内联评论）。

$json = '[{"location":"USA","email":"test@test.com","sex":"male","age":"Unkown","other":null,"profile":{"net":["55","56"],"networks":[{"site_url":"http://site.com","network":"test","username":"mike"},{"site_url":"http://site.com/2","network":"test2","username":"mike2"}]},"name":"Mike Jones","id":111}]';

// "Decode" JSON into (dumb) native PHP object
$data = json_decode($json);

// Get the first item in the array (there is only one)
$item = $data[0];

// Loop over the profile.networks array
foreach ($item->profile->networks as $network) {
    // echos out the site_url,network, and user 
    echo "site_url = " . $network->site_url . PHP_EOL;
    echo "network  = " . $network->network . PHP_EOL;
    echo "user     = " . $network->username . PHP_EOL;
}

// Get "name" at the end
echo "name     = " . $item->name . PHP_EOL;

它应该输出（如果您以HTML格式查看，它将被导入一行......不输出为HTML）。

site_url = http://site.com
network  = test
user     = mike
site_url = http://site.com/2
network  = test2
user     = mike2
name     = Mike Jones

使用PHP从JSON获取数据

5 个答案: