Php 从json文件输出特定变量

我是一个真正的php新手，但我只是想通过实践来学习。请帮助我，告诉我如何加载整个json文件，并仅输出包含以下名称的变量：

假设my file.json包含：

{“类型”：“特征”，“特征”：[ {“类型”：“特征”，“id”：“gf”，“属性”：{“名称”：“美国”}，“几何”：{“类型”：“多边形”，“坐标”：[[[61.210817,35.650072]，[62.230651,35.270664]，[62.984662,35.404041]，[63.193538,35.857166]，[63.982896,36.007957]，[64.546479,36.312073]，[64.746105,37.111818]，[65.588948,37.3052177]，[65.745631,35.11637]，[35.737]，[65.737]，[65.797]，[66.518607,37.362784],[67.075782,37.356144],[67.83,37.144994],[68.135562,37.023115],[68.859446,37.344336],[69.196273,37.151144],[69.518785,37.608997],[70.116578,37.588223],[70.270574,37.735165],[70.376304,38.138396],[70.806821,38.486282],[71.348131,38.258905],[71.239404,37.953265],[71.541918,37.905774],[71.448693,37.065645],[71.844638,36.738171],[72.193041,36.948288],[72.63689,37.047558],[73.260056,37.495257],[73.948696,37.421566],[74.980002,37.41999],[75.158028,37.133031],[74.575893,37.020841],[74.067552,36.836176],[72.920025,36.720007],[71.846292,36.509942],[71.262348,36.074388],[71.498768,35.650563],[71.613076,35.153203],[71.115019,34.733126],[71.156773,34.348911],[70.881803,33.988856],[69.930543,34.02012],[70.323594,33.358533],[69.687147,33.105499],[69.262522,32.501944],[69.317764,31.901412],[68.926677,31.620189],[68.556932,31.71331],[67.792689,31.58293],[67.683394,31.303154],[66.938891,31.304911],[66.381458,30.738899],[66.346473,29.887943],[65.046862,29.472181],[64.350419,29.560031],[64.148002,29.340819],[63.550261,29.468331],[62.549857,29.318572],[60.874248,29.829239],[61.781222,30.73585],[61.699314,31.379506],[60.941945,31.548075],[60.863655,32.18292],[60.536078,32.981269],[60.9637,33.528832],[60.52843,33.676446],[60.803193,34.404102],[61.210817,35.650072]]]}}, {“类型”：“特征”，“id”：“sa”，“属性”：{“名称”：“津巴布韦”}，“几何学”：{“类型”：“多边形”，“坐标”：[[[31.191409，-22.25151]，[30.659865，-22.151567]，[30.322883，-22.271612]，[29.839037，-22.102216]，[29.432188，-22.091313]，[28.794656，-21.639454]，[28.02137，-21.485975]，[27.72727228，-20.851802]，[27.72476537，-20.499027]，[20.399027]，[29]，[26.164791,-19.293086],[25.850391,-18.714413],[25.649163,-18.536026],[25.264226,-17.73654],[26.381935,-17.846042],[26.706773,-17.961229],[27.044427,-17.938026],[27.598243,-17.290831],[28.467906,-16.4684],[28.825869,-16.389749],[28.947463,-16.043051],[29.516834,-15.644678],[30.274256,-15.507787],[30.338955,-15.880839],[31.173064,-15.860944],[31.636498，-16.07199]，[31.852041，-16.319417]，[32.328239，-16.392074]，[32.847639，-16.713398]，[32.849861，-17.979057]，[32.654886，-18.67209]，[32.611994，-19.419383]，[32.772708，-19.715592]，[32.659743，-20.30429]，[32.508693，-20.395292]，[32.244988，-21.116489]，[31.1914051]，[31]

]}

在my output.php中，它应该输出如下名称：

姓名：美国 姓名：津巴布韦

从json文件中，请向我展示一个例子，让我了解php世界并提高我的技能

$data = json_decode($your_json);
$type = $data->type;
var_dump($type); //

---

你可以阅读更多关于在php中解码json和访问值的信息

你的json有问题，你能更新json吗请你测试你的代码吗？我只想回显名称