在PHP中从JSON获取值
我想从JSON中获取特定的值 网址: 我试着像下面一样在PHP中从JSON获取值,php,Php,我想从JSON中获取特定的值 网址: 我试着像下面一样 <?php $data = json_decode('https://api.darksky.net/forecast/92cf27941c6ea888652ba37de4da4044/37.8267,-122.4233', true); echo $data['currently']['temperature']; ?> 但它不起作用。有人能帮我吗?我只想获取当前>温度值 您不能直接调用json url
<?php
$data = json_decode('https://api.darksky.net/forecast/92cf27941c6ea888652ba37de4da4044/37.8267,-122.4233', true);
echo $data['currently']['temperature'];
?>
但它不起作用。有人能帮我吗?我只想获取当前>温度值 您不能直接调用json url。你需要一个文件函数。这是一个样本
$json_url = "http://awebsites.com/file.json";
$json = file_get_contents($json_url);
$data = json_decode($json, TRUE);
echo "<pre>";
print_r($data);
echo "</pre>";
$json\u url=”http://awebsites.com/file.json";
$json=file\u get\u contents($json\u url);
$data=json_decode($json,TRUE);
回声“;
打印(数据);
回声“;
请试试这个
<?php
$your_url = "https://api.darksky.net/forecast/92cf27941c6ea888652ba37de4da4044/37.8267,-122.4233";
$get_data = file_get_contents($your_url);
$data = json_decode($get_data, TRUE);
echo $data['currently']['temperature'];
?>
如果不使用curl或file\u get\u contents 下面的代码片段以获取数据
$url = "https://api.darksky.net/forecast/92cf27941c6ea888652ba37de4da4044/37.8267,-122.4233";
$json_data = file_get_contents($url);
$data = json_decode($json_data, TRUE);
echo $data['currently']['temperature'];
您必须进行以下更改
json\u解码('https://api.darksky.net/forecast/92cf27941c6ea888652ba37de4da4044/37.8267,-122.4233',正确)代码>将尝试解码url字符串,但不会解码结果。为此,您必须执行此url
$url = 'https://api.darksky.net/forecast/92cf27941c6ea888652ba37de4da4044/37.8267,-122.4233';
$result = file_get_contents($url);
$data = json_decode($result, true);
echo $data['currently']['temperature'];
您需要首先使用curl或file\u get\u contents获取页面内容。尝试以下代码
谢谢我错过了文件内容。简单。谢谢。很高兴帮助@RaselAhmed:)
$url = 'https://api.darksky.net/forecast/92cf27941c6ea888652ba37de4da4044/37.8267,-122.4233';
$result = file_get_contents($url);
$data = json_decode($result, true);
echo $data['currently']['temperature'];