如何从json_decode中获取此数组的值？

Question

我正在尝试使用来自json_decode的数组中的一些值，但无法从var_dump获取任何输出（这可能会帮助我弄清楚如何访问特定值）。我也从类似问题的建议中尝试了$ data [0]。

<?PHP
$json = 'callback({"geobytesinternet":"US","geobytescountry":"United States","geobytesregionlocationcode":"USWI","geobytesregion":"Wisconsin","geobytescode":"WI","geobyteslocationcode":"USWILOLA","geobytescity":"Land O Lakes","geobytescityid":"30028","geobytesfqcn":"Land O Lakes, WI, United States","geobyteslatitude":"46.154598","geobyteslongitude":"-89.396896","geobytescapital":"Washington, DC ","geobytestimezone":"-89.3969","geobytesnationalitysingular":"American","geobytespopulation":"278058881","geobytesnationalityplural":"Americans","geobytesmapreference":"North America ","geobytescurrency":"US Dollar","geobytescurrencycode":"USD","geobytestitle":"The United States"});';

$json = str_replace( 'callback(', '', $json);
$json = str_replace( ');', '', $json);
$json = '[' . $json . ']';

$data = json_decode($json);

echo '<br>decoded: ' . $data;

echo '<br><br>var_dump: ' . var_dump($data[0]);

?>

Answer 1

你几乎，函数json_decode()作为输入 json encoded 字符串。由于此行$json = '[' . $json . ']';。

，您的字符串未正确进行json编码

这是正确编码的json字符串

$json = '{"a":1,"b":2,"c":3,"d":4,"e":5}';

如果无法解码json或编码数据深于递归限制，则返回

NULL。请参阅documentation。

修改 所以你的代码应该是这样的：

<?php
$json = 'callback({"geobytesinternet":"US","geobytescountry":"United States","geobytesregionlocationcode":"USWI","geobytesregion":"Wisconsin","geobytescode":"WI","geobyteslocationcode":"USWILOLA","geobytescity":"Land O Lakes","geobytescityid":"30028","geobytesfqcn":"Land O Lakes, WI, United States","geobyteslatitude":"46.154598","geobyteslongitude":"-89.396896","geobytescapital":"Washington, DC ","geobytestimezone":"-89.3969","geobytesnationalitysingular":"American","geobytespopulation":"278058881","geobytesnationalityplural":"Americans","geobytesmapreference":"North America ","geobytescurrency":"US Dollar","geobytescurrencycode":"USD","geobytestitle":"The United States"});';

//Compact a bit the code, note the ');' replacement
//str_replace() can take an array of arguments, so no need to call the function multiple times
$json = str_replace( array('callback(', ');'), '', $json);
//Get elements in a array
$data = json_decode($json);

//See what's inside the array
var_dump($data);

?>

This是一个工作小提琴。

Answer 2

解码后的数据是对象数组，所以请执行以下操作：

$result = $data[0];
echo $result->geobytesinternet;