Question

我这里有代码，可以使用curl获取json数据。现在我想回复所有数据。这将显示无输出。

<?php
$json = file_get_contents('myurl');
$data = json_decode($json);
print_r($data);
?>

这是json来自我的网址：

 ({"Response_Code":"0000","ResultMobilePrefix":["0917","0905","0906","0915","0916","0926","0927","0937","0935","0817","0936","0922","0923",
"0932","0933","0934","0942","0943","0907","0908","0909","0910","0912","0918","0919","0920","
0921","0928","0929","0930","0938","0939","0948","0949","0925","0989","0999","0947","0998","
0946","0975","0977"]});

Answer 1

你需要的只是 echo json_encode（$ json）;

Answer 2

使用true将其转换为数组;

$data = json_decode($json,true);
echo '<pre>';
print_r($data);

Answer 3

以格式使用json_encode()显示json，并使用JSON_PRETTY_PRINT

进行正确格式显示

header('Content-type: application/json');
echo json_encode($json,JSON_PRETTY_PRINT);

Answer 4

在使用json_decode（）之前，您需要首先删除JSON文件上的括号和分号;

<?php
    $json = file_get_contents('myurl');

    //remove the brackets
    $json = str_replace("(", "", $json);
    $json = str_replace(")", "", $json);

    //remove the semicolon
    $json = str_replace(";", "", $json);

    $data = json_decode($json);
    print_r($data);
?>

这有点难看，但希望你明白这个想法，你需要首先删除那个角色

Answer 5

我只想添加Manish上面解释的内容。这就是PHP Docs关于第二个参数的说法：

When TRUE, returned objects will be converted into associative arrays.

否则你只需获得一个stdclass对象

Answer 6

问题是URL没有返回有效的Json

你可以试试

  var_dump($data);

这将返回 null ，因为它不是有效的json，请参阅json_decode

这将起作用

 $json = file_get_contents('myurl');
 $json = preg_replace('/[ ]{2,}|[\t\n\r\(\)\;]/', '', trim($json));
 $data = json_decode($json);
 print_r($data);

如何使用curl在数组中输出json数据

6 个答案: