php - 淘宝手机号归属地api

Question

https://tcc.taobao.com/cc/json/mobile_tel_segment.htm?tel=13001631234
使用这个接口查询归属地信息后php不能够解析json字符串
php

    $ch = curl_init();
    $url = "https://tcc.taobao.com/cc/json/mobile_tel_segment.htm?tel=$mobile"."&t=".time();

    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
    // 执行HTTP请求
    curl_setopt($ch, CURLOPT_URL, $url);
    $res = curl_exec($ch);
    $res = trim(explode('=',$res)[1]);
    $res = iconv('gbk','utf-8', $res);
    var_dump($res);
    $res = json_decode($res, true);
    var_dump($res);

Answer 1

-已解决

通过json_last_error()发现是JSON_ERROR_SYNTAX: ' - Syntax error, malformed JSON'

后来将key接用""扩了起来就可以了

    $res = trim(explode('=',$res)[1]);
    $res = iconv('gbk','utf-8', $res);
    $res = str_replace("'",'"', $res);
    $res = preg_replace('/(\w+):/is', '"$1":', $res);

Answer 2

建议换个接口,或者是请求缺少某参数,返回的不是json数据

Answer 3

对于php来说jsonp转json，并非直接取后就完事了。

下面是官方文档的说明：
<?php

// the following strings are valid JavaScript but not valid JSON

// the name and value must be enclosed in double quotes
// single quotes are not valid 
$bad_json = "{ 'bar': 'baz' }";
json_decode($bad_json); // null

// the name must be enclosed in double quotes
$bad_json = '{ bar: "baz" }';
json_decode($bad_json); // null

// trailing commas are not allowed
$bad_json = '{ bar: "baz", }';
json_decode($bad_json); // null

?>

官方文档入口json-decode