使用JSON响应从PHP文件发出http请求

时间:2012-10-04 13:35:48

标签: php json http

我正在尝试通过http调用php脚本并从我计划进一步处理的地方接收一个json对象。

基本上,代码如下:

<?php
    if ($_SERVER['REQUEST_METHOD'] === 'GET') {
        $version=$_GET["v"];
        $product=$_GET["p"];
        $stream=$_GET["s"];
        $cmd=$_GET["c"];

        $string = file_get_contents("http://localhost:82/releasenote/src/getTSBDetails.php?p=$product&v=$version&s=$stream&c=$cmd");
        print_r($string);
        exit();
    } else {
        print("2");
        $string = file_get_contents('tsbDetails.json');
    }

当在浏览器中直接调用get_file_contents http请求时,输出是json,但在尝试使用上面时没有响应。

2 个答案:

答案 0 :(得分:2)

enter image description here

<?php
        // JSon request format is :
        // {"userName":"654321@zzzz.com","password":"12345","emailProvider":"zzzz"}

        // read JSon input
        $data_back = json_decode(file_get_contents('php://input'));

        // set json string to php variables
        $userName = $data_back->{"userName"};
        $password = $data_back->{"password"};
        $emailProvider = $data_back->{"emailProvider"};

        // create json response
        $responses = array();
        for ($i = 0; $i < 10; $i++) {
            $responses[] = array("name" => $i, "email" => $userName . " " . $password . " " . $emailProvider);
        }

        // JSon response format is :
        // [{"name":"eeee","email":"eee@zzzzz.com"},
        // {"name":"aaaa","email":"aaaaa@zzzzz.com"},{"name":"cccc","email":"bbb@zzzzz.com"}]

        // set header as json![enter image description here][2]
        header("Content-type: application/json");

        // send response
        echo json_encode($responses);
        ?>


  [1]: http://i.stack.imgur.com/I7imt.jpg
  [2]: http://i.stack.imgur.com/XgvOT.jpg

答案 1 :(得分:0)

首先,您应该确保您的变量可以在网址中使用:

    $version=urlencode($_GET["v"]);
    $product=urlencode($_GET["p"]);
    $stream=urlencode($_GET["s"]);
    $cmd=urlencode($_GET["c"]);

然后你应该检查你在$string中读到的值是否有效json。您可以使用this answer

然后,如果你的字符串包含有效的json,你应该echo它。

最后,如果你总是期望你的脚本中有json,你也应该json_encode你的错误处理:

} else {
    echo json_encode("2");
    // $string = file_get_contents('tsbDetails.json');  /* commented out as you don't seem to use it */
}