尽管正在发送参数，但 JSON API 返回错误

debugcn 发表于 Dev

查理

我正在使用 PHP向 JSON API 系统（http://help.solarwinds.com/backup/documentation/Content/service-management/json-api/login.htm）发送请求：

$base = 'https://cloudbackup.management/jsonapi';

$vars = array(
    "jsonrpc" => "2.0",
    "method" => "Login",
    "params" => array(
        "partner" => "partner",
        "username" => "username",
        "password" => "pass",
    ),
    "id" => "1",
);

$ch = curl_init( $base );
curl_setopt($ch, CURLOPT_POST, true);
curl_setopt($ch, CURLOPT_POSTFIELDS, $vars);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
$response = curl_exec($ch);
curl_close($ch);

$output = json_decode($response, true);

但它返回这个数组 $output

Array
(
    [error] => Array
        (
            [code] => -32700
            [data] => 119
            [message] => Parse error: Failed to parse request body: * Line 1, Column 1
  '--------------------------' is not a number.

        )

    [id] => jsonrpc
    [jsonrpc] => 2.0
)

我无法弄清楚为什么它会返回错误，因为我正在发送它在文档中所说的正确参数。

有人可以指出我正确的方向还是我错过了什么？

肖恩·墨菲

将内容类型设置为 application/json 因为 curl 可能默认将其发送为 x-www-form-urlencoded

curl_setopt($ch, CURLOPT_HTTPHEADER, array('Content-Type: application/json'));

还对您的数组进行 JSON 编码：

$jsonDataEncoded = json_encode($vars);

完整重构示例：

$base = 'https://cloudbackup.management/jsonapi';

$vars = array(
    "jsonrpc" => "2.0",
    "method" => "Login",
    "params" => array(
        "partner" => "partner",
        "username" => "username",
        "password" => "pass",
    ),
    "id" => "1",
);

$jsonDataEncoded = json_encode($jsonData);
$ch = curl_init( $base );
curl_setopt($ch, CURLOPT_POST, true);
curl_setopt($ch, CURLOPT_POSTFIELDS, $jsonDataEncoded);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_HTTPHEADER, array('Content-Type: application/json')); 
$response = curl_exec($ch);
curl_close($ch);

$output = json_decode($response, true);

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