使用PHP Guzzle HTTP 6向已经编码的数据发送JSON

2023年10月4日 417次阅读

我正在尝试发送一个POST请求,其中包含带有以下标头的原始
JSON字符串：Content-Type：application / json.

从查看文档,我可以看到我可以做这样的事情……

$data = ['x' => 1, 'y' => 2, 'z' => 3];
$client = new \GuzzleHttp\Client($guzzleConfig);
$options = [
    'json' => $data,
];
$client->post('http://example.com', $options);

我的问题是,当我到达这一点时,$data已经是json_encode’d.

我尝试了以下但它不起作用.

$data = json_encode(['x' => 1, 'y' => 2, 'z' => 3]);
$client = new \GuzzleHttp\Client($guzzleConfig);
$options = [
    'body' => $data,
    'headers' => ['Content-Type' => 'application/json'],
];
$client->post('http://example.com', $options);

我的问题是：我可以将json选项与已编码的数组一起使用吗？或者我有办法简单地设置Content-Type标题吗？

最佳答案 根据guzzle的文档
http://docs.guzzlephp.org/en/latest/request-options.html#json

您可以将已编码的json直接传递给body参数

Note This request option does not support customizing the Content-Type
header or any of the options from PHP’s json_encode() function. If you
need to customize these settings, then you must pass the JSON encoded
data into the request yourself using the body request option and you
must specify the correct Content-Type header using the headers request
option.

This option cannot be used with body, form_params, or multipart

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