[英]Issue converting from cURL to Guzzle 6 with JSON and XML file
我很難將 cURL 轉換為 Guzzle6。 我想通過 JSON 和 XML 文件的內容發送名稱和引用 UUID 以處理到 REST 端點。
卷曲
curl -H 'Expect:' -F
'request={"name":"test", "reference":"870e0320-021e-4c67-9169-d4b2c7e5b9c9"}'
-F 'file=@sample.xml' http://ec2-48-88-173-9.us-east-1.compute.amazonaws.com:8180/rs/process
狂飲
$client = new Client(['debug' => true]);
$request = $client->request('POST',
'http://ec2-48-88-173-9.us-east-1.compute.amazonaws.com:8180/rs/process', [
'multipart' => [
[
'name' => 'data',
'contents' => "{'name':'test','reference':870e0320-021e-4c67-9169-d4b2c7e5b9c9}",
'headers' => ['Content-Type' => 'application/json']
],
[
'name' => 'file',
'contents' => fopen('sample.xml', 'r'),
'headers' => ['Content-Type' => 'text/xml']
],
]
]
);
$response = $request->getBody()->getContents();
另外,我不確定 'name' 字段應該是什么( 'name' => 'data'
)等。
這是相當於 curl 命令的 Guzzle:
$client = new Client(['debug' => true]);
$request = $client->request('POST',
'http://ec2-48-88-173-9.us-east-1.compute.amazonaws.com:8180/rs/process', [
'multipart' => [
[
'name' => 'request',
'contents' => "{'name':'test','reference':870e0320-021e-4c67-9169-d4b2c7e5b9c9}",
],
[
'name' => 'file',
'contents' => fopen('sample.xml', 'r'),
],
]
]
);
$response = $request->getBody()->getContents();
對於file
Guzzle 將指定適當的內容類型,就像 curl 一樣。 第一部分的名稱是request
——來自-F 'request={"name":"test", "reference":"870e0320-021e-4c67-9169-d4b2c7e5b9c9"}'
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