狼吞虎咽；用基本URL、标头和POST设置客户端

Question

我有困难纠正我的语法，我将感谢任何帮助；我试图设置包含基URL和一些必要的头信息（包括自定义安全令牌）的客户端

下一步是向webservice发布：

$baseServiceURL = ['base_uri' => 'http://127.0.0.1:8080/service/v1/ws//something/update']; 
$theHeaders = ['Content-Type' => 'application/json', 'Accept' =>  'application/json', 'token' => 'test-token'];

$updateRequestClient = new Client($baseServiceURL, array(
    "request.options" => array(
       "headers" => $theHeaders
    )
));

//var 1 coming from elsewhere
$varNum2 = $q;
$varNum3 = $w;
$varNum4 = $e;
$varNum5 = $r;
$varNum6 = $t;
$varNum7 = 'me';

// json name/value pairs
$updateBody['name1'] = $varNum1;
$updateBody['name2'] = $varNum2;
$updateBody['name3'] = $varNum3;
$updateBody['name4'] = $varNum4;
$updateBody['name5'] = $varNum5;
$updateBody['name6'] = $varNum6;
$updateBody['name7'] = $varNum7;

//send
$updateRequestResponse = $updateRequestClient->post([ 'body' => json_encode($updateBody) ]);

//response 200??
$responseCode = $updateRequestResponse->getStatusCode();

if ($responseCode == "200") {                                                  

echo ("SUCCESS");

}

如果我将URL更改为string$baseserviceurl=(string)('http://127.0.0.1:8080/service/v1/ws//something/update')；，我会得到：

可捕获的致命错误：传递给guzzleHTTP\client::__construct()的参数1必须是array类型，给定字符串，在第79行的C:\xampp\htdocs\ssquerytool\updatedoctor.php中调用，在第62行的C:\xampp\vendor\guzzleHTTP\guzzle\src\client.php中定义

Answer 1

$baseServiceURL = 'http://127.0.0.1:8080/service/v1/ws//something/update'; 
$theHeaders = ['Content-Type' => 'application/json', 'Accept' =>  'application/json', 'token' => 'test-token'];

$updateRequestClient = new Client(array(
    'base_uri' => $baseServiceURL,
    'headers' => $theHeaders
));

客户端只接受构造函数https://github.com/guzzle/guzzle/blob/master/src/Client.php#l62中的一个参数