如何在php中单元测试curl调用 [英] how to unit test curl call in php

问题描述

如何进行单元测试curl实现？

  public function get（）{
 $ ch = curl_init（$ this-> request-> getUrl（））; 
 
 curl_setopt（$ ch，CURLOPT_HEADER，false）; 
 curl_setopt（$ ch，CURLOPT_RETURNTRANSFER，true）; 
 
 $ result = curl_exec（$ ch）; 
 $ code = curl_getinfo（$ ch，CURLINFO_HTTP_CODE）; 
 $ type = curl_getinfo（$ ch，CURLINFO_CONTENT_TYPE）; 
 curl_close（$ ch）; 
 
 if（！strstr（$ type，'application / json'））{
 throw new HttpResponseException（'JSON response not found'）; 
} 
 
 return new HttpResponse（$ code，$ result）; 
} 
  

我需要测试返回的内容类型，以便它可以抛出异常。

解决方案

正如thomasrutter建议的，创建一个类来抽象cURL函数的使用。

  interface HttpRequest 
 {
 public function setOption（$ name，$ value）; 
 public function execute（）; 
 public function getInfo（$ name）; 
 public function close（）; 
} 
 
类CurlRequest实现HttpRequest 
 {
 private $ handle = null; 
 
 public function __construct（$ url）{
 $ this-> handle = curl_init（$ url）; 
} 
 
 public function setOption（$ name，$ value）{
 curl_setopt（$ this-> handle，$ name，$ value）; 
} 
 
 public function execute（）{
 return curl_exec（$ this-> handle）; 
} 
 
 public function getInfo（$ name）{
 return curl_getinfo（$ this-> handle，$ name）; 
} 
 
 public function close（）{
 curl_close（$ this-> handle）; 
} 
} 
  

现在你可以使用< c $ c> HttpRequest 接口而不调用任何cURL函数。

  public function testGetThrowsWhenContentTypeIsNotJson {
 $ http = $ this-> getMock（'HttpRequest'）; 
 $ http-> expects（$ this-> any（））
  - > method（'getInfo'）
  - > will（$ this-> returnValue不是JSON'））; 
 $ this-> setExpectedException（'HttpResponseException'）; 
 //使用$ http创建测试类，而不是真正的CurlRequest 
 $ fixture = new ClassUnderTest（$ http）; 
 $ fixture-> get（）; 
} 
  

编辑修复了简单的PHP解析错误。 p>

How would you go about unit testing a curl implementation?

  public function get() {
    $ch = curl_init($this->request->getUrl());

    curl_setopt($ch, CURLOPT_HEADER, false);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);

    $result = curl_exec($ch);
    $code = curl_getinfo($ch, CURLINFO_HTTP_CODE);
    $type = curl_getinfo($ch, CURLINFO_CONTENT_TYPE);
    curl_close($ch);

    if (!strstr($type, 'application/json')) {
      throw new HttpResponseException('JSON response not found');
    }

    return new HttpResponse($code, $result);
  }

I need to test the content type returned so that it can throw an exception.

解决方案

As thomasrutter suggested, create a class to abstract the usage of the cURL functions.

interface HttpRequest
{
    public function setOption($name, $value);
    public function execute();
    public function getInfo($name);
    public function close();
}

class CurlRequest implements HttpRequest
{
    private $handle = null;

    public function __construct($url) {
        $this->handle = curl_init($url);
    }

    public function setOption($name, $value) {
        curl_setopt($this->handle, $name, $value);
    }

    public function execute() {
        return curl_exec($this->handle);
    }

    public function getInfo($name) {
        return curl_getinfo($this->handle, $name);
    }

    public function close() {
        curl_close($this->handle);
    }
}

Now you can test using a mock of the HttpRequest interface without invoking any of the cURL functions.

public function testGetThrowsWhenContentTypeIsNotJson() {
    $http = $this->getMock('HttpRequest');
    $http->expects($this->any())
         ->method('getInfo')
         ->will($this->returnValue('not JSON'));
    $this->setExpectedException('HttpResponseException');
    // create class under test using $http instead of a real CurlRequest
    $fixture = new ClassUnderTest($http);
    $fixture->get();
}

Edit Fixed simple PHP parse error.

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