postman 请求webservice接口
时间: 2023-09-03 15:15:57 浏览: 192
你可以使用Postman来发起一个请求到Webservice接口。首先,打开Postman并新建一个Request请求。填写Webservice地址,将其粘贴到地址栏中,例如:【http://xxx:xx/xx/xx/xx?WSDL】。请确保请求方式设置为POST,并在请求头headers参数中添加Content-Type,其值为text/xml;charset=utf-8。请求报文和返回报文都是XML格式,所以你需要将请求报文以XML格式进行编写。具体的请求报文可以参考提供的示例,如下所示:
<?xml version="1.0" encoding="utf-8"?>
<soap:Envelope xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/">
<soap:Body>
<Login xmlns="http://tempuri.org/">
<user>admin</user>
<psd>123456</psd>
</Login>
</soap:Body>
</soap:Envelope>
请将上述XML报文作为参数传递给Webservice接口。这样,你就可以使用Postman发送请求到Webservice接口了。<span class="em">1</span><span class="em">2</span><span class="em">3</span>
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