Question

我试图以一种友好的方式显示错误，但是我总是通过我想摆脱的控制台日志获得错误堆栈跟踪。

该想法是使用任何来源（例如Google表格）在我们的平台中创建销售线索。当线索中提供了无效的电子邮件并将其发布到我们的API时，我将得到想要显示的预期消息，然后显示堆栈跟踪。

我的自定义错误消息是

INVALID FORMAT for email. Object didn't pass validation for format email: as1@mail.

但这就是我得到的：

INVALID FORMAT for email. Object didn't pass validation for format email: as1@mail. What happened: Starting POST request to https://cosmo-charon-production.herokuapp.com/v1/lead/vehicle Received 500 code from https://cosmo-charon-production.herokuapp.com/v1/lead/vehicle?api-key=gIBp04HVdTgsHShJj6bXKwjbcxXTogsh after 62ms Received content "{"code":"SCHEMA_VALIDATION_FAILED","message":"Request validation failed: Parameter (lead) failed sch" INVALID FORMAT for email. Object didn't pass validation for format email: as1@mail. Console logs:

Image showing error displayed in Zapier

我已经在afterResponse中添加了用于ErrorHandling的中间件，就像Zapier文档中提供的示例之一一样。

函数analyzeAndParse()从API接收一个错误对象，并返回一个字符串，并以一种友好的方式翻译了错误消息

const checkForErrors = (response, z) => {
  // If we get a bad status code, throw an error, using the ErrorTranslator
  if (response.status >= 300) {
    throw new Error(analyzeAndParse(response.json))
  }
  // If no errors just return original response
  return response
}

这是在我们的平台中创建销售线索并向我们的API发出请求的代码。

function createLead (z, bundle) {
  const industry = bundle.inputData.industry

  // add product to request based on the inputFields
  leadType[industry].addProductFields(bundle.inputData)

  const requestOptions = {
    url: `${baseUrl}lead/${_.kebabCase(industry)}`,
    method: 'POST',
    body: JSON.stringify(checkSourceForCreate(bundle.inputData)),
    headers: {
      'content-type': 'application/json'
    }
  }

  return z.request(requestOptions).then((response) => {
    if (response.status >= 300) {
      throw new Error(analyzeAndParse(response.content))
    }
    const content = JSON.parse(response.content)
    if (content && content.leads) {
      // get only the last lead from the list of leads
      content.lead = content.leads[0]
      delete content.leads
    }
    return content
  })
}

有什么想法吗？

谢谢！

在Zapier中以友好方式显示错误时出现问题

0 个答案: