Question

出于某些原因，我有一个创建实体记录的javascript，并且会在创建后触发插件。如果pugin抛出异常，它不会显示给用户，因为涉及到javascript。我想知道是否有办法将插件中的异常传递给javascrit，以便它可以将其显示给用户。

感谢。

Answer 1

您是否收到任何http状态代码错误，如40 *，500,50 *，如果您使用Odata，您有回调来获取错误。 Javascript异常可能不会被抛出。

示例：

.ajax({
type: “POST”,
contentType: “application/json; charset=utf-8″,
datatype: “json”,
url: serverUrl + ODATA_ENDPOINT + “/” + odataSetName,
data: jsonEntity,

beforeSend: function (XMLHttpRequest) {
//Specifying this header ensures that the results will be returned as JSON.
XMLHttpRequest.setRequestHeader(“Accept”, “application/json”);

},
success: function (data, textStatus, XmlHttpRequest) {

},

error: function (XmlHttpRequest, textStatus, errorThrown) {

if (errorCallback)

errorCallback(XmlHttpRequest, textStatus, errorThrown);

else

alert(“Error on the creation of record; Error – “+errorThrown);

}

});

}

Answer 2

检查XMLHttpRequest readyState和status应该可以解决这个问题：

function createRecord(eObject, eName) {
    var jsonEntity = window.JSON.stringify(eObject);
    var u = Xrm.Page.context.getServerUrl();
    var ODATA_ENDPOINT = "/XRMServices/2011/OrganizationData.svc";
    var rec = new XMLHttpRequest();
    var p = u + ODATA_ENDPOINT;
    rec.open('POST', p + "/" + eName, false);
    rec.setRequestHeader("Accept", "application/json");
    rec.setRequestHeader("Content-Type", "application/json; charset=utf-8");
    rec.send(jsonEntity);
    if (rec.readyState == 4) {
        if (rec.status == 200) {
            var r = JSON.parse(rec.responseText).d;
            return r;
        }
        else {
            var error;
            try { 
                // To get the exception from the responseXML
                // I get an error when trying to use JSON.parse(rec.response)
                // Unexpected token <
                // I would much rather use JSON.parse 
                // than fiddling with the XML
                // Please let me know if you find how to 
                // get the error message with JSON.parse
                var xmlDoc = rec.responseXML;
                var error = xmlDoc.childNodes[0]
                    .childNodes[0]
                    .childNodes[0]
                    .childNodes[1]
                    .childNodes[0].nodeValue + "."; // The error message
            }
            catch (e) {
                // unable to get error message from response
                error = "General Error"; 
            }
            return error;
        }
    }
    // there is a problem
    // maybe return general error message
    return "General Error";
}

Answer 3

谢谢你们所有人。我确实通过将部分代码修改为以下内容来解决问题：

 rec.send(jsonEntity);
  if (rec.readyState == 4) {
    if (rec.status >= 200) {
        var newRecord = JSON.parse(rec.responseText).d;
        return newRecord;
    }
    else {
           var error;
           var er;
        try { 
              error = rec.responseText.toString();
             error = error.substring(error.lastIndexOf("value\": ")+9,error.lastIndexOf("\""));
             alert(error);
              }
         catch(e) {
                error = rec.responseText.toString();
              error = encodeURIComponent(error);
            er = error.substring(error.lastIndexOf("value%22%3A%20%22")+17,
                             error.lastIndexOf("%22%0D%0A%7D%0D%0A%7D%0D%0A%7D"));
            er = decodeURIComponent(er);
              return er;
           }
    }

将插件中的异常传递给Javascript

3 个答案: