Question

如何使用AJAX提交dojo表单，如果有错误，在错误填写的字段附近打印错误？现在我正在做类似的事情：

dojo.ready(function() {
    var form = dojo.byId("user_profile_form");
    dojo.connect(form, "onsubmit", function(event){
    dojo.stopEvent(event);

    var xhrArgs = {
        form: form,
        handleAs: "json",
        load: function(responseText){
            var result_data = zen.json.getResult(responseText);
            dojo.byId("response").innerHTML = "Form posted.";
        },
        error: function(error){
            // We'll 404 in the demo, but that's okay.  We don't have a 'postIt' service on the
            // docs server.
            dojo.byId("response").innerHTML = "Form posted.";
        }
    }
    // Call the asynchronous xhrPost
    dojo.byId("response").innerHTML = "Form being sent..."
    var deferred = dojo.xhrPost(xhrArgs);
});

但我不知道如何打印错误

Answer 1

有几种方法可以做到这一点。我更喜欢的是订阅IO Pipeline Topics

如有错误，请订阅/dojo/io/error主题。这是一个Growl错误的例子。

dojo.subscribe("/dojo/io/error", function(/*dojo.Deferred*/ dfd, /*Object*/ error){
    // Triggered whenever an IO request has errored.
    // It passes the error and the dojo.Deferred
    // for the request with the topic.

    var responseTextObject = dojo.fromJson(error.responseText)
    var growlMessage = '';

    if (responseTextObject && responseTextObject.message) {
        growlMessage += responseTextObject.message
    } else {
        // Don't Growl the xhr cancelled messages.
        if (error.message == 'xhr cancelled') {
            return;
        }

        growlMessage = error.message
    }

    new ext.Growl({
        message: growlMessage
    });
});

服务器应提供响应中的所有错误详细信息。在此示例中，预期会出现JSON格式的响应，但如果未提供，则仍会显示错误。

如果您想查看好的无效字段样式，请将小部件放在dijit.form.Form

中

Zend Dojo。 Ajax提交dojo表单

1 个答案: