extjs4全局网络异常监听器 [英] extjs4 global network exception listener

问题描述

我想编写一个listener，它将侦听所有网络请求错误，如下所示:

I want to write a listener that will listen to all network requests errors, something like this :

Ext.Ajax.on('requestexception', function(conn, response, options) {
    if (response.status === 555) {
        Ext.Msg.alert('test', 'test');
    }
});

上面的代码仅适用于通过Ext.Ajax.request()进行的请求，如何重写它，以便它也可以用于表单提交，找不到URL错误等.

The above code works only for requests via Ext.Ajax.request(), how to rewrite it so it could work also for form submits, url not found error etc.

在服务器端，我有Spring MVC调度所有请求，如果有error，则返回555的响应状态.

On server side I have Spring MVC that dispatches all requests and if there is any error, the response status of 555 is returned.

form.submit({
     url: dispatcher.getUrl('savePlanRequest'),
     //headers: {'Content-Type':'multipart/form-data; accept-charset=utf-8'},
     scope: this,
     method: 'GET',
     params: {
         scan: scan_id,
         attachments: attachments_id,
         parcels: parcels_id
     },
     success: function(form, action) {
         this.fireEvent('plansaved', this);
         Ext.Msg.alert(i18n.getMsg('success'), i18n.getMsg('gsip.view.plans.NewPlanForm.success_info'))
     },
     failure: function(form, action) {
         console.log('failure');
         //Ext.Msg.alert(i18n.getMsg('failure'), action.result.msg);
     }
 });

推荐答案

这应该有效:

Ext.override( Ext.form.action.Submit, { 
    handleResponse : function( response ) {

        var form = this.form,
            errorReader = form.errorReader,
            rs, errors, i, len, records;

        if (errorReader) {
             rs = errorReader.read(response);
             success = rs.success;
             // Do something if success is false
        }

        this.callParent ( arguments ); 
    }
});

看看精确的handleResponse()方法的源代码，我从上面复制了大部分代码.

Have a look at the source code for the exact handleResponse() method from which I copied most of the code above.

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