如何发送URLRequest参数

Question

我在PHP中为后端发送了参数，但是我有代码，我没有发送参数

var request:URLRequest = new URLRequest(modelLocator.CaminhoServidor+"AnexoDocumentos_Financeiro/asdas/xml.php");
            var loader:URLLoader = new URLLoader();
            loader.dataFormat = URLLoaderDataFormat.VARIABLES;
            request.data = Remessa;
            request.method = URLRequestMethod.POST;
            loader.addEventListener(Event.COMPLETE, RecebeXML);
            loader.load(request);

如何发送3个参数这段代码？

Answer 1

您需要在URLRequest对象中使用“变量”字段：

var request:URLRequest = new URLRequest(url);
request.method = URLRequestMethod.POST;
var variables:URLVariables = new URLVariables();
variables.param1 = ...
variables.param2 = ...
variables.param3 = ...
request.data = variables;

var urlLoader:URLLoader = new URLLoader();
urlLoader.dataFormat = URLLoaderDataFormat.VARIABLES;
urlLoader.addEventListener(Event.COMPLETE, urlLoader_complete);
urlLoader.addEventListener(IOErrorEvent.IO_ERROR, urlLoader_error);
urlLoader.load(request);

此外，定义错误处理程序(如上面的IO_ERROR和HTTP_STATUS和SECURITY_ERROR所示)也是个好主意。