wcf - get fault information from a webinvoke method using WebClient -

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मैं नया WebClient (url) का उपयोग कर रहा हूं। अपलोडस्ट्रेस (... WCF WebInvoke वेब विधि को खोलने के लिए। विधि फेंक रही है एक गलती अपवाद।

क्या रिमोट सर्वर ने त्रुटि लौटाई है: (400) खराब अनुरोध "

क्या मैं एक गलती फेंक रहा हूं? FaultException) इसमें सभी प्रासंगिक जानकारी के साथ।

सादर क्रेग। 

 < Code> try {string result = client.UploadString ("http: // localhost: 8080 / method", "POST", xml.ToString ()); result.dump ();} कैच (वेबएक्सेशन एक्स) {XNamespace exn = "Http://www.w3.org/1999/xhtml"; var त्रुटि = XElement.Load में नोड से (ex.Response.GetResponseStream ())। Descendants (exn + "p") नोड का चयन करें। Value; त्रुटि। डंप ();}

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