ios - NSJSONSerialization parse error - unrecognized selector -

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यह कोड है: 

  NSString * jsonString = @ "[{\" sn \ ": \" ई \ ", \" टी \ ": \" के \ ", \" d \ ": \" मंगल 3-मार्च \ "}, {\" sn \ ": \" F \ ", \ "टी \": \ "वाई 1 \", \ "डी \": \ "मंगल 3-मार्च \"}] "; एनएसडीटा * डेटा = [jsonString डेटाउपकरण एन्कोडिंग: NSUTF8StringEncoding]; NSArray * jsArray = [NSJSONSerialization JSONObjectWithData: डेटा विकल्प: 0 त्रुटि: शून्य]; एनएसएलॉग (@ "जेएसएआरए:% @", जेएसएआरआरए); (जेएसएआरआर में आईडी नौकरी) {एनएसएलॉग (@ "जॉब:% @", जॉब); एनएसएलॉग (@ "% @", [नौकरी एसओ]); }   
 कंसोल में, मुझे यह मिलता है:  
  2014-04-22 15: 40: 46.464 परीक्षण [2442: 60b] जेएसएआरआरए: ({ D = "मंगल 3-मार्च"; sn = E; t = K;}, {d = "मंगल 3-मार्च"; sn = f; t = "y 1";}) 2014-04-22 15:40 : 46.466 टेस्ट [2442: 60 बी] नौकरी: {d = "मंगल 3-मार्च"; sn = E; टी = के; } - [__ एनएससीएफएप्शन एसआईपी]: अज्ञात चयनकर्ता को उदाहरण 0x8fa53b0 *** भेजा गया है। बिना अपवाद अपवाद 'एनएसआईएनआईडीएडीएआरएगमेंट एक्स्प्शन' के कारण ऐप्पलिंग एप्लिकेशन को कारण: '- [__ एनएससीएफडीएन स्नैप]: अपरिचित चयनकर्ता को उदाहरण के लिए भेजा गया 0x8fa53b0'   
 यह वस्तुओं की सरणी और व्यक्तिगत वस्तु को पहचानने लगता है संपत्ति  sn    
 
 
  जेएसएआरआर  शब्दकोशों में शामिल हैं तो  नौकरी  एक  NSDictionary  है।  NSDictionary  में  sn  नाम का कोई तरीका नहीं है यदि आप कुंजी  @ "sn"  के लिए मान चाहते हैं तो आपको:  
  के लिए (jsArray में NSDictionary * job) {NSLog (@ "job:% @ ", काम); एनएसएलॉग (@ "% @", नौकरी [@ "स्न"]); }

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