JSON parsing error in perl -

-

मुझे त्रुटि मिलती है

कार्यान्वयन विफल ... खराब JSON स्ट्रिंग, न तो सरणी, ऑब्जेक्ट, नंबर, स्ट्रिंग या परमाणु, चरित्र ऑफसेट 0 ("स्ट्रिंग का अंत") से पहले) "

जब मैं डीबी से इस स्निपेट को अपने JSON स्ट्रिंग को पार्स करता हूं। <$>

क्या मुझे कुछ याद आ रहा है? जेसन स्ट्रिंग डेटाबेस से है। "एसी", "आरएस 718757": "एजी", "आरएस6557634": "सीसी", "आरएस 995553": "एसी" सीजी "," आरएस 6166 ":" एजी "," आरएस 4925 ":" एए "," आरएस 502843 ":" जीटी "," आरएस 725029 ":" कोई कॉल नहीं "," आरएस 3 9 4872 ":" जीजी "," आरएस 14026 9 5 ":" टीटी "" rs719601 ":" ए.ए. "," rs2374061 ":" एजी "," rs952503 ":" टीटी "," rs1801262 ":" एजी "," rs5215 ":" सीटी "," rs978422 ":" सीसी ", "rs12828016": "जीजी", "rs958388": "एजी", "rs999072": "सीटी", "rs967344": "एजी", "rs2207782": "सीसी", "rs349235": "ए.ए.", "rs1074553 ":" सीटी "," rs1395936 ":" एजी "," GS35220 ":" सीटी "," rs7627615 ":" एजी "," rs727336 ":" एजी "," rs2077774 ":" एसी "," rs8065080 ": "सीसी", "rs1131498": "टीटी", "आरएस 2247870": "कोई कॉल नहीं", "आरएस 803172": "टीटी", "आरएस15412 9 0": "एजी", "आरएस 1414 9 4": "एए", "आरएस 1 9 8045": "कॉल नहीं", "आरएस 2077743": "जी.टी.", "rs2361128": "कोई कॉल", "rs3795677": "एजी", "rs1030687": "सीटी", "rs156318": "जीजी", "rs952768": "सीसी", "rs1363333": " टीटी "," rs7298565 ":" एजी "," rs310929 ":" सीसी "," rs2369898 ":" सीटी "," rs1327118 ":" सीसी "," rs4619 ":" एजी "," rs965323 ":" टीटी " , "rs2887851": "एजी", "rs1862456": "जी.टी.", "rs6759892": "जी.टी.", "rs753381": "एजी", "rs1805034": "सीसी", "rs1812642": "ए.ए." " rs4075254 ":" सीटी "," rs1805087 ":" ए.ए. "," rs532841 ":" सीटी "," rs951629 ":" जीजी "," rs2286963 ":" जीजी "," rs763553 ":" सीटी "," rs1074042 " : "जीजी", "rs2241714": "जीजी", "rs894240": "टीटी", "rs522073": "सीटी", "GS35205": "टीसी", "rs1368136": "टीटी", "rs1426003": " जी जी "," आरएस2016588 ":" कोई कॉल नहीं "," आरएस 621277 ":" कोई कॉल नहीं "," आरजे 727081 ":" जीजी "," आरएस 1392265 ":" एसी "," आरएस 10 9 820 ":" कोई कॉल नहीं "," आरएस 4843075 ": "एजी", "rs156697": "सीसी", "rs11096957": "एसी", "rs1952161": "जीजी", "rs1961416": "एजी", "rs1585676": "जीजी", "rs890910": "टीटी "," Rs171953 ":" एजी "," आरएस 1843026 ":" सीसी "," आरएस1515002 ":" सीसी "," आरएस 7564 9 7 ":" कोई कॉल नहीं "," आरएस 12 9 3153 ":" कॉल नहीं "," आरएस 754257 ":" जीटी "" rs649058 ":" एजी "," rs726957 ":" एजी "," rs728189 " : "सीसी", "आरएस 2216629": "एजी", "आरएस1541836": "सीटी", "आरएस 722 9 52": "टीसी", "आरएस 3742207": "कोई कॉल नहीं", "आरएस 210310" : "सीटी", "आरएस 1105176": "जीजी"} '];

धन्यवाद

आपको शायद इसका उपयोग करना चाहिए: 

  मेरी $ json_geno1 = decode_json ($ VAR1 [0]);   
 क्योंकि  $ var1  वस्तु अब एक सरणी।  
 यह मान्य JSON है (मैं PHP में यह परीक्षण किया है) और मैं इस स्ट्रिंग से कोई समस्या नहीं है।

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