php - Javascript Function to populate a select menu from MySQL -

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मेरे पास यह PHP कोड है जो एक MySQL डाटाबेस से एक मेनू का उपयोग करता है ... 

  & Lt; चयन करें नाम = "इनपुट" id = "इनपुट" & gt; & Lt;? Php $ sql = "चयन करें" तालिका से; $ Rs = mysql_query ($ sql, $ conn) या मर (mysql_error ()); जबकि ($ result = mysql_fetch_array ($ rs)) {गूंज '& lt; विकल्प मान = "'। $ परिणाम [" db_field "]। '" & Gt;'। $ परिणाम ["db_field"]। '& Lt; / option & gt; '; }? & Gt; & Lt; / चयन & gt;   
 जो बिल्कुल ठीक काम करता है लेकिन मुझे किसी भी तरह इसे जावास्क्रिप्ट फ़ंक्शन में प्राप्त करना है।  
 मेरे पास जावास्क्रिप्ट कोड है, जब आप एक बटन क्लिक करते हैं तो यह अधिक टेक्स्ट बॉक्स जोड़ता है और एक अन्य चयन मेनू परन्तु यह किसी भी नए (जोड़ा गया) चयन मेनू में डेटाबेस से डेटा को पॉप्युलेट नहीं करता   "post-text" itemprop = "text"> < P> आप विकल्प को जेसनॉन में json_encode का उपयोग कर कनवर्ट कर सकते हैं (मैं एक PHP प्रोग्रामर नहीं हूं और इसका उपयोग करने के सटीक शब्दों का पता नहीं है)  
 PHP में ऐसा कुछ करना है:  
  इको '& lt; स्क्रिप्ट & gt; var विकल्प जेसन =' .json_encode (mysql_fetch_array ($ rs))। '& Lt; / script & gt;'   
 जावास्क्रिप्ट एन में कुछ करना है (मैं jquery का उपयोग कर रहा हूं):  
  var चुनें = $ ('select.classOfThisSelect'); Var विकल्प = JSON.parse (optionsJSON); के लिए (var i = 0; i & lt; options.length; i ++) $ ('विकल्प')। Attr ({value: options [i]}) .एंडेंड (विकल्प [i])। AppendTo (select);   
 optionsJSON JSON स्ट्रिंग होगा जो विश्व स्तर पर उपलब्ध होगा आप अपने जावास्क्रिप्ट फ़ंक्शन में इसे स्वतंत्र रूप से उपयोग कर सकते हैं  
 नोट: आपको उद्धरण चिह्नों के साथ json_encode को घेरना पड़ सकता है < / P>

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