Populate select menu from PHP/MySQL using javascript array -

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मेरे पास यह कोड है: 

  & lt; script & gt; Var मान = & lt;? Php $ sql = "SELECT * बिलिंग_साज़नामिनलकोड आदेश से कोड एएससी"; $ Rs = mysql_query ($ sql, $ conn) या मर (mysql_error ()); $ नाममात्रिकोड = सरणी (); जबकि ($ परिणाम = mysql_fetch_assoc ($ rs)) {$ नाममात्रकोड [] = $ परिणाम ['कोड']; } गूंज json_encode ($ नाममात्रकोड); ? & Gt ;; & Lt; / स्क्रिप्ट & gt; & Lt; चयन करें नाम = "sagenominalcode" id = "sagenominalcode" & gt; & Lt; script & gt; addOptions (document.getElementById ('sagenominalcode'), मान); & lt; / script & gt; & Lt; / चयन & gt;   
 यह वास्तव में कैसे काम करता है मुझे इसकी ज़रूरत है, लेकिन मैं सोच रहा हूं कि क्या चयन के साथ ही कोड में कॉलम के नाम "नाम" को प्रदर्शित करना संभव है?  
 123  
 123  
 123  
 123 - नाम   
 > 
 
 आप अपने php और अपने जेएस को सख्ती से अलग करना चाहते हैं जिस तरह से कोड लिखा जाता है, वह काम करेगा, लेकिन यह  बहुत ही  भंगुर है, और भयावहता को तोड़ सकता है, जिससे शैली की बग्स हो सकती है। उदाहरण के लिए, यदि आपका डेटाबेस स्कीमा पुन:  कोड  कॉलम को हटा दें, फिर आपका जावास्क्रिप्ट टूट जाएगा, जिसके कारण किसी को भी संदेह नहीं होगा।  
 
 उस ने कहा, अगर आप इसे भी प्रदर्शित करना चाहते हैं चुनिंदा विकल्पों में  नाम  कॉलम में डेटा, आपको उस डेटा को  $ नाममात्रकोड  में भी पास करना होगा अभी, इसमें एक अनुक्रमित सरणी डेटा (शायद एक  int  या  स्ट्रिंग ) होता है, लेकिन आप उसे बदलना चाहेंगे, ताकि इसमें एक इंडेक्टेड सरणी की सरणी हो।  
 तो, पहले:  
  $ nominalcodes = array (1,3,5,78,3,2,234);   
 और बाद में:  
  $ nominalcodes = array (array (1, 'foobar'), सरणी (3, 'लारी'), सरणी (5 , 'बॉब'), ...);   
 इस बिंदु पर, आपको अपने जावास्क्रिप्ट  addOptions  फ़ंक्शन को पुनरिक्षण करना होगा ताकि वह एकल मानों की एक सरणी के बजाय सरणी की सरणी को ठीक से पार्स कर सके ।

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