php - opening a div when a checkbox is checked and no of checkboxes are coming from database -

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  http://www.tutorialrepublic.com/codelab.php?topic=faq&file=jquery-show -वे-डिवेल-चेक-चेकबॉक्स   
 यहां ऊपर दिए गए लिंक में ... एक ऐसा कोड है जहां आप चेकबॉक्स पर क्लिक करते हैं, एक डिव खोला जाता है ... लेकिन चेकबॉक्स होते हैं स्थिर।  
 अब मेरा कोई चेकबॉक्स इस तरह से डेटाबेस से आ रहा है:  
  & lt; php $ sql_activities = "tb_activities से * चुनें '; $ Query_activities = mysql_query ($ sql_activities); $ सक्रिय = 1; जबकि ($ row_activities = mysql_fetch_array ($ query_activities)) {? & Gt; & Lt; इनपुट प्रकार = "चेकबॉक्स" मान = "& lt;? Php echo $ row_activities ['activity_id'];? & Gt;" नाम = "गतिविधियां & lt;? Php $ सक्रिय सक्रिय करें;? & Gt;" & Gt; & lt;? Php echo ""। $ Row_activities ["activity_name"]; ? & Gt; & Lt;? Php $ सक्रिय = $ सक्रिय + 1; }? & Gt;   
 अब जो मैं चाहता हूँ वह है कि अगर पांच चेकबॉक्स डेटाबेस से आ रहे हैं तो पांच div को खोला जाना चाहिए क्योंकि हम प्रत्येक डिवेल पर क्लिक करते हैं .... मैं यह कैसे प्राप्त कर सकता हूं?   
 
 
  जैसा कि आप नीचे एक div  खोलने का तरीका नहीं समझते हैं I परीक्षण किया है तो इस कोड का उपयोग करें और अपना काम करें।):  
  $ sql_activities = "tb_activities से * चुनें '; $ Query_activities = mysql_query ($ sql_activities); $ सक्रिय = 1; जबकि ($ row_activities = mysql_fetch_array ($ query_activities)) {? & Gt; & Lt; इनपुट प्रकार = "चेकबॉक्स" मान = "& lt;? Php echo $ row_activities ['activity_id'];? & Gt;" नाम = "गतिविधियां & lt;? Php $ सक्रिय सक्रिय करें;? & Gt;" वर्ग = "" & gt; & lt;? Php echo ""। $ Row_activities ["activity_name"]; ? & Gt; & Lt;? Php $ सक्रिय = $ सक्रिय + 1; } $ Sql_activities1 = "tb_activities से * चुनें '; $ Query_activities1 = mysql_query ($ sql_activities1); $ Active1 = 1; जबकि ($ row_activities1 = mysql_fetch_array ($ query_activities1)) {? & Gt; & Lt; div class = "& lt;? Php echo $ row_activities1 ['activity_id'];; & gt; बॉक्स" & gt; आपने & lt; मजबूत & gt; & lt; php; php echo $ row_activities1 ['activity_name']; ? & Gt; चेकबॉक्स & lt; / strong & gt; इसलिए मैं यहां हूं & lt; / div & gt; & Lt;? Php $ active1 = $ active1 + 1; }? & Gt; & Lt; स्क्रिप्ट प्रकार = "टेक्स्ट / जावास्क्रिप्ट" src = "http://code.jquery.com/jquery.min.js" & gt; & lt; / स्क्रिप्ट & gt; & Lt; स्क्रिप्ट प्रकार = "टेक्स्ट / जावास्क्रिप्ट" & gt; $ (Document) .ready (function () {$ ('input [type = "checkbox"]')। क्लिक करें (function () {// if ($ (this) .attr ("value") == "red ") {A = $ (this) .attr (" value "); $ ("। "+ A) .toggle ();});}); & Lt; / स्क्रिप्ट & gt; & Lt; शैली प्रकार = "टेक्स्ट / सीएसएस" & gt; .बॉक्स {पैडिंग: 20px; कुछ भी डिस्प्ले मत करो; मार्जिन-शीर्ष: 20px; सीमा: 1px ठोस # 000; } & Lt; / style & gt;   
  mysql एक्सटेंशन बहिष्कृत किया गया है

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