javascript - How to get relative position of a draggable item when dropped? -

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मेरे पास निम्न है: 

  $ (function () {// चित्र खींचें $ ("। Item")। Draggable ({// स्थान हटा दिया गया है जहां स्थिति का पता लगाएं। स्टॉप: फ़ंक्शन (ईवेंट, यूआई) {// शो गिरा दिया गया स्थान। Var Stoppos = $ (this) .position (); $ ( "Div # stop")। पाठ ("रोकें: \ n लेफ्ट:" + Stoppos.left + "\ nTop:" + Stoppos.top);}}})}};   
 जब मैं इसे ड्रॉप करता हूं, तब आइटम की स्थिति प्राप्त होगी, लेकिन यह ब्राउज़र विंडो के सापेक्ष है, न कि वे उस डिवाइन के भीतर हैं जो वे अंदर हैं। मैं उनकी रिश्तेदार स्थिति कैसे प्राप्त कर सकता हूं?  
 संपादन- & gt; यहां html / css है:  
  & lt; शैली प्रकार = "टेक्स्ट / सीएसएस" & gt; कंटेनर {मार्जिन-टॉप: 50 पीएक्स; कर्सर: ले जाने के; } #screen {अतिप्रवाह: छिपा हुआ; चौड़ाई: 500px; ऊँचाई: 500px; दोनों को साफ करो; सीमा: 1px ठोस काला; } & Lt; / style & gt; & LT; बॉडी & gt; & Lt; div वर्ग = "कंटेनर" & gt; & Lt; div id = "स्क्रीन" & gt; & Lt; img id = "productid_1" src = "चित्र / कीपैड लिंक OFF.jpg" वर्ग = "आइटम" alt = "" title = "" / & gt; & Lt; img id = "productid_2" src = "चित्र / कीपैड लिंक्ड ON.jpg" class = "item" alt = "" title = "" / & gt; & Lt; / div & gt; & Lt; / div & gt; & Lt; div id = "stop" & gt; छवि की कमी हो रही है ... & lt; / div & gt; & Lt; / body & gt;    
 
 
 माता-पिता div  स्थिति: रिश्तेदार  दें।  
  #screen {स्थान: रिश्तेदार; ...}

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