javascript - jQuery Traversing -

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मेरे पृष्ठ में यह डिज़ाइन है। 

  & lt; tr id = "master" & gt; & Lt; td id = "row1" & gt; & LT; textbot & gt; & Lt; / टीडी & gt; & Lt; td id = "row2" & gt; & LT; linkbutton & gt; & Lt; / टीडी & gt; & Lt; / टीआर & gt;

मैं लिंकबटन की उसी टीआर पर क्लिक की क्लोनिंग कर रहा हूं और वर्तमान लिंक बटन छिपे होंगे। इसलिए एक नई पंक्ति बनाई जाएगी।

विचार करें कि मुझे तीन बार क्लोन है 

  & lt; tr id = "master" & gt; & Lt; td id = "row1" & gt; & LT; textbot & gt; & Lt; / टीडी & gt; & Lt; td id = "row2" & gt; & LT; linkbutton & gt; & Lt; / टीडी & gt; & Lt; / टीआर & gt; & Lt; tr id = "master" & gt; & Lt; td id = "row1" & gt; & LT; textbot & gt; & Lt; / टीडी & gt; & Lt; td id = "row2" & gt; & LT; linkbutton & gt; & Lt; / टीडी & gt; & Lt; / टीआर & gt; & Lt; tr id = "master" & gt; & Lt; td id = "row1" & gt; & LT; textbot & gt; & Lt; / टीडी & gt; & Lt; td id = "row2" & gt; & LT; linkbutton & gt; & Lt; / टीडी & gt; & Lt; / टीआर & gt;

अब मैं दूसरी पंक्ति के लिंक बटन को दिखाना चाहता हूं।

मैं 

  $ (this) .parents के साथ कोशिश कर रहा हूं ( '# मास्टर: आखिरी')। पिछला ()। बच्चों ('# पंक्ति 2)। दिखाएं ();

लेकिन इसका काम नहीं कर रहा है सहायता pls टीआईए।

आप ID वाले तत्वों को क्लोन नहीं कर सकते / नहीं कर सकते, क्योंकि आईडी एक दस्तावेज़ में अद्वितीय होना यद्यपि एक ब्राउज़र खुशी से आपको एक ही आईडी को दो बार दूँगा, जावास्क्रिप्ट और jQuery इस के लिए आपको माफ नहीं करते हैं और चीजों की अपेक्षा के अनुरूप काम नहीं करेगा। समूह तत्वों का उचित तरीका वर्गों से होता है।

इसलिए यदि आप अपना कोड इस पर स्विच करते हैं: 

  & lt; tr class = "master" & gt; & Lt; td वर्ग = "पंक्ति 1" & gt; & LT; textbot & gt; & Lt; / टीडी & gt; & Lt; td वर्ग = "पंक्ति 2" & gt; & LT; linkbutton & gt; & Lt; / टीडी & gt; & Lt; / टीआर & gt;

आपका चयनकर्ता इस तरह दिख सकता है: 

  $ ('tr.master')। Eq (1) .find ('td.row2') ।दिखाना();

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