JQuery Datepicker is not working when appending new row - JQuery UI










0















I have a datepicker in a table. the <tr> appends when button clicked.



It's work fine when populate 1st <tr> but when 2nd <tr> created on button click, datepicker is not working.



Check My Fiddle










share|improve this question


























    0















    I have a datepicker in a table. the <tr> appends when button clicked.



    It's work fine when populate 1st <tr> but when 2nd <tr> created on button click, datepicker is not working.



    Check My Fiddle










    share|improve this question
























      0












      0








      0








      I have a datepicker in a table. the <tr> appends when button clicked.



      It's work fine when populate 1st <tr> but when 2nd <tr> created on button click, datepicker is not working.



      Check My Fiddle










      share|improve this question














      I have a datepicker in a table. the <tr> appends when button clicked.



      It's work fine when populate 1st <tr> but when 2nd <tr> created on button click, datepicker is not working.



      Check My Fiddle







      jquery jquery-ui jquery-ui-datepicker






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 15 '18 at 12:13









      HuzaifaHuzaifa

      347




      347






















          1 Answer
          1






          active

          oldest

          votes


















          2














          This code should do the trick. You need to remove the id from the date input. Since, datepicker uses the id internally and this must be unique. Also, in case you still need ids on each field. I added a counter, so that you could create unique ids for each row.



          // Variable that we will be using to create unique ids.
          var counter = 0;

          $('#addnew').click(function ()

          //Creating <tr> for cheque details
          var tr = $("<tr><td class='banks'><input type="text" class="form-control Product" name="Product"+ counter +"" id="Product"+ counter +"" placeholder="Product"></td>"
          + "<td><div>"
          + "<div class="input-group date">"
          + "<div class="input-group-addon"><i class="fa fa-calendar"></i>"
          + "</div>"
          + "<input type="text" class="form-control Date" name="ChqDate" id="Datet"+ counter +"" placeholder="Date">"
          + "</div><!-- /.input group -->"
          + "</div></td>"
          + "<td><input type="text" name="Price" class="form-control" id="Pricet"+ counter +"" placeholder="Price" /></td>"
          + "<td><button data-itemId="0" type="button" class="btn btn-danger removeRow">remove</button></td></tr>");

          //append <tr> into table <tbody>

          $('#example1 tbody').append(tr);
          $(tr).find('.Date').datepicker(
          autoclose: true
          );

          counter++;

          );


          Hope it works.






          share|improve this answer

























          • Thanks brother it works fine. :-)

            – Huzaifa
            Nov 15 '18 at 13:01












          • I am glad I was of help! :)

            – Alain Cruz
            Nov 15 '18 at 13:02










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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          This code should do the trick. You need to remove the id from the date input. Since, datepicker uses the id internally and this must be unique. Also, in case you still need ids on each field. I added a counter, so that you could create unique ids for each row.



          // Variable that we will be using to create unique ids.
          var counter = 0;

          $('#addnew').click(function ()

          //Creating <tr> for cheque details
          var tr = $("<tr><td class='banks'><input type="text" class="form-control Product" name="Product"+ counter +"" id="Product"+ counter +"" placeholder="Product"></td>"
          + "<td><div>"
          + "<div class="input-group date">"
          + "<div class="input-group-addon"><i class="fa fa-calendar"></i>"
          + "</div>"
          + "<input type="text" class="form-control Date" name="ChqDate" id="Datet"+ counter +"" placeholder="Date">"
          + "</div><!-- /.input group -->"
          + "</div></td>"
          + "<td><input type="text" name="Price" class="form-control" id="Pricet"+ counter +"" placeholder="Price" /></td>"
          + "<td><button data-itemId="0" type="button" class="btn btn-danger removeRow">remove</button></td></tr>");

          //append <tr> into table <tbody>

          $('#example1 tbody').append(tr);
          $(tr).find('.Date').datepicker(
          autoclose: true
          );

          counter++;

          );


          Hope it works.






          share|improve this answer

























          • Thanks brother it works fine. :-)

            – Huzaifa
            Nov 15 '18 at 13:01












          • I am glad I was of help! :)

            – Alain Cruz
            Nov 15 '18 at 13:02















          2














          This code should do the trick. You need to remove the id from the date input. Since, datepicker uses the id internally and this must be unique. Also, in case you still need ids on each field. I added a counter, so that you could create unique ids for each row.



          // Variable that we will be using to create unique ids.
          var counter = 0;

          $('#addnew').click(function ()

          //Creating <tr> for cheque details
          var tr = $("<tr><td class='banks'><input type="text" class="form-control Product" name="Product"+ counter +"" id="Product"+ counter +"" placeholder="Product"></td>"
          + "<td><div>"
          + "<div class="input-group date">"
          + "<div class="input-group-addon"><i class="fa fa-calendar"></i>"
          + "</div>"
          + "<input type="text" class="form-control Date" name="ChqDate" id="Datet"+ counter +"" placeholder="Date">"
          + "</div><!-- /.input group -->"
          + "</div></td>"
          + "<td><input type="text" name="Price" class="form-control" id="Pricet"+ counter +"" placeholder="Price" /></td>"
          + "<td><button data-itemId="0" type="button" class="btn btn-danger removeRow">remove</button></td></tr>");

          //append <tr> into table <tbody>

          $('#example1 tbody').append(tr);
          $(tr).find('.Date').datepicker(
          autoclose: true
          );

          counter++;

          );


          Hope it works.






          share|improve this answer

























          • Thanks brother it works fine. :-)

            – Huzaifa
            Nov 15 '18 at 13:01












          • I am glad I was of help! :)

            – Alain Cruz
            Nov 15 '18 at 13:02













          2












          2








          2







          This code should do the trick. You need to remove the id from the date input. Since, datepicker uses the id internally and this must be unique. Also, in case you still need ids on each field. I added a counter, so that you could create unique ids for each row.



          // Variable that we will be using to create unique ids.
          var counter = 0;

          $('#addnew').click(function ()

          //Creating <tr> for cheque details
          var tr = $("<tr><td class='banks'><input type="text" class="form-control Product" name="Product"+ counter +"" id="Product"+ counter +"" placeholder="Product"></td>"
          + "<td><div>"
          + "<div class="input-group date">"
          + "<div class="input-group-addon"><i class="fa fa-calendar"></i>"
          + "</div>"
          + "<input type="text" class="form-control Date" name="ChqDate" id="Datet"+ counter +"" placeholder="Date">"
          + "</div><!-- /.input group -->"
          + "</div></td>"
          + "<td><input type="text" name="Price" class="form-control" id="Pricet"+ counter +"" placeholder="Price" /></td>"
          + "<td><button data-itemId="0" type="button" class="btn btn-danger removeRow">remove</button></td></tr>");

          //append <tr> into table <tbody>

          $('#example1 tbody').append(tr);
          $(tr).find('.Date').datepicker(
          autoclose: true
          );

          counter++;

          );


          Hope it works.






          share|improve this answer















          This code should do the trick. You need to remove the id from the date input. Since, datepicker uses the id internally and this must be unique. Also, in case you still need ids on each field. I added a counter, so that you could create unique ids for each row.



          // Variable that we will be using to create unique ids.
          var counter = 0;

          $('#addnew').click(function ()

          //Creating <tr> for cheque details
          var tr = $("<tr><td class='banks'><input type="text" class="form-control Product" name="Product"+ counter +"" id="Product"+ counter +"" placeholder="Product"></td>"
          + "<td><div>"
          + "<div class="input-group date">"
          + "<div class="input-group-addon"><i class="fa fa-calendar"></i>"
          + "</div>"
          + "<input type="text" class="form-control Date" name="ChqDate" id="Datet"+ counter +"" placeholder="Date">"
          + "</div><!-- /.input group -->"
          + "</div></td>"
          + "<td><input type="text" name="Price" class="form-control" id="Pricet"+ counter +"" placeholder="Price" /></td>"
          + "<td><button data-itemId="0" type="button" class="btn btn-danger removeRow">remove</button></td></tr>");

          //append <tr> into table <tbody>

          $('#example1 tbody').append(tr);
          $(tr).find('.Date').datepicker(
          autoclose: true
          );

          counter++;

          );


          Hope it works.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 15 '18 at 12:33

























          answered Nov 15 '18 at 12:22









          Alain CruzAlain Cruz

          2,1003920




          2,1003920












          • Thanks brother it works fine. :-)

            – Huzaifa
            Nov 15 '18 at 13:01












          • I am glad I was of help! :)

            – Alain Cruz
            Nov 15 '18 at 13:02

















          • Thanks brother it works fine. :-)

            – Huzaifa
            Nov 15 '18 at 13:01












          • I am glad I was of help! :)

            – Alain Cruz
            Nov 15 '18 at 13:02
















          Thanks brother it works fine. :-)

          – Huzaifa
          Nov 15 '18 at 13:01






          Thanks brother it works fine. :-)

          – Huzaifa
          Nov 15 '18 at 13:01














          I am glad I was of help! :)

          – Alain Cruz
          Nov 15 '18 at 13:02





          I am glad I was of help! :)

          – Alain Cruz
          Nov 15 '18 at 13:02



















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