i have php script render 3 form elements each item in database. example, output this:
<!-- entry 1 --> <div data-role="fieldcontain"> <label for="1-name"></label> <input type="text" name="1-note" id="1-note" data-mini="true" /> </div> <div data-role="fieldcontain"> <label for="1-age"></label> <input type="text" name="1-note" id="1-note" data-mini="true" /> </div> <div data-role="fieldcontain"> <label for="1-description"></label> <input type="text" name="1-note" id="1-note" data-mini="true" /> </div> ... <!-- entry n --> <div data-role="fieldcontain"> <label for="n-name"></label> <input type="text" name="n-note" id="n-note" data-mini="true" /> </div> <div data-role="fieldcontain"> <label for="n-age"></label> <input type="text" name="n-note" id="n-note" data-mini="true" /> </div> <div data-role="fieldcontain"> <label for="n-description"></label> <input type="text" name="n-note" id="n-note" data-mini="true" /> </div>
then on main page, use ajax add php-generated content page...
$(document).on('pageshow', function(){ $("#menu1").bind("change", function() { $.ajax({ type: "get", url: "includes/ajax_get_entries.php", data: "table="+$("#menu1").val(), success: function(html) { $("#wrap").html(html).selectmenu('refresh', true); } }); }); });
... putting php content in between wrapper:
<div data-role="fieldcontain" id="wrap"></div>
adding page works fine, jquery mobile not update style.
how can make update style?
i cannot use add content site , hide if using .show() , .hide(), need response user on table need use.
you can call
$("#wrap").trigger('create');
in success callback.
also see question jquery mobile: markup enhancement of dynamically added content
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