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Updating jQuery Tablesorter plugin after removing a row from DOM

由 杀马特。学长 韩版系。学妹 提交于 2019-11-26 12:44:46
问题 I have some code at the moment that hides a row that is deleted and then removes it using the .remove() function. However I\'m having difficulty is making it remain \"deleted\" as every time I refresh the table sorted pager plugin or the filter plugin addon I\'m using.. the deleted rows re-appear as they are of course cached. Current code just simple with widget update at the moment $(\'.deleteMAP\').live(\"click\", function(){ $(this).closest(\'tr\').css(\'fast\', function() { $(this).remove

Unable to position pager (navigation bar) above jqGrid

由 為{幸葍}努か 提交于 2019-11-26 11:53:41
According the the jqGrid documentation, I should be able to place the pager above or below the jqGrid by moving the pager div. Unfortunately, the pager always renders below the grid. <div id="pager"></div> <table id="list"> <tr> <td /> </tr> </table> The jqGrid configuration (related to the pager) looks like: pager: '#pager', pginput: false, pgbuttons: false, Any suggestions? Oleg You should use toppager:true jqGrid option instead. You don't need define <div id="pager"></div> and use pager: '#pager' parameter. The id of the pager from the top of jqGrid will be "list_toppager" (id of the table

Get focused View from ViewPager

由 一世执手 提交于 2019-11-26 07:24:08
问题 i use the ViewPager for switching views with left/right swipe. The ViewPager needs an Adapter, so I\'ve built this one: public class ListViewPagerAdapter extends PagerAdapter { protected static final String TAG = \"ListViewPagerAdapter\"; protected static final int NUM_VIEWS = 3; protected final Activity mActivity; public ListViewPagerAdapter(Activity activity) { mActivity = activity; } @Override public int getCount() { return NUM_VIEWS; } @Override public void startUpdate(View container) {}

Unable to position pager (navigation bar) above jqGrid

由 北城余情 提交于 2019-11-26 02:38:50
问题 According the the jqGrid documentation, I should be able to place the pager above or below the jqGrid by moving the pager div. Unfortunately, the pager always renders below the grid. <div id=\"pager\"></div> <table id=\"list\"> <tr> <td /> </tr> </table> The jqGrid configuration (related to the pager) looks like: pager: \'#pager\', pginput: false, pgbuttons: false, Any suggestions? 回答1: You should use toppager:true jqGrid option instead. You don't need define <div id="pager"></div> and use

Replace Fragment inside a ViewPager

由 时光毁灭记忆、已成空白 提交于 2019-11-26 01:19:23
问题 I\'m trying to use Fragment with a ViewPager using the FragmentPagerAdapter . What I\'m looking for to achieve is to replace a fragment, positioned on the first page of the ViewPager , with another one. The pager is composed of two pages. The first one is the FirstPagerFragment , the second one is the SecondPagerFragment . Clicking on a button of the first page. I\'d like to replace the FirstPagerFragment with the NextFragment. There is my code below. public class FragmentPagerActivity
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