javascript - tablesorter pager plugin - show all entries as default -


i'm using tablesorter version: http://tablesorter.com/docs/

having relevant html:

 <select class="pagesize form-control text-center">         <option value="20">20</option>         <option value="30">30</option>         <option value="40">40</option>         <option value="all" selected="selected">alle</option>  </select>

and js:

function tablesorter() {     $("#tableoverview")     .tablesorter()     .tablesorterpager({container: $("#pager")});    var rows = $('table.tablesorter')[0].config.totalrows;   $('select.pagesize').find('option:contains("all")').val(rows); }

now want displayed items of table default / on loading. if not default , have choose after loading page, works fine this: 

var rows = $('table.tablesorter')[0].config.totalrows; $('select.pagesize').find('option:contains("all")').val(rows);

setting size in defaults in jquery.tablesorter.pager.js high value works too, causes error if want sort table (nothing shown then).

so there way set default size of paging entries of table?

all need different set value big number 9999 in both selected option value , in pager size option (demo):

html

<select class="pagesize">     <option value="10">10</option>     <option value="20">20</option>     <option value="30">30</option>     <option value="40">40</option>     <option selected="selected" value="9999">all</option> </select>

javascript

$('table')     .tablesorter({         widthfixed: true,         widgets: ['zebra']     })     .tablesorterpager({         container: $("#pager"),         size: 9999 // pick number larger table     });

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