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Author Topic: pass parameter via $ _POST  (Read 14653 times)
impperio
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Posts: 11


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« on: October 25, 2012, 10:39:40 AM »

Gentlemen need help have the value
of the selected row datagrid to call page PDF print.
How poss pass the parameter via $ _POST NOT $_GET
below example of functional code in $ _GET


function print () {
var row = $ ('# dg'). datagrid ('getSelected');
var minha_variavel = [window.open ('impress.php? cCodalu =' + row.codalub)]
};
I appreciate any help in advance
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Kevin
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« Reply #1 on: November 17, 2012, 06:51:36 PM »

I'm no expert, but you could try the following;

Code:
function print () {
   var row = $ ('# dg'). datagrid ('getSelected');
 
   $.post('impress.php?&cCodalu =' + row.codalub,{id:'dummy'},function(result){
      if (result.success){
         ...success processing...
      } else {
         ...failed processing...
      }
   },'json');
}
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impperio
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« Reply #2 on: November 21, 2012, 07:40:01 AM »

Thanks for the posts, I decided in a way that can serve to someone more below the codes



1- <a href="#" class="easyui-linkbutton" plain="true" iconCls="icon-print" plain="true" onClick="print_it()">Imprimir</a>   
 
2- function print_it(){
var row = $('#dg').datagrid('getSelected');
var oDados
oDados = document.getElementById('cSeqlot'); 
oDados.value = row.seqlotB;
$('#form-printer').submit();
}; 
 

3- <form action="impress.php" method="post" id="form-printer">
     <input type="hidden"  id="cSeqlot"  name="cSeqlot" value="cSeqlot"/> 
     </form>         
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