jqgrid編輯操作如何向服務器發送變量？

Question

我認爲，jqgrid作爲變量發送時，你編輯一個行，列名稱，但在我創建這個網格似乎不工作，除非我在php中犯了一個錯誤的地方。 jqgrid將哪些方法發送到服務器？

這是我的代碼。

的Index.html

<script type="text/javascript"> 

$(function(){ 

    var mygrid = $("#list"); 

    mygrid.jqGrid({ 
    url:'example1.php', 
    editurl: 'edit.php', 
    datatype: 'xml', 
    mtype: 'GET', 
    colNames:['Inv No','Date', 'Amount','Tax','Total','Notes'], 
    colModel :[ 
     {name:'invid', 
     index:'invid', 
     width:55 
     }, 
     {name:'invdate', 
     index:'invdate', 
     width:90, 
     editable:true 
     }, 
     {name:'amount', 
     index:'amount', 
     width:80, 
     align:'right', 
     search:true , 
     stype:'select', 
     searchoptions:{value:':All;8:8.00;6:6.00'}, 
     editable:true 
     }, 
     {name:'tax', 
     index:'tax', 
     width:80, 
     align:'right', 
     editable:true 
     }, 
     {name:'total', 
     index:'total', 
     //width:80, 
     align:'right', 
     sortable:true, 
     editable:true, 
     editrules:{required:true}, 
     edittype:'select', 
     editoptions:{value:"5:55.00;2:25.00"} 
     }, 
     {name:'note', 
     index:'note', 
     width:150, 
     search:true , 
     align:'center', 
     editable:true, 
     editrules:{required:true, edithidden:true}, 
     hidden:true, 
     edittype:'textarea', 
     editoptions: {rows:"12",cols:"10"} 
     } 
    ], 
    pager: '#pager', 
    emptyrecords: "Nothing to display", 
    recordtext: '{0} - {1} of {2}', 
    rowNum:9, 
    rowList:[7,9,11], 
    viewrecords: true, 
    caption: 'My first grid', 
    sortname: 'invid', 
    sortorder: 'desc', 
    loadonce : true 

    }); 
    //Search button 
    $("#bsdata").click(function(){ mygrid.jqGrid('searchGrid', {sopt:['eq'],top:300,caption:"test searching"}); }); 
    // Search toolbar. 
    mygrid.jqGrid('filterToolbar', {stringResult: true, searchOnEnter: false, defaultSearch : "eq"}); 
    //NavBar 
    mygrid.jqGrid('navGrid','#pager',{edit:false,add:false,del:false,search:true}); 

    // Edit Row 

    $("#bedata").click(function(){ 
     var gr = mygrid.jqGrid('getGridParam','selrow'); 
     if(gr != null) 
     mygrid.jqGrid('editGridRow',gr,{height:480,reloadAfterSubmit:false}); 
     else alert("Please Select Row");   
    }); 

    // Add row 

    $("#badata").click(function(){ 
     mygrid.jqGrid('editGridRow',"new",{height:480,reloadAfterSubmit:false}); 
    }); 

    // Delete row 

    $("#bddata").click(function(){ 
     var gr = mygrid.jqGrid('getGridParam','selrow'); 
     if(gr != null) 
      mygrid.jqGrid('delGridRow',gr,{reloadAfterSubmit:false}); 
     else 
      alert("Please Select Row to delete!"); 
    }); 

}); 


</script> 

和edit.php

<?php 

// connect to the database 
$dbhost = "localhost"; 
$dbuser = "user"; 
$dbpassword = "user123"; 
$database = "test"; 
$tablename = "invheader"; 
$db = mysql_connect($dbhost, $dbuser, $dbpassword) 
or die("Connection Error: " . mysql_error()); 

mysql_select_db($database) or die("Error conecting to db."); 
//mysql_set_charset('utf8',$database); 
mysql_query("SET NAMES 'utf8'"); 


if($_GET['oper']=='add') 
{ 

} 
if($_GET['oper']=='edit') 
{ 
    $id  = mysql_real_escape_string($_GET['id']); 
    $date = mysql_real_escape_string($_GET['invdate']); 
    $amount = mysql_real_escape_string($_GET['amount']); 
    $tax = mysql_real_escape_string($_GET['tax']); 
    $total = mysql_real_escape_string($_GET['total']); 
    $note = mysql_real_escape_string($_GET['note']); 

    $sql = "UPDATE ".$tablename." SET invdate = '".$date."', client_id = '5' , amount = '".$amount."', tax = '".$tax."', total = '".$total."' , "; 
    $sql.= "note = '".$note."'"; 
    $sql.= " WHERE invid = ".$id; 

    echo $sql; 
$result=mysql_query($sql) or die(mysql_error()); 

mysql_close($db); 


} 
if($_GET['oper']=='del') 
{ 

} 



?> 

感謝您的時間。

Answer 1

嗯，終於我使用了Tamper數據firefox的插件，並意識到從jqgrid發送的數據通過POST方法發送，而不是GET，因爲我在mtype：label中設置它。

因此，我將edit.php文件中的所有$ _GET更改爲$ _POST並工作。