空網格返回到JQgrid 4.1.1與jquery 1.5.2

我有真正的困難獲取數據顯示在我的網格。查詢可以使用名字，姓氏或「SERIALNUMBER」空網格返回到JQgrid 4.1.1與jquery 1.5.2

的HTML看起來retricve數據，如

<script src="lib/jquery/jquery-1.5.2.min.js" type="text/javascript" ></script> 
    <script src="lib/jquery/jquery-ui.min.js" type="text/javascript"> </script> 
    <script src="lib/jqgrid/i18n/grid.locale-en.js" type="text/javascript"></script> 
    <script src="lib/jqgrid/jquery.jqGrid.min.js" type="text/javascript"></script> 
    <div id="search"> 
    <table id="list"></table> 
    <div id="pager" ></div> 
     </div>

數據庫和PHP都是當地現在和mysqli的查詢得到每次都有正確的數據。

的Javascript看起來像

 
$('#list').jqGrid({ 
url:'http://localhost/ajax-search.php', 
datatype: 'json', 
postData: { 
user: function() { return jQuery("input#username").val(); }, 
pass: function() { return jQuery("input#password").val(); }, 
firstname: function() { return jQuery("#firstname_label").val(); }, 
lastname: function() { return jQuery("#lastname_label").val(); }, 
number: function() { return jQuery("#number_label").val(); } 
}, 
mtype: 'POST', 
colNames:['id','First','Last', 'Number'], 
colModel:[ 
{name: 'id', index:'id',width:10,hidden:false}, 
{name: 'targetfirstname', index:'targetfirstname',width:25}, 
{name: 'targetlastname', index:'targetlastname',width:25}, 
{name: 'number',  index:'number', width:25}, 
], 
rowNum:10, 
width:340, 
height: 25, 
setGridHeight:45, 
shrinkToFit:true, 
rowList:[10,20,30], 
imgpath: 'lib/jquery/themes/base/images', 
pager: $('#pager'), 
sortname: 'id', 
viewrecords: true, 
sortorder: "desc", 
caption:"Edit records", 
});

的PHP/MySQL的（在準備使用statemenst）看起來像

 
$response = new stdClass(); 
$response->page =  $page;   // current page 
$response->total =  $total_pages; // total pages 
$response->records = $numrows;  // total records 
       $stmt2->execute() 
       || fail('MySQL execute', $db->error); 
       $stmt2 -> bind_result($seltargetfirstname,$seltargetlastname,$selnumber) 
       || fail('MySQL bind_result', $db->error); 

$i=0; 
while($row = $stmt2->fetch()) { 
    fwrite($fh2,"\n I is $i \n"); 
    $response->rows[$i]['id']=$i; 
    $response->rows[$i]['cell']=array($i,$seltargetfirstname,$seltargetlastname,$selnumber); 
    $i++; 
    fwrite($fh2, "\nTO ENCODER $i $seltargetfirstname $seltargetlastname $selnumber\n"); 
} 

echo json_encode($response);

我看到的是

 
[1] NO Apache/PHP/JSON encode or MYSQL errors 
[2] I can write out the returned MYSQLI fetch values to a file, all look good 
[3] I generate a row number as I dont use/need these in my schema 
[4] The grid generates in JQGRID and I can sort and see my query re-execute but NO data 
    in the grid at all..

知道的任何想法。 .....

來源

2011-07-03 ivan bishop

使用firebug的網絡控制檯（用於Firefox），同時你開始請求，看看你是否從服務器獲取JSON。告訴我們之後會發生什麼。

希望這會有所幫助。歡呼聲

來源

2011-07-03 23:20:03

空網格返回到JQgrid 4.1.1與jquery 1.5.2

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