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데이터가 그리드에 표시되는 데 어려움이 있습니다. 쿼리는 이름, 성 또는 '의 serialNumber'빈 그리드가 Jquery 1.5.2로 JQgrid 4.1.1로 반환되었습니다.
html로 보이는하여 데이터를 retricve 할 수 있습니다
<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>
데이터베이스와 PHP 지금은 LOCAL하고 MYSQLi 쿼리가 점점
같은 매번 정확한 데이터. PHP는/MYSQL (사용 준비업자가)
$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);
과 같은
$('#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",
});
처럼
자바 스크립트 내가 감사
[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..
어떤 아이디어입니다 참조 무엇 보인다. .....