Oracle 데이터베이스에서 jqgrid를 사용하고 있습니다. 근본적인 문제는 데이터베이스 테이블에 정보를 표시하는 것뿐만 아니라 모든 것이 정확하다는 것입니다. 나를 도울 수있는 모든 코드를 남겨두고 있습니다. "총"2 :Oracle 데이터베이스가있는 jQGrid
{ "페이지"
book.php
<?php
//database setting
$dbuser='system';
$dbpassword='manager';
$database='orcl';
$page = 2;
// get the requested page
$limit = 10;
// connect to the database
$db = oci_connect($dbuser, $dbpassword, $database) or die("Connection Error: " . oci_error());
$result = oci_parse($db, 'SELECT COUNT(*) AS count from aaz');
oci_execute($result);
$row = oci_fetch_array ($result);
$count = $row[0];
if($count > 0)
{
$total_pages = ceil($count/$limit);
}
else {
$total_pages = 0;
}
if ($page > $total_pages) $page=$total_pages;
$start = $limit*$page - $limit; // do not put $limit*($page - 1)
$SQL = "SELECT * FROM aaz where rownum <=$limit ORDER BY id";
$result = oci_parse($db, $SQL) or die("Couldn t execute query.".oci_error());
oci_execute($result);
$responce->page = $page;
$responce->total = $total_pages;
$responce->records = $count;
$i=0;
while($row = oci_fetch_array($result,OCI_ASSOC))
{
$responce->rows[$i]['id']=$row[ID];
$responce->rows[$i]['cell']=array($row[ID],$row[FIRSTNAME],$row[LASTNAME],$row[PHONE],$row[EMAIL]);
$i++;
}
echo json_encode($responce);
?>
index.php를
<html>
<head>
<title>jQGrid example</title>
<!-- Load CSS--><br />
<link rel="stylesheet" href="css/ui.jqgrid.css" type="text/css" media="all" />
<!-- For this theme, download your own from link above, and place it at css folder -->
<link rel="stylesheet" href="css/ui-lightness/jquery-ui-1.8.23.custom.css" type="text/css" media="all" />
<!-- Load Javascript -->
<script src="js/jquery-1.7.2.min.js" type="text/javascript"></script>
<script src="js/jquery-ui-1.8.23.custom.min.js" type="text/javascript"></script>
<script src="js/i18n/grid.locale-pt.js" type="text/javascript"></script>
<script src="js/jquery.jqGrid.min.js" type="text/javascript"></script>
</head>
<body>
...
<table id="datagrid"></table>
<div id="navGrid"></div>
<p><script language="javascript">
jQuery("#datagrid").jqGrid({
url:'book.php',
datatype: 'json',
colNames:['ID','Nome', 'Apelido', 'Telefone','Email'],
colModel:[
{name:'ID',index:'ID', width:155,editable:false,editoptions:{readonly:true,size:10}},
{name:'FIRSTNAME',index:'FIRSTNAME', width:180,editable:true,editoptions:{size:10}},
{name:'LASTNAME',index:'LASTNAME', width:190,editable:true,editoptions:{size:25}},
{name:'PHONE',index:'PHONE', width:160, align:"right",editable:true,editoptions:{size:10}},
{name:'EMAIL',index:'EMAIL', width:160, align:"right",editable:true,editoptions:{size:10}}
],
rowNum:10,
rowList:[10,15,20,25,30,35,40],
pager: '#navGrid',
sortname: 'no',
sortorder: "asc",
height: 210,
viewrecords: true,
caption:"Example"
});
jQuery("#datagrid").jqGrid('navGrid','#navGrid',{edit:true,add:true,del:true});
</script>
</body>
</html>
내가 할 대답은 book.php이다 2 "레코드": "12", "행": [{ "id": "1", "셀": [ "1", "fname1", "lname1", "(000) 000-0000" [email protected] "]}, {"id ":"2 ","cell ": ["2 ","fname1 ","lname1 ","(000) 000-0000 ","[email protected] " "}}, {"id ":"3 ","셀 ":" "3", "fname1", "lname1", "(000) 000-0000", "[email protected]" "신분증" "4", "fname1", "lname1", "(000) 000-0000", "[email protected]"]}, { "id": "5", " "cell": [ "5", "fname1", "lname1", "(000) 000-0000", "[email protected]"]}, { "id": "6" "6", "fname1", "lname1", "(000) 000-0000", "[email protected]"]}, { "id": "7", "cell": [ "7" ("000"000-0000 ","[email protected] "]}, {"id ":"8 ","셀 ": ["8 ","fname1 ","lname1 " ","(000) 000-0000 ","[email protected] "]}, {"id ":"9 ","셀 ":" "9", "fname1", "lname1", "(000 ("000"000-0000 ","[email protected] "]}, {"id ":"10 ","cell ":"10 ","fname1 ","lname1 " , "[email protected]"]}]}
나를 도와 줄 사람이 있습니까? 나는, 행운을 돕기 위해 희망
SELECT * FROM
(
SELECT/* + FIRST_ROWS (n) */*, ROWNUM rnum FROM (our query)
WHERE ROWNUM <= limit-sup
)
WHERE rnum > = limit-inf;
을 :
Plz 도움이 필요합니다. – zmki