我是jqgrid的新手。我做了很多关于如何在不使用GET方法的情况下从数据库中调用数据的研究，但没有成功。
我有以下 HTML 代码：
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>jqGrid UI</title>
<link rel='stylesheet' type='text/css' href='http://code.jquery.com/ui/1.10.3/themes/redmond/jquery-ui.css' />
<link rel='stylesheet' type='text/css' href='http://www.trirand.com/blog/jqgrid/themes/ui.jqgrid.css' />
<script src="http://code.jquery.com/jquery-1.10.1.min.js"></script>
<script type='text/javascript' src='http://www.trirand.com/blog/jqgrid/js/jquery-ui-custom.min.js'></script>
<script type='text/javascript' src='http://www.trirand.com/blog/jqgrid/js/i18n/grid.locale-en.js'></script>
<script type='text/javascript' src='http://www.trirand.com/blog/jqgrid/js/jquery.jqGrid.js'></script>
<script>
$(document).ready(function () {
$("#list_records").jqGrid({
url: "getGridData.php", // I want to change this to be on the same page to call the data from the database, do I have to use the data: varname?
datatype: "json",
mtype: "GET", // should I change this to data type: local if I want to change the url?
colNames: ["User Id", "User Name", "First Name", "Last Name"],
colModel: [
{ name: "invid",align:"right"},
{ name: "invdate"},
{ name: "amount"},
{ name: "tax"}
],
pager: "#perpage",
rowNum: 10,
rowList: [10,20],
sortname: "invid",
sortorder: "asc",
height: 'auto',
viewrecords: true,
gridview: true,
caption: ""
});
});
</script>
</head>
<body>
<table id="list_records"><tr><td></td></tr></table>
<div id="perpage"></div>
</body>
</html>
我在下面有php mysql代码：
<?php
$conn = mysql_connect("localhost", "root", "") or die("Connection Error: " . mysql_error());
mysql_select_db("jqgrid") or die("Error connecting to db.");
$page = $_GET['page'];
$limit = $_GET['rows'];
$sidx = $_GET['sidx'];
$sord = $_GET['sord'];
if(!$sidx) $sidx =1;
$result = mysql_query("SELECT COUNT(*) AS count FROM invheader");
$row = mysql_fetch_array($result,MYSQL_ASSOC);
$count = $row['count'];
if( $count > 0 && $limit > 0) {
$total_pages = ceil($count/$limit);
} else {
$total_pages = 0;
}
if ($page > $total_pages) $page=$total_pages;
$start = $limit*$page - $limit;
if($start <0) $start = 0;
$SQL = "SELECT * FROM invheader ORDER BY $sidx $sord LIMIT $start , $limit";
$result = mysql_query( $SQL ) or die("Couldn't execute query.".mysql_error());
$i=0;
while($row = mysql_fetch_array($result,MYSQL_ASSOC)) {
$responce->rows[$i]['id']=$row['invid'];
$responce->rows[$i]['cell']=array($row['invdate'],$row['amount'],$row['tax'],$row['total']);
$i++;
}
echo json_encode($responce);
?>
这完全有效。但是，我不想使用 getmethod。我想把它全部放在一个php页面中。有人可以给我一个示例代码，说明我最好使用数组如何实现这一目标。提前感谢！
更新
我现在已经设法了解如何按照此处给出的说明将 php 查询结果转换为 javascript http://www.dyn-web.com/tutorials/php-js/json/array.php
我现在的问题是我无法将转换后的变量使用 jqgrid。我想得到与 http://trirand.com/blog/jqgrid/jqgrid.html#t95 页面上相同的结果，唯一的区别是我使用转换后的 php 结果到 javascropt 而不是创建静态数据。
下面是我的脚本代码：
var mydata = <?php echo json_encode($dataitems)?>;
jQuery("#list4").jqGrid({
data: mydata,
datatype: "local",
height: 250,
colNames:['Inv No','Date', 'Client', 'Amount','Tax','Total','Notes'],
colModel:[
{name:'id',index:'id', width:60, sorttype:"int"},
{name:'invdate',index:'invdate', width:90, sorttype:"date"},
{name:'name',index:'name', width:100},
{name:'amount',index:'amount', width:80, align:"right",sorttype:"float"},
{name:'tax',index:'tax', width:80, align:"right",sorttype:"float"},
{name:'total',index:'total', width:80,align:"right",sorttype:"float"},
{name:'note',index:'note', width:150, sortable:false}
],
multiselect: true,
caption: "Manipulating Array Data"
});
虽然这不是非常有效，但你可以做这样的事情
$data = array();
while($row = mysql_fetch_assoc($result)) $data[] = $row;
echo '<script>var data =' . json_encode($data) . ';</script>';
此时，您将拥有一个包含所有数据的 JSON 对象。这将使页面加载时间更长，但可以在没有 AJAX 的情况下执行您要求的操作
当我使用它时，mysql_函数已被弃用。科普西德改用mysqli_