调用服务器上的数据,而不使用在jqGrid的GET方法 [英] Call data from the server without using GET Method on jqgrid
问题描述
我是新来jqGrid的。我做了很多关于如何调用从数据库中的数据的研究,而无需使用GET方法都没有成功。
我有以下HTML code:
<!DOCTYPE HTML>
< HTML>
< HEAD>
<元字符集=utf-8>
<冠军> jqGrid的UI< /标题>
<链接相对=样式类型=文本/ CSSHREF =HTTP://$c$c.jquery.com/ui/1.10.3/themes/redmond/jquery-ui.css/>
<链接相对=样式类型=文本/ CSSHREF =HTTP://www.trirand.com/blog/jqgrid/themes/ui.jqgrid.css/>
&所述;脚本的src =的http://$c$c.jquery.com/jquery-1.10.1.min.js>&所述; /脚本>
<脚本类型=文/ JavaScript的SRC =HTTP://www.trirand.com/blog/jqgrid/js/jquery-ui-custom.min.js'>< / SCRIPT>
<脚本类型=文/ JavaScript的SRC =HTTP://www.trirand.com/blog/jqgrid/js/i18n/grid.locale-en.js'>< / SCRIPT>
<脚本类型=文/ JavaScript的SRC =HTTP://www.trirand.com/blog/jqgrid/js/jquery.jqGrid.js'>< / SCRIPT>
<脚本>
$(文件)。就绪(函数(){
$(#list_records)。jqGrid的({
网址:getGridData.php,//我想改变这是同一个页面,从数据库中调用数据的,我必须使用的数据:varname的?
数据类型:JSON,
MTYPE:GET,//我应该改变这种数据类型:本地,如果我想改变的网址是什么?
colNames:用户ID,用户名,名,姓],
colModel:
{名字:因维,对齐:右},
{名字:invdate},
{名字:量},
{名字:税}
]
寻呼机:#perpage
的rowNum:10,
rowList:[10,20],
sortname:因维,
排序顺序:ASC
高度:自动,
viewrecords:真正的,
gridview的:真正的,
标题:
});
});
< / SCRIPT>
< /头>
<身体GT;
<表ID =list_records>< TR>< TD>< / TD>< / TR>< /表>
< DIV ID =perpage>< / DIV>
< /身体GT;
< / HTML>
和我有下面的PHP MySQL的code:
< PHP
$康恩=的mysql_connect(localhost的根,)或死亡(连接错误:mysql_error());
mysql_select_db(jqGrid的)或死亡(错误连接到数据库。);
$页= $ _GET ['页'];
$上限= $ _GET ['行'];
$ SIDX = $ _GET ['SIDX'];
$ SORD = $ _GET [SORD'];
如果(!$ SIDX)$ SIDX = 1;
$结果= mysql_query(SELECT COUNT(*)作为计数从invheader);
$行= mysql_fetch_array($结果,MYSQL_ASSOC);
$数= $行['数'];
如果($计数大于0和放大器;&安培; $限制大于0){
$ TOTAL_PAGES = CEIL($计数/ $限制);
} 其他 {
$ TOTAL_PAGES = 0;
}
如果($页> $ TOTAL_PAGES)$页= $ TOTAL_PAGES;
$启动= $限制* $页 - $限制;
如果($开始℃,)$开始= 0;
$ SQL =SELECT * FROM invheader ORDER BY $ SIDX $ SORD LIMIT $开始,$限制;
$结果= mysql_query($的SQL)或死亡(无法执行查询。mysql_error());
$ i = 0;
而($行= mysql_fetch_array($结果,MYSQL_ASSOC)){
$ responce->行[$ i] ['身份证'] = $行['因维'];
$ responce->行[$ i] ['细胞'] =阵列($行['invdate'],$行['量'],$行['税'],$行['总']) ;
$ I ++;
}
回声json_en code($性反应);
?>
这是完全正常的。不过,我不希望使用GetMethod的。我想把这一切在一个PHP页面。能有人给我我如何能做到这一点preferably使用数组的样品code。在此先感谢!
更新
我现在已经设法找到了关于我如何PHP的查询结果转换为JavaScript的按照此处给出的指令的 http://www.dyn-web.com/tutorials/php-js/json/array.php
我现在的问题是,我无法使用转换变量到jqGrid的。我想有相同的结果,它是此网页。 http://trirand.com/blog上/jqgrid/jqgrid.html#t95 ,唯一不同的是我用,而不是建立一个静态的数据转换PHP成果转化javascropt。
下面是我的脚本code:
VAR MYDATA =<?PHP的回声json_en code($ dataitems)取代;
jQuery的(#list4)。jqGrid的({
数据:MYDATA,
数据类型:本地,
高度:250,
colNames:['INV无,日期,客户,金额,税法,合计,注意]
colModel:
{名字:'身份证',索引:'身份证',宽度:60,sorttype:INT},
{名字:invdate,索引:invdate',宽度:90,sorttype:日期},
{名称:名称,索引:'名',宽度:100},
{名称:'量',索引:'量',宽度:80,对齐:正确的,sorttype:浮动},
{名称:'税',索引:'税',宽度:80,对齐:正确的,sorttype:浮动},
{名称:'总',索引:'总',宽度:80,对齐:正确的,sorttype:浮动},
{名字:'注意',索引:'注意',宽度:150,排序:假}
]
多选:真正的,
标题:操控阵列数据
});
虽然这不是非常有效,你可以做这样的事情
$数据=阵列();
而($行= mysql_fetch_assoc($结果))$数据[] = $行;
回声'<脚本> VAR数据='。 json_en code($的数据)。 ';< / SCRIPT>';
在这一点上,你将有一个包含所有数据的JSON对象。这将使一个较长的页面加载,但会做你的要求没有Ajax
而我在这, mysql_职能是去precated 。 Copnsider切换到的mysqli _
I'm new to jqgrid. I did alot of research on how to call the data from the database without using GET method without success.
I have below HTML code:
<!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>
And I have the php mysql code below:
<?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);
?>
This is fully working. However, I don't want to use the getmethod. I want to put it all in a php page. Can someone give me a sample code on how can I achieve this preferably using an array. Thanks in advance!
UPDATE
I have now managed to find out on how I can convert php query results to javascript by following the instruction given here http://www.dyn-web.com/tutorials/php-js/json/array.php
My problem now is I'm unable to use the converted variable into jqgrid. I wanted to have the same result that is on this page http://trirand.com/blog/jqgrid/jqgrid.html#t95, only difference is I used converted php results into javascropt instead of creating a static data.
Below is my script code:
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"
});
While this isn't terribly efficient you can do something like this
$data = array();
while($row = mysql_fetch_assoc($result)) $data[] = $row;
echo '<script>var data =' . json_encode($data) . ';</script>';
At this point you will have a JSON object that contains all your data. This will make for a longer page load but will do what you asked without AJAX
While I'm at it, mysql_ functions are deprecated. Copnsider switching to mysqli_
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