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我是新來使用jqGrid,並且在嘗試使用內聯編輯功能時出現問題。這裏是我的代碼: 我的第一個網格jqGrid - 內聯編輯無法在編輯後將結果保存到數據庫
<link rel="stylesheet" type="text/css" media="screen" href="css/ui-lightness/jquery-ui-1.8.16.custom.css" />
<link rel="stylesheet" type="text/css" media="screen" href="css/ui.jqgrid.css" />
<style type="text/css">
html, body {
margin: 0;
padding: 0;
font-size: 75%;
}
</style>
<script src="js/jquery-1.5.2.min.js" type="text/javascript"></script>
<script src="js/i18n/grid.locale-en.js" type="text/javascript"></script>
<script src="js/jquery.jqGrid.min.js" type="text/javascript"></script>
<script type="text/javascript">
$(function(){
var lastsel2
$("#list").jqGrid({
url:'example.php',
datatype: 'xml',
mtype: 'GET',
colNames:['id','name', 'status'],
colModel :[
{name:'id', index:'id', width:55},
{name:'name', index:'name', width:90, editable: true},
{name:'status', index:'status', width:80, align:'right', editable: true},
],
onSelectRow: function(id){
if(appid && appid!==lastsel2){
jQuery('#list').restoreRow(lastsel2);
jQuery('#list').editRow(id,true);
lastsel2=id;
}
},
editurl: "example.php",
pager: '#pager',
rowNum:10,
rowList:[10,20,30],
sortname: 'appid',
sortorder: 'desc',
viewrecords: true,
gridview: true,
caption: 'My first grid'
});
});
</script>
</head>
<body>
<table id="list"><tr><td/></tr></table>
<div id="pager"></div>
</body>
問題是,它可以編輯該行,當我點擊它,也可以保存後暫時我按回車。但是當我重新加載網格時,它仍然在我之前顯示之前的數據,這意味着它不會更新數據庫。我想知道爲什麼以及如何解決它?
下面是使用example.php代碼:
<?php
include("dbconfig.php");
$page = $_GET['page'];
$limit = $_GET['rows'];
$sidx = $_GET['sidx'];
$sord = $_GET['sord'];
if (!$sidx)
$sidx = 1;
$db = mysql_connect("$dbhost", "$dbuser", "$dbpassword") or die("Connection Error: " . mysql_error());
mysql_select_db("$database") or die("Error connecting to db.");
$result = mysql_query("SELECT COUNT(*) AS count FROM app");
$row = mysql_fetch_array($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 id, name, status FROM app";
$result = mysql_query($SQL) or die("Couldn't execute query." . mysql_error());
header("Content-type: text/xml;charset=utf-8");
$s = "<?xml version='1.0' encoding='utf-8'?>";
$s .= "<rows>";
$s .= "<page>" . $page . "</page>";
$s .= "<total>" . $total_pages . "</total>";
$s .= "<records>" . $count . "</records>";
// be sure to put text data in CDATA
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$s .= "<row id='" . $row['id'] . "'>";
$s .= "<cell>" . $row['id'] . "</cell>";
$s .= "<cell>" . $row['name'] . "</cell>";
$s .= "<cell>" . $row['status'] . "</cell>";
$s .= "</row>";
}
$s .= "</rows>";
echo $s;
?>
謝謝你的幫助,但我做了你所說的,它仍然不起作用。我想知道jqgrid會在數據編輯完成後自動更新到數據庫嗎?如果沒有,我需要調用saveRow方法來創建它還是其他? – user878729
@ user878729:在用戶按下「Enter」後,editRow調用'saveRow'實習生。可能你在服務器端*在URL「example.php」下有問題*(參見你使用的editurl參數)。 jqGrid只是將修改後的數據發送到URL和Web服務器(您的PHP代碼在'example.php'下)負責將修改後的數據保存到數據庫中。 – Oleg
我檢查了example.php,發現我沒有sql語句從saveRow獲取數據,並執行sql。我想這是問題所在。你能告訴我我能做到嗎?謝謝! – user878729