Loading PHP file that loads records from a database which is already divided into pages
loading an external php file is cheap easy
<html>
<head>
<script type="text/javascript" src="jquery-1.6.3.js"></script>
<script type="text/javascript">
$(document).ready(function(){
// load home page when the page loads
$("#result").load("tablerecord.php");
});
</script>
</head>
<body>
THE LIST
<div id="result">
</div>
</body>
however....
$maxsecCode = "SELECT * FROM t_board";
$maxResult = mysql_query($maxsecCode);
$maxRows = mysql_num_rows($maxResult);
print "max rows is " . $maxRows . "<br>";
$linkNum = ceil($maxRows/10开发者_Go百科);
print $linkNum . "<br>";
for ($i=1; $i<=$linkNum; $i++) {
echo "<a href='tablerecord.php?page=".$i."'>".$i."</a> ";
}
the external php file loads records from a database, and divides the pages by 10, producing multiple links in the process.
if i click one of the pages i will be redirected to menu.php?page="whatever_page_i_clicked"!
i want to browse through the pages without leaving the main page.
is there any way to solve this problem?
If your number of records is less than a gazillion, it would be easier and cleaner to just retrieve all records, return them in a JSON string, and then use a paged table like datatables to show it.
EDIT: To turn a PHP object into a JSON string, just use json_encode() like this:
// some code to get your data into $myDataObject should go here, because
// you don't want to return the query result object; it's too bulky. Instead
// convert your result object to either a simple array of rows or fields, or into
// an object with fieldnames as attributes. Then...
$myJSONString = json_encode($myDataObject);
echo $myJSONString;
加载中,请稍侯......
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