<code>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>jQuery.pager.js Test</title>
<link href="Pager.css" rel="stylesheet" type="text/css" />
<script src="jquery-1.2.6.min.js" type="text/javascript"></script>
<script src="jquery.pager.js" type="text/javascript"></script>
<script type="text/javascript" language="javascript">
$(document).ready(function() {
$("#pager").pager({ pagenumber: 1, pagecount: 15, buttonClickCallback: PageClick });
});
PageClick = function(pageclickednumber) {
$("#pager").pager({ pagenumber: pageclickednumber, pagecount: 15, buttonClickCallback: PageClick });
$("#result").html("Clicked Page " + pageclickednumber);
}
</script>
</head>
<body>
$query = "select name from student";
$result = mysql_query(Query);
while($row=mysql_fetch_array($result)){
$student_name = $row['name'];
?>
<h1 id="result"><?php echo $student_name; ?></h1>
<? }
?>
<div id="pager" />
</body>
</html>
</code>
For my above code am not geting the student name , if i remove the pager script , the student name will appear, may i know , why, where i made mistake..
i thing i have to pass somthing to html() , but i am not sure..
while($row=mysql_fetch_array($result)){ $student_name = $row['name']; }
should be
$student_name = NULL;
while($row=mysql_fetch_array($result)){ $student_name .= $row['name']; }
Related
i'm working on a project and below is the following code i am using to make a bar chart.
the db table named bar contains, id, factor, score, goal_1, goal_2, goal_3 attributes.
i am not getting an error, but the chart is not showing and blank page is displayed to me.
<?php
$connect = mysqli_connect("localhost","root","","chart");
$query = "SELECT * FROM bar";
$result = mysqli_query($connect, $query);
$chart_data = '';
while ($row = mysqli_fetch_array($result)) {
$chart_data .="{ factor:'".$row["factor"]."', goal_1:'".$row["goal_1"]."', goal_2:'".$row["goal_2"]."', goal_3:'".$row["goal_3"]."'},";
}
$chart_data = substr($chart_data, 0, -2);
?>
<html>
<head>
<title>bar chart project</title>
<link rel="stylesheet" href="//cdnjs.cloudflare.com/ajax/libs/morris.js/0.5.1/morris.css">
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.0/jquery.min.js"></script>
<script src="//cdnjs.cloudflare.com/ajax/libs/raphael/2.1.0/raphael-min.js"></script>
<script src="//cdnjs.cloudflare.com/ajax/libs/morris.js/0.5.1/morris.min.js"></script>
</head>
<body>
<div class="container" style="width:900px;">
<div id="chart"></div>
</div>
</body>
</html>
<script type="text/javascript">
Morris.bar({
element: 'chart',
data: [<?php echo $chart_data; ?>],
xkey: 'factor',
ykeys: ['goal_1','goal_2','goal_3'],
labels: ['goal_1','goal_2','goal_3'],
hideHover: 'auto',
});
</script>
Try specifying the columns within the query
$query = "SELECT factor, goal_1, goal_2, goal_3 FROM bar";
and add a this to your code
<div id="chart" class="chart" style=<your preferred dimensions></div>
I have a problem using flot to plot data from a database.
using json_encode in php I have tried to get the plot data but I have been unsuccessful.
Below is my code, kindly assist
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>`enter code here`
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<title>Flot Examples</title>
<link href="./flot/layout.css" rel="stylesheet" type="text/css">
<!--[if lte IE 8]><script language="javascript" type="text/javascript" src="../excanvas.min.js"></script><![endif]-->
<script language="javascript" type="text/javascript" src="./flot/jquery.js"></script>
<script language="javascript" type="text/javascript" src="./flot/jquery.flot.js"></script>
</head>
<body>
<h1>Database Plotting</h1>
<div id="placeholder" style="width:600px;height:300px;"></div>
<?php
$server = "localhost";
$user="user";
$password="password";
$database = "database";
$connection = mysql_connect($server,$user,$password);
$db = mysql_select_db($database,$connection);
$query = "SELECT x_value, y_value FROM table";
$result = mysql_query($query);
while($row = mysql_fetch_assoc($result))
{
$dataset1[] = array($row['x_value'],$row['y_value']);
}
?>
<script type="text/javascript">
$(function () {
var dataset1 = <?php echo json_encode($dataset1,); ?>;
$.plot($("#placeholder"),[dataset1] );
});
</script>
</body>
</html>
The ouput from the json-encode dataset1 is thus the following [["1380111342","0"],["1380111679","0"],["1380112099","20"],["1380112109","300"],["1380112114","40"],["1380112120","56"],["1380112127","36"],["1380112132","40"],["1380112138","78"]]
Try change this line:
$.plot($("#placeholder"), [dataset1] );
To that:
$.plot($("#placeholder"), [JSON.parse(dataset1)]);
var dataset1 = <?php echo json_encode($dataset1,); ?>;
Lose the comma after "$dataset1" and the semi-colon at the end.
This should work just fine:
var dataset1 = <?php echo json_encode($dataset1); ?>
Hi i am able to get json data as a list in html page but i like to get as a list of link. Because when i click on the link it will show details.
What i have done in tenantlistmob.php
<?php
include('connection.php');
$result = mysql_query("SELECT * FROM tenanttemp");
while ($row = mysql_fetch_assoc($result))
{
$array[] = array($row['TenantFirstName']);
}
echo json_encode($array);
?>
Then my html page is
<!DOCTYPE HTML>
<html>
<link rel="stylesheet" href="../jasmine-device_2/styles/main.css" />
<head>
<script type="text/javascript" src="http://code.jquery.com/jquery-latest.min.js"></script>
<script type="text/javascript" language="javascript">
$(document).ready(function() {
jQuery.getJSON("tenantlistmob.php", function (jsonData) {
jsonData= eval(jsonData);//get json array
for (i = 0; i < jsonData.length; i++)//iterate over all options
{
for ( key in jsonData[i] )//get key => value
{
$("#getname").append($("<li></li>").html(jsonData[i][key]), document.all ? i : null);
}
}
});
});
</script>
</head>
<body>
<form name="index">
<div id="getname"></div>
</form>
</body>
</html>
The output is
Humayun
Sahjahan
Bayezid
Bayezid
Asaduzzaman
Mouri
I am getting the TenantFirstName as a list.But i like to get it as a list of link.Because when i click on a name it will show details of that name. How can i do both work(as a list of link and when click on a link it will show details in a html page query from mysql database)?
Please help.
use table like below
<!DOCTYPE HTML>
<html>
<link rel="stylesheet" href="../jasmine-device_2/styles/main.css" />
<head>
<script type="text/javascript" src="http://code.jquery.com/jquery-latest.min.js"></script>
<script type="text/javascript" language="javascript">
$(document).ready(function() {
jQuery.getJSON("tenantlistmob.php", function (jsonData) {
$("#user_spec").html("");//clear old options
jsonData= eval(jsonData);//get json array
for (i = 0; i < jsonData.length; i++)//iterate over all options
{
for ( key in jsonData[i] )//get key => value
{
//$("#user_spec").get(0).add(new Option(jsonData[i][key],[key]), document.all ? i : null);
//$("#getname tbody").append($("<li></li>").html(jsonData[i][key]), document.all ? i : null);
var tblRow = "<tr>" + "<td>" + "<a href='#'>" + jsonData[i][key] + "</a>" + "</td>" + "</tr>"
$(tblRow).appendTo("#getname tbody");
}
}
});
});
</script></head>
<body>
<form name="index">
<div>
<table id="getname" border="1">
<thead>
<th>Name</th>
</thead>
<tbody>
</tbody>
</table>
</div>
</form>
</body>
</html>
basically I want php to decide which version of the site pieces to load
something along the lines of:
<?php
$sql="SELECT _1 FROM Player_Registry WHERE Player_Name = $_SESSION[user_name]";
$result_1=mysql_query($sql);
if($result_1 < 10) { ?>
<script type="text/javascript">
function setdescription_1() {
document.getElementById('title_box').innerHTML = 'Alchemist';
}
</script>
<?php
}
else {
?>
<script type="text/javascript">
function setdescription_1() {
document.getElementById('title_box').innerHTML = 'Master Alchemist';
}
</script>
<?php
}?>
issue is that they both run so even if I set the condition to true I still get the second.
The complete source which is identical to what is being viewed-as-source as far as I can tell but none of the php shows up in the site preview (it is actually there though because my other computer can view the site without seeing the php but when it views source it can see it which it shouldn't be able to):
<head>
<title>Blank</title>
<link rel="icon" type="image/png" href="Pictures/favicon.png">
<meta name="description" content="" />
<meta name="keywords" content="" />
<meta name="author" content="" />
<meta name="owner" content="" />
<meta name="copyright" content="" />
<meta http-equiv="content-type" content="text/html;charset=UTF-8" />
<link rel="stylesheet" href="TreeStyle.css" type="text/css" />
</head>
<body>
<div style="width:100%; height:12.5%;">
<p style="font-size: 300%;"><b>Skill Tree: Alchemist</b></p>
</div>
<div style="width:100%; height:75%;">
<?php
$sql="SELECT _1 FROM Player_Registry WHERE Player_Name = $_SESSION[user_name]";
$result_1=mysql_query($sql);
if($result_1 < 10)
{
?>
<script type="text/javascript">
function setdescription_1()
{
document.getElementById('title_box').innerHTML = 'Alchemist';
document.getElementById('desc_box').innerHTML = 'Turn surroundings into base element costs.';
}
</script>
<?php
}
else
{
?>
<script type="text/javascript">
function setdescription_1()
{
document.getElementById('title_box').innerHTML = 'Master Alchemist';
document.getElementById('desc_box').innerHTML = 'Equiped Philosopher's Stone allows bypass of element costs.';
}
</script>
<?php
}
$sql="SELECT _2 FROM Player_Registry WHERE Player_Name = $_SESSION[user_name]";
$result_2=mysql_query($sql);
if($result_2 < 10)
{
?>
<script type="text/javascript">
function setdescription_2()
{
document.getElementById('title_box').innerHTML = 'Learn Runes';
document.getElementById('desc_box').innerHTML = 'Create Runes and Cores up to Lv.<?php echo "$result_2" ?>';
}
</script>
<?php
}
else
{
?>
<script type="text/javascript">
function setdescription_2()
{
document.getElementById('title_box').innerHTML = 'Comprehension';
document.getElementById('desc_box').innerHTML = 'Create Runes and Cores up to Lv.10';
}
</script>
<?php
}
?>
<script type="text/javascript">
function cleardescription()
{
document.getElementById('title_box').innerHTML = 'Skill Name';
document.getElementById('desc_box').innerHTML = 'This is the skill description.';
}
</script>
<table>
<tr>
<td></td>
<td><img src="Alchemist.png" /></td>
<td><img src="Blank_Tile.png"/></td>
<td><a href='AddSkillPoints.php?skill=_2' onmouseover="setdescription_2()" onmouseout="cleardescription()"><img src="Learn_Runes.png"/></td>
etc..............
</table>
</div>
<div style="width:100%; height:12.5%;">
<b><p id="title_box" style="font-size: 150%;">Skill Name</p></b><br />
<p id="desc_box">This is the skill description.</p>
</div>
</div>
</body>
</html>
I do not believe you get both if your PHP is actually parsed.
If it is not parsed, you will see the code you pasted exactly as you pasted it if you look at the source of the web page
Make sure the php is parsed - if not, then that is the issue.
A script tag will generally not interfere with parsing unless you have to post processing that does not allow the tags - for example if you use some kind of framework where you are not supposed to enter php but instead some tokenised code
Here is a better example - note I moved the script to the head but did not optimise really - so the php can be more elegant if you get both values in one call and create an object array
<?PHP
$sql="SELECT _1 FROM Player_Registry WHERE Player_Name = $_SESSION[user_name]";
$result_1=mysql_query($sql);
$sql="SELECT _2 FROM Player_Registry WHERE Player_Name = $_SESSION[user_name]";
$result_2=mysql_query($sql);
%><html>
<head>
<title>Blank</title>
<link rel="icon" type="image/png" href="Pictures/favicon.png">
.
.
<meta http-equiv="content-type" content="text/html;charset=UTF-8" />
<link rel="stylesheet" href="TreeStyle.css" type="text/css" />
<script type="text/javascript">
var result1 = parseInt("<?PHP echo $result_1; ">,10);
var skillLevel = (result1>=10?'Master ':'')+'Alchemist';
var skillDesc = result1>=10?"Equipped with Philosopher´s Stone allows bypass of element costs.":"Turn surroundings into base element costs.";
var runeLevel = result2>=10?'Comprehension':'Learn Runes';
var runeDesc = 'Create Runes and Cores up to Lv.'+result2;
window.onload=function() {
setdescription_1();
}
function setdescription_1() {
document.getElementById('title_box').innerHTML = skillLevel;
document.getElementById('desc_box').innerHTML = skillDesc;
}
function setdescription_2() {
document.getElementById('title_box').innerHTML = runeLevel;
document.getElementById('desc_box').innerHTML = runeDesc;
}
function cleardescription() {
document.getElementById('title_box').innerHTML = 'Skill Name';
document.getElementById('desc_box').innerHTML = 'This is the skill description.';
}
</script>
</head>
I was trying to create a notifications system for a website with the following code :
setInterval(function() {
$.post('notifications.php', {
email: 123
}, function(data) {
});
}, 30000);
And the notifications.php :
$userid = $_SESSION[username];
$notifications = array();
$sql = "SELECT appreciation FROM ms_notifications WHERE email = '$userid' AND new = '1'";
$res = mysql_query($sql) or die(mysql_error());
if (mysql_num_rows($res)) {
while ($r = mysql_fetch_object($res)) {
$notifications[] = $r->appreciation;
}
$nb_result = mysql_num_rows($res);
}
$sql = "UPDATE ms_notifications SET new = '0' WHERE email = '$userid'";
mysql_query($sql) or die(mysql_error());
echo $nb_result;
Problem is that, as I'm new to jquery/Js, I don't know how to put the result inside of a div. For now it stays on the top of the page, above everything.
I tried that, inside the data function but not working... :
$('#test_app').html(data);
I guess it's a very stupid question but help would be really appreciated ! Thank you very much.
UPDATE : Here the HTML Code
<?php
session_start();
include ('connection/database.php');
include ('login_script.php');
include ('notifications.php');
if($logged_in){
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>My Home</title>
</head>
<body>
<link rel="stylesheet" type="text/css" href="ms.css" />
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.4.4/jquery.min.js" type="text/javascript" charset="utf-8"></script>
<script src="js/font_i/specimen_files/easytabs.js" type="text/javascript" charset="utf-8"></script>
<!-- FONT SCRIPT-->
<script type="text/javascript" charset="utf-8">
$(document).ready(function() {
$('#container').easyTabs({defaultContent:1});
});
</script>
<script type="text/javascript" charset="utf-8">
var scrollSpeed = 70; // Speed in milliseconds
var step = 1; // How many pixels to move per step
var current = 0; // The current pixel row
var imageHeight = 505; // Background image height
var headerHeight = 59; // How tall the header is.
//The pixel row where to start a new loop
var restartPosition = -(imageHeight - headerHeight);
function scrollBg(){
//Go to next pixel row.
current -= step;
//If at the end of the image, then go to the top.
if (current == restartPosition){
current = 0;
}
//Set the CSS of the header.
$('#main_logo').css("background-position","0 "+current+"px");
}
//Calls the scrolling function repeatedly
var init = setInterval("scrollBg()", scrollSpeed)
</script>
<!-- NOTIFICATIONS_SCRIPT !!!!!! -->
<script type="text/javascript" charset="utf-8">
setInterval(function() {
$.post('notifications.php', {
email: 123
}, function(data) {
});
}, 30000);
</script>
<style>
body{
font-family:SommetRoundedLight;
}
</style>
</head>
<body>
<div id="container">
<div id="header_contain">
<div id="test_app">
</div>
</div>
</div>
</body>
</html>
<?php }
else{
echo '<script language="javascript" type="text/javascript">
<!--
window.location.replace("connexion.php");
-->
</script>';
} ?>
Thanks again
Use .append() instead of .html()
Instead of
$('#test_app').html(data);
just use:
$('#test_app').html('<div>' + data + '</div>');
Seems to simple to be it. Am I missing something?