*I know MYSQL is depricated. I am just using it as a learning tool for now.
UPDATED QUESTION:
I updated my question to make a lil more sense...
How can I display the output array (json data in PHP file) on my HTML page assuming I wanted to add it to the div id rvotes?
JavaScript
<script>
$(document).ready(function () {
$('.answer').click(function (e) {
var color = $(this).attr("data-color");
$.ajax({
type: 'POST',
url: 'mm.php',
data: { color: color},
dataType: 'json',
cache: false,
success: function(showVotes) {
$('#rvotes').html(row[0]);
},
error: function (jqXHR) {
}
})
})
});
</script>
PHP
function showVotes()
{
$sql = "SELECT * FROM mms";
$result = mysql_query($sql) or die(mysql_error());
$showresult = mysql_query("SELECT * from mms") or die("Invalid query: " . mysql_error());
$response = array();
while($row = mysql_fetch_assoc($showresult))
$results[] = $row;
echo json_encode($results);
}
ADDING my HTML code
<div id="wrapper">
<div id="red" data-color="red" class="answer">
<img src="images/red.jpg" width="100%" />
</div>
<div id="blue" data-color="blue" class="answer">
<img src="images/blue.jpg" width="100%" />
</div>
<div id="green" data-color="green" class="answer">
<img src="images/green.jpg" width="100%" />
</div>
<div id=rvotes>
TEST
</div>
<div id=bvotes>
TEST
</div>
How can I display the output from the array back at my HTML page?
I think this is similar to this question.
Get data from php array - AJAX - jQuery
You cannot access the php array as it is in your ajax. You need to represent it first as JSON by passing your php array to json_encode() function and echo it.
echo json_encode($results);
And it will be passed to your ajax callback as parameter.
In you success ajax callback,
success: function(showVotes) {
$('#rvotes').html(showVotes[0]);
},
Personally, I would have my php spit out pre-rendered html like so:
$sql = "SELECT * FROM mms";
$result = mysql_query($sql) or die(mysql_error());
//I cleaned up redundant queries here...
//echo top border
echo "<table border='1'>
<tr>
<th>Color</th>
<th>Votes</th>
</tr>";
while($row = mysql_fetch_assoc($showresult))
while($row = mysql_fetch_assoc($showresult)
{
echo "<tr>";
echo "<td>" . stripslashes($row['color']) . "</td>";
echo "<td>" . stripslashes($row['votes']) . "</td>";
echo "</tr>";
}
After it spits out html, just do this:
<script>
$(document).ready(function () {
$('.answer').click(function (e) {
var color = $(this).attr("data-color");
$.ajax({
type: 'POST',
url: 'mm.php',
data: { color: color},
success: function (showVotes) {
$("#votes").html(showVotes); //this edits the text inside the div to be whatever your php spits out
},
error: function (jqXHR) {
}
})
})
});
</script>
<div id="votes></div>
Related
I successfully fetch the data from the database. The problem is that, I want to display the results using Ajax request. The results/Output which displayed twice, I mean the *first output which displayed through Ajax (#the bottom of index.php), and the Second output displayed through PHP ECHO (**#**the bottom of the page). What can I do to get a single output through Ajax without adding another file and without refreshing the page?
index.php
<head>
<script src="https://code.jquery.com/jquery-3.1.1.js"></script>
<script type="text/javascript" src="javas.js"></script>
</head>
<body>
<div id="table_content"></div>
<?php
include ("db.php");
error_reporting(~E_NOTICE);
function ShowForm($AnswerCommentId) {
echo '<form id="myForm">';
echo '<input id="user" name="user" />';
echo '<textarea id="text" name="text"></textarea>';
echo sprintf('<input id="ParentId" name="ParentId" type="hidden" value="%s"/>', ($AnswerCommentId));
echo '<button type="button" OnClick=SendComment()>Comment</button>';
echo '</form>';
}
$query="SELECT * FROM `comm` ORDER BY id ASC";
$result = mysqli_query($conn,$query);
if (isset($_REQUEST['AnswerId'])) $AnswerId = $_REQUEST['AnswerId'];
else $AnswerId = 0;
$i=0;
while ($mytablerow = mysqli_fetch_row($result)){ $mytable[$i] = $mytablerow; $i++; }
function tree($treeArray, $level, $pid = 0) {
global $AnswerId;
foreach($treeArray as $item){
if ($item[1] == $pid){
echo sprintf('<div class="CommentWithReplyDiv" style="margin-left:%spx;">',$level*60);
echo sprintf('<p>%s</p>', $item[2]);
echo sprintf('<div>%s</div>', $item[3]);
if ($level<=2) echo sprintf('Reply', $item[0]);
if ($AnswerId == $item[0]) echo sprintf('<div id="InnerDiv">%s</div>', ShowForm($AnswerId));
echo '</div><br/>';
tree($treeArray, $level+1, $item[0]); // Recursion
}
}
}
tree($mytable,0,0);
?>
Comment
<div id="MainAnswerForm" style="display:none;width:40%; margin:0 auto;"><?php ShowForm(0); ?></div>
<div id="AfterMainAnswerForm"></div>
<script>
$(document).ready(function(){
$("#Link").click(function () {
$("#InnerDiv").remove();
$("#MainAnswerForm").slideToggle("normal");
return false;
});
});
</script>
</body>
Page.php
<?php
include ("db.php");
$user = $_POST['user'];
$text = $_POST['text'];
$ParentId = $_POST['ParentId'];
$action = $_POST['action'];
if ($action=="add") $query= mysqli_query($conn,"INSERT into `comm` VALUES (NULL,'{$ParentId}','{$user}','{$text}',NOW())");
?>
Javas.js
function show_messages(){
$.ajax({
url: "index.php",
cache: false,
success: function(html){
$("#table_content").html(html);
}
});
}
function AnswerComment (id){
$.ajax({
type: "POST",
url: "index.php",
data: "AnswerId="+id,
success: function(html){
$("#table_content").html(html);
}
});
}
function SendComment (){
var user1 = $("#user").val();
var text1 = $("#text").val();
var ParentId1 = $("#ParentId").val() + "";
$.ajax({
type: "POST",
url: "page.php",
data: "user="+user1+"&text="+text1+"&ParentId="+ParentId1+"&action=add",
success: function(html){
show_messages();
}
});
return false;
}
OK I think its more simple than you think
while you're using .html() it should change all the content of your div but while you're using <div id="table_content"></div> then the <?php ?> your code will show both - one appended to the div and under it the one which echoed by php ... so instead of using
<div id="table_content"></div>
<?php ?>
just wrap the <?php ?> inside the div
<div id="table_content">
<?php ?>
</div>
Alternative way: by using jquery wrapAll() but while Id must be unique you'll need to .remove() the #table_content element first then wrap all your code inside a new #table_content div .. see the example below
$(document).ready(function(){
$('#table_content').remove();
$('.CommentWithReplyDiv').wrapAll('<div id="table_content"></div>');
});
#table_content{
margin : 50px;
background : red;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="table_content"></div>
<div class="CommentWithReplyDiv">1</div>
<div class="CommentWithReplyDiv">2</div>
<div class="CommentWithReplyDiv">3</div>
<div class="CommentWithReplyDiv">4</div>
I am trying to fix this issues from last two days but :(
Now I have a index.php page where I am displaying all the data of adds from database.
this is my index page
<html>
<head>
<script>
$(document).ready(function(){
var overlayclass = $("<div id='overlayclass'></div>");
//pagination starts
$("#pagination li:first")
.css({'color' : '#FF0084'}).css({'border' : 'solid #dddddd 10px'});
$("#page").load("Controller/pagination_data_all_adds.php?page=1");
//Pagination Click
$("#pagination li").click(function(){
//CSS Styles
$("#pagination li")
.css({'color' : '#FF0084'})
.css({'border' : 'none'});
$(this)
.css({'border' : 'solid #dddddd 10px'})
.css({'color' : '#0063DC'});
//Loading Data
var pageNum = this.id;
$("#page").load("Controller/pagination_data_all_adds.php?page=" + pageNum);
});
//pagination ends
//for closing the div
$(".close").click(function () {
$(".divpopup").hide();
overlayclass.appendTo(document.body).remove();
return false;
});
//for close image
$(".x").click(function () {
$(".divpopup").hide();
overlayclass.appendTo(document.body).remove();
return false;
});
//onclick on button display div and overlay with button value and post the value to the index page where I can use it to select the specific adds data from database
$(':submit').live('click',function () {
var id = $(this).val();
overlayclass.show();
overlayclass.appendTo(document.body);
$(".divpopup").show('slow');
$.ajax({
url: 'index.php',
data:'button=' + $(this).val(),
dataType: 'json',
succss: function(data)
{
$('.divpopup').html('');
alert(data);
$('.divpopup').append('button:'+data);
}
});
return false;
});
});
</script>
</head>
<body>
<div class="divpopup">
<div class="divcontent">
<img src="http://www.catalyst-gc.com/images/icon_close_window.jpg" alt="icon_close_window"
class="x" id="imgclose" />
<div class="detail_data">
<?php
print "<h1>".$button."</h1>";
?>
</div>
Close
</div>
</div>
<div id="page"></div>
<ul id="pagination">
<?php
//Pagination Numbers
for($i=1; $i<=$pages; $i++)
{
echo '<li id="'.$i.'"-->'.$i.'';
}
?>
</ul>
<!--Pagging display ends -->
</body>
</html>
This is my pagination_data_all_adds page which will be load through jQquery load function in index page. inside page div. and the button which is inside table will be click to display the popupdiv with id of adds and than I will use it to display the rest of the data.
<?php include_once('dbconn.php');
$per_page = 6;
if($_GET)
{
$page=$_GET['page'];
}
$start = ($page-1)*$per_page;
$sql = "select * from adds order by add_id limit $start,$per_page";
$result = mysql_query($sql);
echo "<table cellspacing='0'>";
echo "<thead>";
echo "<th></th>";
echo "<th>Advertise</th>";
echo "<th>Title</th>";
//echo "<th>Description</th>";
echo "<th>Price</th>";
echo "</thead>";
while ($adds = mysql_fetch_array($result)) {
echo "<tbody>";
echo "<tr class='even'>";
$image_link = $adds["image"];
echo "<td><img class='image' src=adds_images/".$image_link."></td>";
echo "<td><b>".$adds["add_type"] . "</b></td>";
echo "<td><b>".$adds["add_title"] ."</b></td>";
echo "<td><b>".$adds["add_price"] ."</b></td>";
echo "<td>";
echo "<form method='post' action=''>"
echo "<button class='button' value='".$adds['add_id']."' type='submit'>Detail!</button>";
echo "</form>";
echo"</td>";
echo "</tr>";
echo "</tbody>";
}
echo "</table>";?>
Now the problem is when I click on the button which is page div it display nothing but when I look at Firebug response it shows there. It does not display when I click the button.
Your dataType in ajax call is json.
$.ajax({
url: 'index.php',
data:'button=' + $(this).val(),
dataType: 'json', <--------------------------------- JSON
But you are returning html.
echo "<table cellspacing='0'>";
echo "<thead>";
echo "<th></th>";
echo "<th>Advertise</th>";
..........................
..........................
Hence your call is actually failing (Which you could have spotted had you used fail).
So answer is return json or change data type.
N.B.
LOTS AND LOTS OF deprecated function. use .on instead of live. use .done instead of success. use mysqli/pdo instead of mysql.
if($_GET)
{
$page=$_GET['page'];
}
You need to use isset or empty that too on approriate index. $_GET is always set. So checking its existance is futile. You need to check the index. i.e.
if(isset($_GET['page']))
again echoing what #itachi has said, pass json data from your php. your js script is waiting for json object yet you passing html file.
to use your code the way it is comment out this line from your JavaScript //dataType: 'json',.
i would also encourage you to use jquery dataTables to save time. take a look at it here
I'm trying import images to HTML using PHP, but NivoSlider not loaded that.
I looked for the cause of the problem.
I am printing a alert message of response and the right.
Here is the HTML and AJAX query:
<div id="workcontent" class="pcontent" style="display:none;">
<div class="slider-wrapper theme-default">
<div id="slider" class="nivoSlider">
</div>
</div>
<script>
$(document).ready(function() {
var wl = $('#worklist div');
wl.on('click', function(){
var name = $(this).attr('id');
console.log(name);
$.ajax({
url: 'read.php',
type: 'POST',
data: { data : name }
}).done(function (response) {
$('#slider').prepend(response);
alert(response);
});
});
});
</script>
<div id="back"></div>
<div id="backcontainer">
<div id="back">
Back
</div>
</div><!--End backcontainer-->
</div><!--End content-->
And here is the other jQuery:
<script>
$(document).ready(function() {
$('#slider').nivoSlider(function(){alert('OK');});
});
</script>
This alert don't show! ):
Finally, here is the PHP code:
<?php
if (isset($_POST["data"])){
if ($_POST["data"] == "") {
echo "data ures";
} else {
$data = $_POST["data"];
$fname = "content/".$data."/*.*";
$files = glob($fname);
for ($i=0; $i<count($files); $i++)
{
$num = $files[$i];
echo '<img src="'.$num.'" data-thumb="'.$num.'">';
}
}
} else {
echo "nem jott data";
}
?>
Sorry for my bad english
NivoSlider doesn't take a function as an argument.
Also .nivoSlider() is probably called before the AJAX call returns it's response.
A better solution would be:
$(document).ready(function() {
$.ajax({
url: 'read.php',
type: 'POST',
data: { data : name }
}).done(function (response) {
$('#slider').prepend(response).nivoSlider( {} );
});
});
Now you can be fairly sure #slider contains the images from the response body so NivoSlider can act on them.
When using jQuery AJAX in cakePHP 1.3, the AJAX response includes the full layout rather than just the array with dataset from the model. How to I prevent this whole page from being rendered? I have tried to use $this->autoRender = false; and $this->layout = 'ajax', but this did not solve the problem. It actaully produced no response in the returned data.
My controller action is:
public function search() {
if (empty($this->data)) {
} else {
// $this->autoRender = false;
// $this->layout = 'ajax';
$request = $this->data;
$this->set('data', $this->Event->search($request['Event']['search']), 'host');
}
}
And my view is:
<!-- app/views/events/search.ctp -->
<h1>Corporate Events</h1>
<form method="post" action="search">
<input type="text" name="search" id="search_box" class='search_box'/>
<input type="submit" value="Search" class="search_button" /><br />
</form>
<p><?php echo $html->link("Add Event", "/events/add"); ?>
<table>
<tr>
<th style='width:100px'>Id</th>
<th style='width:100px'>Title</th>
<th style='width:100px'>Host</th>
<th style='width:100px'>Description</th>
<th style='width:100px'>Action</th>
</tr>
<div id= "results"></div>
<?php foreach ($data as $event): ?>
<tr>
<td><?php echo $event['id']; ?></td>
<td>
<?php echo $html->link($event['event_title'],'/events/view/'.$event['id']);?>
</td>
<td><?php echo $event['host']; ?></td>
<td><?php echo $event['description']; ?></td>
<td>
<?php echo $html->link(
'Delete',
"/events/delete/{$event['id']}",
null,
'Are you sure?'
)?>
<?php echo $html->link('Edit', '/events/edit/'.$event['id']);?>
</td>
</tr>
<?php endforeach; ?>
</table>
<script type="text/javascript">
$(function() {
$(".search_button").click(function() {
var searchString = $("#search_box").val();
var data = {'data[Event][search]':searchString};
if(searchString) {
$.ajax({
type: "POST",
url: "/events/search",
data: data,
beforeSend: function(html) { // this happens before actual call
$("#results").html('');
$("#searchresults").show();
$(".word").html(searchString);
},
success: function(html){ // this happens after we get results
$("#results").show();
$("#results").append(html);
}
});
}
return false;
});
});
</script>
Thanks!
I don't understand your question well, but this will help you with displaying to your view without the layout being echoed wholly.
Put this on the javascript inside the view,
<script type="text/javascript">
$(function() {
$(".search_button").click(function() {
var searchString = $("#search_box").val();
if(searchString) {
$.ajax({
url: "/events/search/" + searchString,
beforeSend: function() { // this happens before actual call
$("#results").html('');
$("#searchresults").show();
$(".word").html(searchString);
},
success: function(data){ // this happens after we get results
$("#results").show();
$("#results").append(data);
}
});
}
return false;
});
});
</script>
Then use the code below for your controller,
public function search($data=null) {
$this->autoRender = false;
//your code here;
return $data;
}
I would recommend having another function do the results and rendering them. Make a function called update_results. Following examples assumes you have request handler and jquery in place.
Simple Example:
//Controller
function search()
{
//search filters
$locations = $this->Event->Location->find('list',array('order'=>'Location.name'));
//set vars for view
$this->set(compact( 'locations'));
}//search
function update_results()
{
//location filter
if(!empty($this->data['Event']['location_id']))
$conditions['Event.location_id']=$this->data['Event']['location_id'];
$results=$this->Event->find('all',array('conditions'=>$conditions,'limit'=>15));
//debug($conditions);
$this->set('results',$results);
}//update result
//Views
//search.ctp
<?php echo $form->create('Event',array('action'=>'search'));
echo $this->Form->input('Event.location_id', array('label'=>'Location','empty'=>TRUE));
echo $js->submit('Find Trainings', array('update' => '#trainingresults','url'=> array('controller'=>'events', 'action'=>'update_results')));
echo $form->end(); ?>
<?php
echo $html->div('trainingresults',null,array('id'=>'trainingresults'));
echo '</div>';
?>
//update_results.ctp
<?php
$timestamp=date('r');
echo "<h4>Search Results ($timestamp)</h4><br/>";
if(empty($results))
{
echo "<p class='well'><strong>No Trainings found. Try different search criteria.</strong></p>";
}
else
{
foreach($results as $display)
{
//Event Info
$location=$display['Location']['name'];
echo $location;
}//for
}//else
?>
hello friends i am making a live search with the help of ajax and mysql. the problem is that when we write something suppose a alphabet "a" then it gives live results but when i put a space and then write any alphabet suppose "a" then results does no shows up . how can we ignore the spaces in the search ?? my code is
ajax search.php
<script type="text/javascript">
$(document).ready(function(){
$(".search").keyup(function()
{
var searchbox = $(this).val();
var dataString = 'searchword='+ $.trim(searchbox);
if(searchbox=='')
{
$('#display123').hide();
}
else
{
$.ajax({
type: "POST",
url: "friends/search1.php",
data: dataString,
cache: false,
success: function(html)
{
$("#display123").html(html).show();
}
});
}return false;
});
});
</script>
mysql search1.php
<?php include($_SERVER["DOCUMENT_ROOT"]."/sexy/include/connection.php");
if($_POST)
{
$q=$_POST['searchword'];
$sql_res=mysql_query("select * from users_profile where fname like '%$q%' or Email like '%$q%' order by rand() LIMIT 10");
while($row=mysql_fetch_array($sql_res))
{
$fname=$row['fname'];
$lname=$row['lname'];
$img=$row['profile_pic'];
$country=$row['country'];
$uid=$row['uid'];
$re_fname='<b>'.$q.'</b>';
$re_lname='<b>'.$q.'</b>';
$final_fname = str_ireplace($q, $re_fname, $fname);
$final_lname = str_ireplace($q, $re_lname, $lname);
?>
<div class="display_box" >
<span><img src="<?php echo $img; ?>" width="50" height="50" style="float:left; margin-right:6px" /></span>
<div class="searchtext"><a href="profile.php?f_id=<?php echo $uid;?>"><?php echo $final_fname; ?> <br/>
<span style="font-size:10px; color:#999999"><?php echo $country; ?></span></a></div></div>
<?php
}
}
else
{
}
?>
You can trim whitespaces before using the variable in query, like:
$q= trim($_POST['searchword']); //remove whitespaces from front and end