PHP - Moved: Showing Loading Image On Ajax Call

This topic has been moved to Other Libraries and Frameworks.

http://www.phpfreaks.com/forums/index.php?topic=319682.0

Please help me in this guys.......
I am using MVC model for my application .

I am stuck with the ajax call.

this is my controller
application/controller/admin/index.php

function getCity()
   {
         $city=$this->model->getCity();

         foreach($this->city as $key => $value)
         {
            echo '<option>'.$value['name'].'</option>' ;
         }
   }

this is my model
application/model/admin_model.php




public function getCity()
   {
      if($_POST['id'])
      {
      $cid=$_POST['id'];
      $sth = $this->db->prepare('SELECT id, name FROM unit WHERE cid= :cid');
      $sth->execute(array(':cid' => $cid));
      return $sth->fetchAll();
      }
   }


this is my view

<form id="city" action="<../application/admin/regstr/" method="post">
   
   
   Country
    <select name="country" class="country">
   <option value="" selected="selected">--Select Taluka--</option>
   <option value="1">India</option>
        <option value="2">USA</option>
   <option value="3">China</option>
    </select>
    City
    <select name="city" class="city">
   <option selected="selected">--Select City--</option>
   
     </select>
   
   <input type="submit" />
</form>


and this is my javascript for Ajax

$(document).ready(function()
{
   
   $(".country").change(function()
   {
      var id=$(this).val();
      var dataString = 'id='+ id;

      $.ajax
      ({
         type: "POST",
         url: "admin/getCity",
         data: dataString,
         cache: false,
         success: function(html)
         {
            $(".city").html(html);
         }
      });
   
   });
   
});



But I am not getting the output. I have tested it with firebug in response tab. Instead of getting only option values html code as output  I am getting the whole HTML page content (i.e. content of view page admin/index.php.


Please help me out.

Hello guys
is it possible to create a single php file to contain all the functions to be called via AJAX and  get AJAX to call the specific function required rather than having to create individual pages for each AJAX request you need to make?

Many Thanks

Hi   I have an AJAX call that populates a select menu based on the selection from a previous menu.  All works on on desktop browser, Safari, Firefox IE etc, however when using iOS the second select menu isn't populating.   the AJAX request is as follows  

<script>
function getXMLHTTP() { //fuction to return the xml http object
        var xmlhttp=false;
        try{
            xmlhttp=new XMLHttpRequest();
        }
        catch(e)    {
            try{
                xmlhttp= new ActiveXObject("Microsoft.XMLHTTP");
            }
            catch(e){
                try{
                xmlhttp = new ActiveXObject("Msxml2.XMLHTTP");
                }
                catch(e1){
                    xmlhttp=false;
                }
            }
        }

        return xmlhttp;
    }



    function getTeam(strURL) {

        var req = getXMLHTTP();

        if (req) {

            req.onreadystatechange = function() {
                if (req.readyState == 4) {
                    // only if "OK"
                    if (req.status == 200) {
                        document.getElementById('away-team').innerHTML=req.responseText;
                    } else {
                        alert("There was a problem while using XMLHTTP:\n" + req.statusText);
                    }
                }
            }
            req.open("GET", strURL, true);
            req.send(null);
        }

    }
        function getGP(strURL) {

        var req = getXMLHTTP();

        if (req) {

            req.onreadystatechange = function() {
                if (req.readyState == 4) {
                    // only if "OK"
                    if (req.status == 200) {
                        document.getElementById('gp').innerHTML=req.responseText;
                    } else {
                        alert("There was a problem while using XMLHTTP:\n" + req.statusText);
                    }
                }
            }
            req.open("GET", strURL, true);
            req.send(null);
        }

    }
</script>
 
 

  and the trigger for it...

<select name="year" class="txtfield" onChange="getGP('find_gp_ajax.php?season='+this.value)">

  Any ideas why the call won't work on iOS?

Hey guys,

I'm making an Ajax call to my database and retrieving a newsletters HTML code as the result. I then prepend it into a div area called code_content. I use some php str_replace to add some links also.

When the content loads into the div, the user can then add further content to it via Ajax calls, appending articles and that.

What I'm trying to figure out is how can I then retrieve all the newly edited contents of this code_content div soni can send it to my Ajax function to write to te db?

Can anyone figure out why this works in IE and Firefox but not Safari?  Any help would be great.

Code: [Select]

<!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>Concept Builders</title> <link href="_css/main.css" rel="stylesheet" type="text/css" /> <link rel="stylesheet" href="_css/megamenu.css" type="text/css" media="screen" />

<script src="_scripts/pagebuilder.js" type="text/javascript"></script> <script src="_scripts/jquery-1.7.1.min.js" type="text/javascript"></script> <script src="_scripts/jquery.nivo.slider.pack.js" type="text/javascript"></script> <script src="_scripts/jquery.easing.1.3.js" type="text/javascript"></script> <script src="js/megamenu.js" type="text/javascript"></script>


<?php

include 'dbconfig.php';
include 'dbopen.php';

$plan=$_GET["plan"];

$sql = "SELECT * FROM homes WHERE plan_id= '".$plan."'";

$result = mysql_query($sql);

while ($row = mysql_fetch_array($result)) {
$price=$row['price'];
$sqft=$row['sqft'];
$beds=$row['bedrooms'];
$baths=$row['baths'];
$amenities=$row['amenity_1'].", ".$row['amenity_2']; }

setlocale(LC_MONETARY, 'en_US');
$price=money_format('%(#6.0n', $price);


$plan_details= "<table>
   
        <tr>
      <td width='182' class='propDetailsHeadings'>Price:</td>
      <td width='75'>".$price."</td>
        </tr>
    <tr>
      <td class='propDetailsHeadings'>SQ Feet:</td>
      <td>".$sqft."</td>
        </tr>
 <tr>
      <td class='propDetailsHeadings'>Amenities:</td>
      <td></td>
        </tr>
<tr>
    <td>".$amenities."</td>
      <td></td>
        </tr>
</table>";?>

<script language="javascript" type="text/javascript">


function loadPage()
{
    var ajaxDisplay = document.getElementById("plan_details");
ajaxDisplay.innerHTML = <?PHP echo $plan_details;?>;
   
}

</script>

</head>

<body onload="loadPage()">

Hi,

         This is my first post in this forum and am a PHP beginner. I have written a script in php and need to use echo to see the values of variables etc. However I dont get the output on the screen while using echo . I am using Xampp server with Apache.

The startup.html file code is

<!DOCTYPE HTML>
<html>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
        <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js" type="text/javascript"></script>
        <title>Test Page</title>
   
        <script>
             $.ajax(
                    {
                        url: "http://localhost:8081/test.php",
                        type: 'POST',
                        dataType: 'text',
                        data: {latitude: '7.15588546', longitude: '81.5659984458'},
                        success: function (response)
                        {
                            alert(response);
                        },
                        
                        error: function (response)
                        {
                            alert(response);
                        }
                        
                    });
      
        </script>

        </head>
    <body>
    <p>hello</p>
    </body>
</html>

 

and the test.php code is

<?php

$lat = $_REQUEST['latitude'];
$lon = $_REQUEST['longitude'];

echo 'latitude- '.$lat . ', longitude- ' . $lon;
?>

Please help

regards

 

Sanjish

I have a page that functions like this:   The user begins by loading ajaxtest.php, then using a drop-down menu, selects a user, after which a DIV on ajaxtest.php is populated with the user's selction to getuser.php?q=John%Doe   Here's a visual representation of what is currently happening, along with the part I don't understand.   getuser.php consists largely of a a large HTML table which interacts through PHP with my SQL database.  Several of the fields in this table are editable, and after the user changes one of these fields, he can press an "update" button, which calls an AJAX script in update.js and runs update.php, which contains all my SQL queries to update the database, without the user being redirected to this update.php page. All of this is processing perfectly - the user makes some sort of edit, hits "update", and the database gets updated without any redirection or refresh.   Here's what I can't figure out though.  Once the user makes a change, and hits "update," the database updates, but the end user doesn't see those changes that he made.  I can't just do a simple page refresh, because the action is taking place on getuser.php, but getuser.php is loaded inside of a #DIV on ajaxtest.php.  If I use something like window.location.reload(), this just refreshes ajaxtest.php and puts the user back to the starting point of having to select a user.  This is not what I want.  I don't want to refresh ajaxtest.php, I want to "refresh" ajaxtest.php with getuser.php?q=John%Doe insidethe DIV on ajaxtest.php.   It seems like this is very common - several forums have this capability, where a user has a page loaded, adds a comment, and sees that comment immediately without a full page refresh.  I just don't know how to do it, and I've tried at lengths to search the web for an answer with no luck.   Can someone walk me through how to do this?

I am trying to implement lazy loading into a project with pagination already set up. If I was to go to my website and after the .com, type in "api_courselist.php?page=1" I would receive the first 20 results of my query. If I was to change that to page=2, it would retrieve the next 20 and so on. My issue is I have no idea how to implement that into my java script/ajax file. I have it set up so that when the user scrolls to the bottom of the page, a div will make its self visible with the new page populated inside of it and IT will keep on pulling pages in till There is no more content.

jQuery(function ($) {
jQuery(document).ready(function() {
    var is_loaded = true;
    jQuery(window).scroll(function() {
        if(jQuery(window).scrollTop() == jQuery(document).height() - jQuery(window).height()) {


            jQuery('div#loadMore').show();
            if(is_loaded){
                is_loaded = false;
                jQuery.ajax({
                    url: "api_courselist.php?page=1",
                    success: function(html) {
                        is_loaded = true;
                        if(html){
                            jQuery("#infiscroll").append(html);
                            jQuery('div#loadMore').hide();
                        }else{
                            jQuery('div#loadMore').replaceWith("<center><h1   style='color:red'>End of Content !!!!!!!</h1></center>");
                        }
                    }
                });
            }
        }
    });

});
});

Where is my "Progr" return value?   This is not my first day using php but I don't code in the language that often. I don't understand why my dynamic variable is not returned in AJAX GET? Essentially, I need to give the client a simple UI so that they can copy files to a printer and I wanted to provide a little unfiltered feedback so they knew whether operation was a success. My feedback was just going to be output of exec() function though cleaned up slightly (perhaps replace line breaks with html break tags). PHP version is 5.2.17.17 and cannot be upgraded. Web Server Apache.   PHP dumbed down code:   exec("LoadZPL_PURL.bat", $output);

Hello All,

I have a PHP web application which will refresh itself(ajax calls connecting to the server and get the latest data) periodically. These also update the Database-based session handler class. i.e. There is NO UPDATE to the session data but the timestamp is constantly updated.


Our problem is that garbage collection does not kick in as it looks at the difference between timestamp and session_gc.maxlifetime. So, if  and the user is not interacting with the application.

Now my question is how can I force the timeout even though refreshing happens but the user is not interacting with the application and there are "phantom" session updates made by these ajax calls.

Please let me know. Thanks.

Hello all

I have a php and jquery/ajax call I am using to create a file/send an email. I get all the contents from my textboxes but I want a message to display on the screen afterwards being success or fail. In my .ajax call My call is exiting right before the ‘success:’  part. Why is my call not succeeding? Any tips/help will be appreciated

Thank you

Html page just a form with a submit button
$(document).ready(function () {
    var form = $("#formData");
    
    $(form).on("submit", function (event) {
        event.preventDefault();
        $.ajax({
            type: "post",
            url: "file.php",
            data: $(this).serialize(),
            beforeSend: function () {
                $("#status").html('<span style="color:red;">Loading everything actually creating a .TXT file of contents...</span>')  //--works
            },
            success: function (data) { 
                var responseMsgType = "alert-" + data.type;
                var responseMsgText = data.message;
                // message below
                var alertBox = '<div class="alert ' + responseMsgType + ' alert-dismissable"><button type="button" class="close" ' +
                    'data-dismiss="alert" aria-hidden="true">&times;</button>' + responseMsgText + '</div>';
                if (responseMsgType && responseMsgText) {
                    $(form).find(".successArea").html(alertBox);
                    $(form)[0].reset();
                }
            }
        });
        return false;
    });
<?php
$controls = array('txtName' => 'Name', 'txtEmail' => 'Email', 'txtSubject' => 'Subject', 'txtMessage' => 'Message');
try {
	if(count($_POST)==0) { throw new \Exception ('Contact Form Message is empty'); }

	$headers = array('Content-Type: text/plain; charset="UTF-8";',
					 'From: ' . $email,
					 'Reply-To: ' . $email,
					 'Return-Path: ' .$email);

	$emailMessage = "You have a new message from your contact form" . PHP_EOL;
	$emailMessage .= "-------------------------------------------------------" . PHP_EOL;

	foreach($_POST as $key => $value){
        if(isset($controls[$key])) {
            $emailMessage .= "$controls[$key]: $value". PHP_EOL;
        }
    }
$mailMsg = "email6.txt";
	if(file_exists($filename) == false) {
		$fh = fopen($mailMsg, "w");
		fwrite($fh, $headers);
		fwrite($fh, $emailMessage);
		fclose($fh);
	} else {
		$fhexists = fopen($filename, "a");
		fwrite($fhexists, $content);
		fclose($fhexists);
	}

	$responseMessage = array("type" => "success", "message" => $successMessage);
}
catch (\Exception $ex) {
	$responseMessage = array("type" => "errorM", "message" => $errorMessage);
}
if(!empty($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest') {
	$encodedJSON = json_encode($responseMessage);
	header("Content-Type: application/json");
	echo $encodedJSON;
} else {
	echo $responseMessage["message"];
}

 

Hi All,

I am a newbie to php and I have this issue. I would appreciate if anyone can help out. I tried a few things but was not able to come up with a solution.

From reading few codes and learning from it, I created this function in query.php and am calling this function in uk.php file, but the problem is that it loads and shows the data in table, but then it stops loading page from there on and the footer which is in html code after the place where I am calling this function does not load.

I tried removing return false; command, but with that it loads the footer, but it shows the table after the footer.

I want the table with data to show in between header and footer.

Can you guys let me know what I need to do to achieve that?

File Name: query.php

 

function uk()
{
	$output = '';
	$result = db_query("SELECT * FROM lecture where groups='uk'");

	if(mysqli_num_rows($result) > 0)
	{
		$output .= '<div id="style2" style="overflow-x:auto;">
						<table>
							<tr>
								<th class="text-center">Date</th>
								<th class="text-center">Title</th>
								<th class="text-center">Venue</th>
								<th class="text-center">Duration</th>
								<th class="text-center">Size (MB)</th>
								<th class="text-center">Link</th>
							</tr>';
		while($row = mysqli_fetch_array($result))
		{
			$output .= '
				<tr>
					<td>'.$row["mydate"].'</td>
					<td>'.$row["title"].'</td>
					<td>'.$row["venue"].'</td>
					<td>'.$row["duration"].'</td>
					<td>'.$row["size"].'</td>
					<td><a href="../'.$row["path"].$row["file_name"].'">Save</a></td>
				</tr>
			';
		}
		echo $output;
	}
	else
	{
		echo 'Data Not Found';
	}
}

 

File Name: uk.php

 

Header html code here

Some body text

<?php
include '../includes/query.php';
uk();
return false;
?>

Footer html code here

 

 

This topic has been moved to Third Party PHP Scripts.

http://www.phpfreaks.com/forums/index.php?topic=311501.0

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=306845.0