PHP - Ajax Call Not Working In Safari

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?

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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?

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 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"];
}

 

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.

Hi, I am testing the below JS code in various browsers and it works just fine on my Windows laptop running Firefox but INOP on IE (same laptop) and also from my Mack which runs Firefox and Safari. I'm on the latest versions of all browsers - can someone advise what would prevent the JS function from opening the new page in those browsers mentioned?   Note: I am not blocking JS nor popups - this just a test to see how the function works but now I'm curious as to what's really happening... thx! Here's the code:

<!doctype html>
<html>
<head>
<meta charset="utf-8">

<!DOCTYPE html>
<html>
<head>

<script type="text/javascript">
function process() {
    window.open('http://www.w3schools.com/js/', '_blank'); self.blur();
    }
</script>
  	
</head>
  

<body onUnload="process();">
<p>&nbsp;</p>
<p>Close the browser or enter another URL in window and press enter.</p>
<p>In doing either of those actions, you should be redirected to http://www.w3schools.com/js</p>
</body>
</html>
</head>

<body>
</body>
</html>

This topic has been moved to Ajax Help.

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It give an the link like 

 

http://localhost/aps/undefined

 

Object not found! The requested URL was not found on this server. The link on the referring page seems to be wrong or outdated. Please inform the author of that page about the error. If you think this is a server error, please contact the webmaster. Error 404 localhost Apache/2.4.35 (Win32) OpenSSL/1.1.0i PHP/7.2.11

//The php code to fetch data

 

//The php code to fetch data
<?php
include('db.php');
$query = '';
$output = array();
$query .= "SELECT * FROM users ";
if(isset($_POST["search"]["value"]))
{
    $query .= 'WHERE first_name LIKE "%'.$_POST["search"]["value"].'%" ';
    $query .= 'OR last_name LIKE "%'.$_POST["search"]["value"].'%" ';
}
if(isset($_POST["order"]))
{
    $query .= 'ORDER BY '.$_POST['order']['0']['column'].' '.$_POST['order']['0']['dir'].' ';
}
else
{
    $query .= 'ORDER BY id DESC ';
}
if($_POST["length"] != -1)
{
    $query .= 'LIMIT ' . $_POST['start'] . ', ' . $_POST['length'];
}
$statement = $connection->prepare($query);
$statement->execute();
$result = $statement->fetchAll();
$data = array();
$filtered_rows = $statement->rowCount();
foreach($result as $row)
{
    $sub_array = array();
    $sub_array[] = $row["first_name"];
    $sub_array[] = $row["last_name"];
    $sub_array[] = '<button type="button" name="update" id="'.$row["id"].'" class="btn btn-warning btn-xs update">Update</button>';
    $sub_array[] = '<button type="button" name="delete" id="'.$row["id"].'" class="btn btn-danger btn-xs delete">Delete</button>';
    $data[] = $sub_array;
}
$output = array(
    "draw"              =>  intval($_POST["draw"]),
    "recordsTotal"      =>  $filtered_rows,
    "recordsFiltered"   =>  get_total_all_records(),
    "data"              =>  $data
);
echo json_encode($output);
?>

//javasrcipt jquery
var dataTable = $('#user_data').DataTable({
        "processing":true,
        "serverSide":true,
        "order":[],
        "ajax":{
            url:"fetch.php",
            type:"POST"
        },
        "columnDefs":[
            {
                "targets":[0, 3, 4],
                "orderable":false,
            },
        ]

    });

 

Somebody please help me. The below given code is not working.      

This topic has been moved to Ajax Help.

http://www.phpfreaks.com/forums/index.php?topic=354053.0

I had this working, but when I try and get fancy and use AJAX the data doesn't display. I think this is a PHP problem though.

My code for the select form including AJAX code

Code: [Select]
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en-GB">
<head>
<title>AJAX Example</title>
<link rel="stylesheet" type="text/css" href="Form.css" media="screen" />
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script>
<script type="text/javascript">
 $(document).ready(function(){
 $("tr:odd").addClass("odd");
 });
</script>
<script type="text/javascript">
function showPlayers(str)
{
var xmlhttp;
if (str=="")
  {
  document.getElementById("DataDisplay").innerHTML="";
  return;
  }
if (window.XMLHttpRequest)
  {// code for IE7+, Firefox, Chrome, Opera, Safari
  xmlhttp=new XMLHttpRequest();
else
  {// code for IE6, IE5
  }
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
xmlhttp.onreadystatechange=function()
  {
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
    document.getElementById("DataDisplay").innerHTML=xmlhttp.responseText;
    }
  }
xmlhttp.open("GET","Query.php?category_id="+str,true);
xmlhttp.send();
}
</script>
        </head>

<body>

<h1">AJAX Example</h1>

<?php
#connect to MySQL
$conn = @mysql_connect( "localhost","username","pw")
or die( "You did not successfully connect to the DB!" );

#select the specified database
$rs = @mysql_SELECT_DB ("MyDB", $conn )
or die ( "Error connecting to the database test!");
?>
<form name="sports" id="sports">
<legend>Select a Sport</legend>
<select name="category_id" onChange="showPlayers(this.value)">
<option value="">Select a Sport:</option>
<?php
$sql = "SELECT category_id, sport FROM sports ".
"ORDER BY sport";

$rs = mysql_query($sql);

while($row = mysql_fetch_array($rs))
{
  echo "<option value=\"".$row['category_id']."\">".$row['sport']."</option>\n  ";
}
?>
</select>

</form>
<br />
<div id="DataDisplay"></div>

</body>
</html>


             
                       
 
Query.php


<?php
#get the id
$id=$_GET["category_id"];

#connect to MySQL
$conn = @mysql_connect( "localhost","username","pw")
or die( "Error connecting to MySQL" );

#select the specified database
$rs = @mysql_SELECT_DB ("MyDB", $conn )
or die ( "Could not select that particular Database");
#$id="category_id";
#create the query
$sql ="SELECT *
FROM sports
INNER JOIN players ON sports.category_id = players.category_id
WHERE players.category_id = '".$id."'";


echo $sql;

#execute the query
$rs = mysql_query($sql,$conn);
#start the table code
echo "<table><tr><th>Category ID</th><th>Sport</th><th>First Name</th><th>Last Name</th></tr>";
#write the data
while( $row = mysql_fetch_array( $rs) )
{
echo ("<tr><td>");
echo ($row["category_id"] );
echo ("</td>");
echo ("<td>");
echo ($row["sport"]);
echo ("</td>");

echo ("<td>");
echo ($row["first_name"]);
echo ("</td>");

echo ("<td>");
echo ($row["last_name"]);
echo ("</td></tr>");
}
echo "</tr></table>";
mysql_close($conn);
?>


I think the problem is either with this part in the AJAX
Code: [Select]
xmlhttp.open("GET","Query.php?category_id="+str,true);

or most likely in my Query.php code

when I wasn't using AJAX and using POST it worked fine, but adding the AJAX stuff and GET it doesn't work.

When I echo out the SQL the result is

Code: [Select]
SELECT * FROM sports INNER JOIN players ON sports.category_id = players.category_id WHERE players.category_id = ''

so the category_id is not being selected properly and that is the primary key/foreign key in the MySQL table which connects the JOIN.

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