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Found 23 results

  1. 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 The link on the referring page seems to be wrong or outdated.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"
  2. Hi all, Bit of a dilema. If I add, update items through the Modal form, all works well. But, if I delete an item, everything works fine, but the window stays greyed out. Even if I check a checkbox and delete that way it works fine. The Delete part of the Ajax: $(document).on("click", ".delete", function() { var id=$(this).attr("data-id"); $('#id_d').val(id); }); $(document).on("click", "#delete", function() { $.ajax({ url: "includes/save.php", type: "POST", cache: false, data:{ type:3, id: $("#id_d").val() }, success: function(dataResu
  3. //my controller <?php namespace App\Http\Controllers; use Illuminate\Http\Request; use DB; class homeController extends Controller { public function index() { $employee = DB::table('employee')->orderBy('id','desc')->get(); $department = DB::table('department')->orderBy('id','desc')->get(); return view('index', ['employee' => $employee , 'department' => $department]); } } //my routes Route::get('index','homeController@index'); //my view using blade temmplating engine @foreach($employee as $emp) <div class="employee"&
  4. HI all, I am building a php application and i am wanting to use notifications. At the moment i am just wanting to understand the methodology behind this. I dont necessarily have a use case for these notifications yet but i am going to base it off of the following: A user submits something to the database, another user gets a notifications that this has happened. I see the notifications being something appearing in the header bar (saying "you have a notification"...) I know that i will need to use ajax for this and JS/JQ but i am not really sure where to start with thi
  5. i'm trying to make a notification tab work but do not seem to get it right. The dropdown is working fine but the ajax call to newfriends.php is not working right, when viewed with firebug there are no results to be seen in the dropdown.Quite confusing. (note the dropdown menu is located in header and can only be displayed if the session is initialised) here is the ajax used in jquery: function load_notifications(view=''){ $.ajax({ url: "notification/new_friends.php", method: "POST", data:{view:"view"}, dataType:"json", success: function(data){
  6. i was working on a mini project that imports csv file to the database through ajax and it's working fine <?php if(!empty($_FILES["marks_file"]["name"])) { $connect = mysqli_connect("localhost", "root", "", "dbname"); $output = ''; $allowed_ext = array("csv"); $extension = end(explode(".", $_FILES["marks_file"]["name"])); if(in_array($extension, $allowed_ext)) { $file_data = fopen($_FILES["marks_file"]["tmp_name"], 'r'); fgetcsv($file_data); while($row = fgetcsv($file_data)) { $name = mysqli_real_escape_string($connect, $row[0]);
  7. I'm getting "Uncaught SyntaxError: Unexpected end of input" (VM4193:1) on this AJAX call: MakeRequest : function( blah, blah ) { $.post( blah-blah-url, obj, function( data ) { var result = null; try { result = JSON.parse( data ); } catch (err) { } SOMETHING.HandleResult( result ); }) .fail( function() { }); }, ... the odd thing is, I get this error in Chrome before anything has actually happened in the result. What I mean is: when I put the breakpoint on result=null, the error happens right there. The parse
  8. Here's what I'm trying to do. I retrieve results using php foreach loop. Within that loop results, I want to show a div(.show-details) on mouse over only. How do I do that? This is my code so far. <style> .show-details { display: none; } </style> $get_records = $db->prepare("SELECT * FROM records WHERE record_id = :record_id"); $get_records->bindParam(':record_id', $record_id); $get_records->execute(); $result_records = $get_records->fetchAll(PDO::FETCH_ASSOC); if(count($result_records) > 0){ foreach($result_records as $row) { $get_reco
  9. Hi, I have been reading up on ajax using jquery, and theres something I can't seem to find, i have a form that gets the field values from mysql when the page loads, what is the best way to use ajax to edit this form? Obv the ajax request will execute the mysql update statement but how do i get the edited data back into the form, ready to be worked with again. thanks, hope i made sense, just looking for a little direction. MsKazza
  10. I have made an ajax call (using file activedirectory.php) to active directory to retrieve information such as name,thumbnail photo,mail etc. the ajax call is made on click of a name in a leaderboard ,to fetch only that person's info by matching the post['id']. The problem is that I am unable to retrieve any information and the console gives me an error saying : SyntaxError: JSON.parse: unexpected character at line 1 column 1 of the JSON data the error type makes it clear that it is unable to pull any data whatsoever. .php file that has been used here :(activedirectory.php) --
  11. Hi there, I was trying to do a live table edit for our site. Everything seemed to be going ok, till i realised that its not actually updating the database, there are no error messages, you have to refresh the page to see it hasn't worked. I'm at a loss as to why was hoping someone might be able to point me in the right direction. thanks, MsKazza tableedit.php table_edit_ajax.php
  12. Hello I've been trying to fix this problem for around 3 weeks; so what I want is to be able to send a picture and being able to display it in another page. It send it to the server, but still it doesn't show it. Here is my code: <?php require_once('../Connections/connection.php'); ?> <?php $editFormAction = $_SERVER['PHP_SELF']; if (isset($_SERVER['QUERY_STRING'])) { $editFormAction .= "?" . htmlentities($_SERVER['QUERY_STRING']); } if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "add_post")) { $tiempocotejo= time(); $insertSQL = sprintf("INSERT INTO posts (
  13. i've to populate google charts with dynamic data. the data looks llike. +----+-------+---------+-------+-----------+---------+------+ | id | name | physics | maths | chemistry | biology | sst | +----+-------+---------+-------+-----------+---------+------+ | 1 | Name1 | 10 | 25 | 35 | 42 | 62 | | 2 | Name2 | 80 | 45 | 45 | 45 | 25 | | 3 | Name3 | 63 | 25 | 63 | 36 | 36 | | 4 | Name4 | 82 | 36 | 75 | 48 | 42 | | 5 | Name5 | 45 | 45 | 78 | 25 | 24 | | 6 | Name6 | 36
  14. New to Ajax, and BAD with the scripting below, bare with me. Looking to select a drop down and push all values to the query string at the end. Works perfectly with ONE select element. I just have no idea how to separate the variables. The code below is putting whatever form element I change into the first "str" variable. if possible if one of the elements below is changed, all three of variables the data is placed in the query string in the correct place if that makes sense. Form <form> <table style="width:100%"> <tr> <td style="width:85px;"
  15. I've wrote code for sending array to another php file with mysql, everything works, but but i can't add CHANGE function to my code, to make code react on changes (checked/uchecked checkbox) <div id="checkboxes"> <input id="chkbx_0" type="checkbox" name="first" checked="checked" />Option 1 <input id="chkbx_1" type="checkbox" name=second" />Option 2 <input id="chkbx_2" type="checkbox" name="third" />Option 3 <input id="chkbx_3" type="checkbox" name="fourth" checked="checked" />Option 4 </div> <script> $(document).ready(
  16. Hi, I have a form for news articles. It has a title and a body which is fine. I also have a search box that allows the user to search for a memeber and click their name when it appears. This moves a div with a data-id into another div. The purpose of this is to tag them in the article. I am able to post all of the simple stuff but how would i post these value. I am assuming that i would need to create an array of said values but i am struggling to get them showing in post at all. Here is the code i have so far <?php function searchForPeople($searchVal, $exclude = '0'){
  17. Background: I'm comparing 2 styles of Ajax: 1.) "jquery style" 2.) "ActiveXObject Microsoft.XMLHTTP style" Question: Is one better (faster, more cross-browser compliant) than the other? My experience: Both seem equally fast. The Microsoft style is a bit longer, but I don't have to load jquery.js to my page! Code Examples: Jquery style on my PHP page: function getInfo(ProductNumber){ $.ajax({ url:'Ajax-PHP-Page.php?ProductNumber='+ProductNumber, success: function(html) { document.getElementById("my_div").value = ''; document.getElementById("my_div").value = ht
  18. I have created a registration page to access my website. After the user registrate himself should appear an alert saying that the registration was OK and a redirect to main.php page... however for some reason if I create an insert statement the alert and the redirect don't appear... If I remove the insert the alert and the redirect works... why? This is part of the code of my 3 files: registration.php (ajax call) $('#submit').click(function() { var username2 = $('#uname2').val(); var password2 = $('#psw2').val(); $.ajax({ url
  19. I've been following this tutorial on how to use ajax I have to admit I'm new at it. I could use jquery but I hate it. I would like to know if someone can help me figure out why my code is reading Uncaught TypeError: Cannot set property 'onreadystatechange' of undefined. when I fire off the Friend request button or block button. ? <?php error_reporting(E_ALL); ini_set('display_errors', '1'); include_once('includes/check_login_status.php');?> <?php $u = ""; $profile_pic = ""; $profile_pic_btn = ""; $avatar_form = ""; $avatar = ""; $date_added = ""; $joindate = ""; $lastsessio
  20. Hi all ! In my previous question asked today I said that I am using dropdown lists for selecting country, state, city and pin. The initial lists are blank and use the selection of country to trigger the loading of states and choosing a state triggers the loading of cities and so I am using ajax for this purpose - more specifically the $ajax() function of jquery. In a normal call to a php page, the integrity is maintained via sessions, and csrf is prevented via tokens embedded in the form, but how do I take care of these when data is being passed through the ajax call ? Any other se
  21. i'm trying to make a notification tab work but do not seem to get it right. The dropdown is working fine but the ajax call to newfriends.php is not working right, when viewed with firebug there are no results to be seen in the dropdown.Quite confusing. (note the dropdown menu is located in header and can only be displayed if the session is initialised) here is the ajax used in jquery: function load_notifications(view=''){ $.ajax({ url: "notification/new_friends.php", method: "POST", data:{view:"view"}, dataType:"json", success
  22. his is my code for a simple catalog store (no cart and checkout), a listing page that will change the items depending on the category (or subcategory). This is my db structure: prod_id Primary bigint(20) AUTO_INCREMENT user_id int(11) cat_id int(11) subcat_id int(11) prod_titulo varchar(250) prod_descripcion varchar(250) The file that is currently listing all the products (store.php) has this script: $(document).ready(function(){ $('#listing_store').empty(); $.ajax({ url: 'store-app/db_query.php', type:
  23. I'm having an issue related to creating forms within a foreach loop. As of now, I have 3 forms in a div, each with their own datepicker instance and submit button ( also using a hidden input for the pageID) The datepickers are unique and working independently, but when it comes to posting the data via ajax and handling it in a php script, It's not inserting data into my db because the names of my two inputs are not unique. I don't really know how to make them unique on the form and then handle those unique names back in the script. Only one form will be submitted at a tim
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