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  1. 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
  2. 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'){
  3. 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
  4. 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=''){ $.ajax({ url: "notification/new_friends.php", method: "POST", data:{view:"view"}, dataType:"json", success: function(data){
  5. Hey guys, i am trying to create a contact form that by passes the page refresh and uses Jquery and Ajax to direct to a php file server side. So far i have everything work fine. until the final part where i pass the json data type into PHP and look to catch an error, success, or completion value. Currently i have an error coming back with the following: SyntaxError: Unexpected token < error due to a parsererror condition I do not see any errors in the console of of chrome, and i have no idea why i am getting this syntax error. These does not seem to be any conflicting d
  6. Hi Guys, I have created a calendar view with this plug in. http://fullcalendar.io/ The calendar view will show calendar view by Start Date and End Date. So I have some problem with some of the event, the start date is from 01-January-2015 until 15-Jan-2015, but in between actually there is a break. For example, 01-January-2015 until 03-January-2015 13-January-2015 until 15-January-2015 So, because of the start date and the end date is so long, so in the calendar view, it block the calendar from 01/Jan until 15/Jan. Its look weird. Actually i have actual dates column in my database
  7. i was working on a mini project that imports csv file to the database through ajax and it's working fine <?php if(!empty($_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]);
  8. 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
  9. Basically I have a very long form select option drop down list. What I would like to do is that for each option "value", I want to remove the spaces and instead add dash to it. Doing each option value by hand would take me a very long time, so I thought I might try jquery for short cut. If you have another method, let me know. Here's what I have so far. Doesn't seem to work. <script> $(document).ready(function () { $('#model').change(function(){ var str = $(this).val(); str = str.replace(/\s+/g, "-"); }); }); </scrip
  10. 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
  11. 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
  12. 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) --
  13. 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
  14. 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 (
  15. Hi, I am absolutly beginner in PHP, AJAX. In my Wordpress page I have following scene: 1. input box for name 2. combobox from country-> combo for cities depend of selected country Button for search in MySQL database After submit the database return result in a table (user, country, city, mobile, email, etc.) with 2 editable (mobile, email) input column and I put Update button in every row. My questions: 1) If I change one data on the editable column how can I know the number of row? 2) How can I run the update script without refresh page? Thanks for your answer. //UserI
  16. Hi Peeps So basically im creating a "customer page" i have 2 tabs, 1 is customer info which is the primary and the invoicing tab I have a date picker in the invoicing tab which im hoping will populate a table below. I tried having the form post to php but this resets the page which then goes to the customer info tab so im guessing this will have to be done with jquery and ajax. could someone please point me in the right direction to do this? here is the current page for better explanation: All help is greatly appreciated in advance Mooseh
  17. 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
  18. 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;"
  19. 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(
  20. 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
  21. 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
  22. 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
  23. 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
  24. 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:
  25. 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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