This form retrieves results from a mysql database. It works fine but when I click submit it shows the result only for a second. When I hold down submit the result stays there. How do I get the result to stay on the screen after I hit submit? Even better how do I get it to spit out results without having to hit a submit button or press enter?
View 15 RepliesI am having no luck with my post form to mysql. It is only posting the first letter of the word/phrase for the text input fields to the mysql database. It is a photo and text post form. The photo and only first letter of text are posting to database. I am using Jqtouch, jquery, and phonegap.
Here is my form fields:
Here is my java script:
I have built search form that it retrieves information from database. for ex: when i click axtar (search) button it retrieves normally but when i click ENTER button via keyboard instead of axtar (search) button but it only displays results with white blank page
Here is web page address. [url]
I have a listings page which displays results using images and text which are stored in a MySQL database, all is working fine. The images are pulled into the page from the database and each listing can have up to a maximum of 10 images. Some will have all 10, some probably one or two, so the page is coded such that if there's an image in the field, it displays a thumbnail, if there isn't then it shoves a "Â " in instead.
The results page has one main photo (image 1) and then a series of thumbnails (images 2-10) below it and I need to be able to set this up so that when someone clicks on image 2 it shows the full-size version of that image in the div where image1 sits, and the same for however many remaining images there are.
So my question is, how easy it is to do this with javascript and does anyone have any idea how to code it?
The main photo sits in a div whose id is "mainphoto" the remainder all sit in a div whose id is "thumbpix".
The below coding is the javascript coding in which i have used AJAX coding also.its only pagenavigation coding.
function change(which)
{
if(which=="page1")
{
document.getElementById(which).style.borderBottomColor="white";
document.getElementById("page2").style.borderBottomColor="#778";
document.getElementById("page3").style.borderBottomColor="#778";
document.getElementById("page4").style.borderBottomColor="#778";
[Code]...
I need to create three dropdown boxes (box 1:colour, box 2: price, box 3:brand) and when the user presses "GO" they are returned with the correct information, pulled from corresponding fields in my MYSQL database.
For example, colour: silver, price £100-£200, brand:samsung...should return 5 items on the page. I do not need the dropdown boxes to populate depending on ther previous selection - they are static, so to speak. If possible, I would like the information to be retuned to my page without the URL changing, which leads me to think I will need to be ausing ajax....?
I'm new to ajax. I've made a script that should return results from database but it keeps telling me just undefined. Code what I have:
Code:
var ajax = null;
function showRes(wut) {
if(wut.length == 0) {
document.getElementById("results").innerHTML = "";
return;
} if(window.XMLHttpRequest) {
// IE7+, ff, chrome, opera, safari
ajax = new XMLHttpRequest();
}else if(window.ActiveXObject) {
// retards
ajax = new ActiveXObject("Microsoft.XMLHTTP");
} else {
alert("Your "browser" does not support AJAX!");
return;
} var url = "search.php?q=" + wut;
url = url + "&sid=" + Math.random();
ajax.open("GET", url, true);
ajax.send(null);
document.getElementById("results").innerHTML = ajax.reponseText;
}
could someone please point me in the right direction for creating an AJAX page that displays "live" data from a mysql database? I need the page to update its self every time data changes in a mysql table.
View 1 Replies View RelatedAm looking to find the easiest way to populate an order form that customers can email through to us.
I have created a database with 2 fields, id and name.
I would like to be able to have a form on my page with 3 fields, id name and qty.
I want to link the id and name fields on my form to my database so that when a customer starts to type a product id or description it will show a list of available matches. underneath which once clicked will populate the mssing field.
So for example is i have 2 products id 998 green apples and 999 red apples.
Typing in 999 in the id field of the form would populate the name field with red apples and typing red apples in the name field would populate the id field with 99.
if someone didn't know which apples they wanted they could type apples in the name field and underneath it would show both red and green which when clicked will populate both fields.
I am looking for an Ajax combobox (drop down, not a text box) auto complete script. I searched for it on Google and only result which seems similar to what I am looking for, is dhtmlxCombo. And that one is too complicated for me. What I basicly need is:
When the client types on the drop down, the records appear from MySQL database and when the client chooses one and submits the form, I want the ID (not the text written) to be posted.
I have a form.On the form I have a drop down box that allows the user to select a data set to work with (A or B). The table loads just fine and all is right with the world. No problems there.What I need is for the user to be able to edit the data set and have those changes written directly back to the mySQL database. They also need the option of deleting a record entirely.I know how to do all the database queries, updates, etc. I just don't know how to write the ajax forms to accomplish this.
View 4 Replies View RelatedBasically the user would see a button on a form that would be something like "add contact"...then a window opens that shows a search window (want to implement something like [URL]) and it would show all the contacts available in the contacts table...then the user selects the contact to add (via a link).
I think i could get it this far, but how would i then pass that $row to the main form so that i can use the fields from that $row in the contacts table on my form?
I have a table that has parts with multiple characteristics. Part Number, Height, Width, Capacity, Price, Etc. I would like to have a drop downs for all of these values, when a user selects one of these values it will filter the results into a list. Ex. user selects a part that has a Height of 6 and width of 10 it will only list those parts. As I said earlier, I have a script that will give me all the parts, however I am not able to make the onchange list parts with multiple values.
View 9 Replies View RelatedI've the following AJAX code:
[Code]...
This code works to insert a MySQL record to a Database using AJAX. It works fine on IE, but it's not working on FF or Chrome. When I test it on FF/Chrome, i just get the text "Just a second..." and it doesn't advance from there.
Ajax and jQuery is something I'm new to, but PHP, Javascript, and MySQL I'm pretty well seasoned with.For starters, I want to create a simple form where there are these fields:Name (text field)E-mail (text feld)Zip Code (text field)Cell Phone Number (text field)Phone Carrier (dropdown menu)Thing is, I want to check the database to see if an e-mail address has already been entered. I'd like for that to happen after they enter the email address and tab or click out of the e-mail text field. For instance, after they enter an e-mail address and tab to the Zip Code field, the script queries the database to see if that e-mail exists already. If it does, it notifies the user and greys out the submit button. If not, they're allowed to finish filling the form out and submit. It'll be a huge time saver if this Ajax script is in place.
View 1 Replies View RelatedI am using Roshan's Ajax dropdown code for my two ajax dropdown menus. All is working fine, but when I want to use $_POST in the submitted form with FireFox, its not working...??? Its working fine in IE.
[Code]...
I want to do is let the user input a file,then read the contents send it to the server, loop over the inputs to query MySQL, return the results and display it in a user friendly,scrollable table?
View 1 Replies View RelatedThis is a combination of jQuery and a PHP problem so I'm not 100% certain where to post it. I am working on method to reoder a list that comes from a database. When the lists is displayed, it will be ordered by a "sort" column. The user can drag/drop the list to change the order, and I want the order to update the sort values in the database.I found a script that provides a decent starting point, but I am not getting the results I want.
1: It is able to reorder the columns, and save the results which is good.
2: It is saving the results as "0,1,2,3" rather than reordering the numbers from the "sort" column. (< this is what I need fixed)
EG: Name > Sort
Apple3
Orange 5
Pear 13
Kiwi 105
If I reorder these in the reverse order using the script, I end up with the following in the database.
Apple 3
Orange 2
Pear 1
Kiwi 0
Yet what I want is,
Apple 105
Orange 13
Pear 5
Kiwi 3
how I can get it to work with the existing "sort" values, rather than serializing and starting at 0. Here are the script files I am currently working with for testing.
DATABASE
Code:
CREATE TABLE `menu` (
`id` int(11) NOT NULL auto_increment,[code]....
how to connect with MYSQL database using javascript..
View 3 Replies View RelatedI want to build-up ajax based friends system. though i no nothing about ajax (would like to learn but) actually i have form where user can add his friend. on very next page of this page he/she can see what he/she is entered and where
they should probably see the add and remove button(which i want to build up using ajax)
how to start for this? till now i have code where user can write his friend name and to the next page he/she can see the value he has written. on submit button it all goes in to database. means i have devloped php form for this.
I want to ask you is it possible to use jquery dialog with php. Using this tutorial [URL] I want to call a php script with will delete a row into Mysql database.
View 2 Replies View RelatedBelow i have attached my code. This code is working.....but i want some modification in that code such that when i select 'vegetable' or 'fruits', it should give list of that selected one in second list, BUT THAT list should be stored in mysql database.
In the below program , i have hard-coded that second list, but i want second list to be dynamic from database.
Code:
I am trying to execute a query in mysql using javascript to check the username isnt taken, but dont know how.
View 3 Replies View RelatedI'm triyng to use the $.ajax jquery function to insert some dates in a database, using php.I have the page that insert the data called : sign.php and in the page I have the form, and the ajax scritp.
$("#submit_sign").click(function(data){
$.ajax({
type: "POST",
url: ("sign.php"),
[code]....
if I use the page without ajax it works, if I use the script jquery it works but doesn't insert the dates in the field, it send only white field. can't pass the variable POST beetween the ajax script and the php script.
I am working on a project for which I need an autocomplete form which uses the value from a mysql database. I've seen many examples about using local data.
View 1 Replies View Relatedi have a image that need user click on it and get the coordinates and store the coordinates to mysql database by using php. Here my code:
Code:
<script type="text/javascript">
function(ev){
var pos=zxcPos(this.img);
var mse=[ev.clientX-pos[0],ev.clientY-pos[1]];
var ary=this.ary;
if (ary.length>4)
[Code]...