Insert Data Into The Mysql Database Using Ajax

Oct 19, 2009

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]...

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JQuery :: Insert Data In Database Mysql With Php?

Jun 15, 2009

I'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.

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Mar 11, 2011

I'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.

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Jun 6, 2010

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Jul 11, 2010

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Jul 13, 2011

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Here is my form fields:

Here is my java script:

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Nov 13, 2011

I have a .load that executes a php mysql query from another page. This query displays all the data of the database. Each result is wrapped around a div.

<div id ="post<? echo $row['post_id']; ?>">

the id of the div depends on the post id of the result. How do I put the data of $row['post_id'] in my javascript?

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Output Data From Mysql Database In Javascript?

Mar 16, 2004

Not sure if this is the correct forum as it covers data retrieved from mysql in a .js file...

I need to grab the last 5 rows from my database (2 fields per row) and list all 5 in a table but from within a .js file that I can externally link to.

I can manage the php / mysql and the output to a .php page but how can this be contained within a .js file?

I believe it is possible but after 2 days of searching on google I'm appealing for assistance.

I realise this could be done with a php include or iframe but that isn't an option and the only way available to import this data is via an external javascript script.

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Oct 10, 2010

I'm trying to display data from mysql database using getJSON

this is the code JQuery Code

Code:

//get data from database
$.getJSON('getCategories.php', parseCat);
// display data

[Code]....

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DIV Tag :: Cannot Insert The Form Data Into The Database?

Apr 28, 2011

I have a problem with the DIV tag (which I think is related with the Javascript).I cannot insert the form data into the database.

Code:
<script type="text/javascript">
var member_prices = new Array();
member_prices["no"]=200.00;[code]........

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Apr 28, 2011

I have a problem with the DIV tag (which I think is related with the Javascript). I cannot insert the form data into the database.

[Code]...

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Apr 8, 2011

I am writing a web page that accesses Mysql database data through PHP then dynamically displayes the Mysql data when the user mouses over different parts of an imagemap. I use Javascript OnLoad function to load the Mysql data into the header and then I use the Onmouseover function to change the display dynamically.

Here is my question; when I mouse-over a part of an image the data in my side bar changes (exactly what I want) but I can't seem to make the data linkable. For example, when a user mouse-overs a certain part of the image, on the side bar it displays the companies web adr. How do I make that data a link? On the initial load the default data displayed is linkable but because I use <div> </div> to change the data dynamically, I loose the likability.

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Sep 17, 2009

I'm using AJAX for all the updates to the web site I'm building and it works very well. But now I face the need to:

1) Upload a file (an image or video) to a folder in the server.

2) Insert a row in MySQL with the name of the file plus some other information (year and a comment).

The user would upload the file and fill in the information in the same window, and should have a single message (if possible) saying that the file was added or not (meaning both the upload and the insert).I have been doing some googling and found information as how to simulate and AJAX upload, since it seems AJAX does not support file uploads (hope I'm wrong). This is what I found: [URL]

So following that example I could upload the file, and I'd have all the information from the form, in the PHP, and I could synchronize the upload and the insert, but then I can't figure out how to send the OK from the PHP to a Javascript (i guess) to update the html and display the message in the web site.

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Sep 17, 2009

I'm using AJAX for all the updates to the web site I'm building and it works very well. But now I face the need of:

1) Upload a file (an image or video) to a folder in the server

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The user would upload the file and fill in the information in the same window, and should have a single message (if possible) saying that the file was added or not (meaning both the upload and the insert).

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Jun 22, 2011

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Jun 20, 2009

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.

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i am trying to create a form that where the page will not refresh if the form is submitted specially if the user inputs any errors in the form my problem is i do not know how to send multiple variables using jquery to a php file i saw somewhere that in order to do this you need to concatenate the variables you will be sending this is the code i am using

<html>
<head>
<style type="text/css">
#add(display: none;}

[code]....

the jquery successfully sends the variable to the php file the problem is once i insert it into a database all the variables i sent are concatenated for example in the column stud_no once you send the form this is what will be inserted "123123name=asdasd"

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[Code]...

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Jan 21, 2010

i want to retrieve data from mysql database using ajax and php. my code is below which does not work

here is index.html

<!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>

[Code].....

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Dec 13, 2011

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I have checked the internet for this and all of the examples show a form select/combobox where you select a "User" and then it populates the information automatically in a div below.

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Mar 11, 2009

How do I insert the following data into a php file using AJAX

function AddCity()
{
itemID = document.form.ItemID.value;
countryID = document.form.refCountryID.value;
regionID = document.form.refRegionID.value;

[Code]....

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Jan 30, 2011

I want to add a comment system after my article, <form id="postform" class="postform"> is written into a MYSQL_QUERY result circle. but after post a comment, the current posted comment will be showed in all the <div class="post_comment"></div>, how to modify jquery ajax part so that the current posted comment only be showed in its own <div class="post_comment"></div>? [code]...

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Sep 2, 2009

Im having the weirdest problems with my ajax. Now first off i have horrible coding convention!! I have one Ajax.js file which handles all my ajax requests and one ajax.php file which handles retrieving data from a database. The problem I am having is when ever I call a certain method which contains a if statement and runs the respective ajax function, it sometimes evaluates the if statement wrong. The weirdest thing of all is when I use FireFox's firebug and step through the javascript code to see why its evaluating the if statement wrong, it evaluates it correctly.

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I am pretty certain that it is because I took out the <form action="insert.php"> because otherwise the page would redirect but if that is the case I don't know my way around it and if it isn't the problem I'm not sure what is.

But here is my code:

Form:

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