Update database using Ajax and Jquery - php

I have a table with multiple rows that lists records from my database.
These records are projects' information and in each row, I have drop down list to modify the status of the project.
To do so, I used Ajax because I hate to refresh the whole page after update.
This is the function I created to do the update:
function call(){
var projid=$('#projid').val();
var projstatus=$('#projstatus').val();
var dataall={'projid':projid, 'projstatus':projstatus};
$.ajax({
type: 'POST',
url: "stsproj.php",
data: dataall,
success: function (data) {
}
});
}
And below is my drop down list:
<?php do { ?>
<td>
<form action="<?php echo $editFormAction; ?>" method="post" name="form2" id="form2">
<input type="hidden" name="MM_update" value="form2" />
<input type="hidden" name="projid" id="projid" value="<?php echo $row_projlist['projid']; ?>" />
<select name="projstatus" id="projstatus" class="select1" onchange="call()">
<option value="<?php echo $row_status['projstatus'];?>"><?php echo $row_status['sts'];?></option>
<option value="1">Awaiting</option>
<option value="2">Ongoing</option>
<option value="3">Finishing</option>
</select>
</form>
</td>
<?php }while($row_projlist = $projlist->fetch(PDO::FETCH_ASSOC)); ?>
My problem is the following:
When I update the status of the first project, it works but when I try to do it with other projects, it doesn't work.
To be more specific, the parameters of the first project are sent always (this is what firebug says).
Please help!

Your problem is due to duplicate ids. You don't need to use ids(actually do not use id for automatic list generation. Id names must be unique). Remove call function from your select box and use below javascript;
You can use such js to handle that;
$(function() {
$("select[name='projstatus']").change(function() {
var projid = $(this).parent("form").find("input[name='projid']").val();
var projstatus = $(this).val();
var dataall = {'projid':projid, 'projstatus':projstatus};
$.ajax({
type: 'POST',
url: "stsproj.php",
data: dataall,
success: function (data) {
}
});
});
});
You can see working example for form manipulating part here : http://jsfiddle.net/cubuzoa/SYf8s/

The previous answer will probably solve your problem, but I think it can be simplified.
for the form...
<form>
<select class="select1" onchange="call(<?=$row_projlist['projid']?>)">
<option value="<?=$row_status['projstatus']?>"><?=$row_status['sts']?></option>
<option value="1">Awaiting</option>
<option value="2">Ongoing</option>
<option value="3">Finishing</option>
</select>
</form>
and the javascript
function call(id)
{
$.post('stsproj.php',{'projid':id, 'projstatus':$('#projstatus').val()} )
.done(function(data){
//do something with data
});
}

Related

Updating form fields using AJAX and PHP

Problem: How can I update a form's select input values and text input fields based on a MySQL query after select input's onchange event is fired?
What I've tried:
I have tried to use AJAX with post and get data types, calling a php file that runs the query and echoes the results. Nothing displays. Any errors I have gotten along the way are usually small things that result in server 500 error. I have placed console.log statements in the function that runs the JQuery AJAX request. The change event was detected, the ajax success was called. I also tried using .load(), with GET and POST, no luck either. I have other features that implement AJAX, and I've tried modifying them to fit this scenario and have been unsuccessful.
I also tried to only use a select input that when changed would use AJAX request and .load function to display the other inputs which would be formatted on the php side and echoed to page with selected and values reflecting the db result.
What I want:
I would like a simple example of a form with a select input with three options, text type input, and a submit button. The form is a client backend form to send updates to the MySQL db. Each input represents a filed in the db. The idea is that when the user changes the select inputs selected value, a query is done that uses the selected value for only returning one result. Each field of that one records values in db should now be reflected in the form. First, tell me if this is the correct way to approach this problem, and if not show me how you would.
Example index.php:
<form action="editForm.php" method="POST" enctype="multipart/form-data">
<select id="contact_name" name="contact_name" placeholder="Select Contact" required>
<option value="John Smith">John Smith</option>
<option value="Jane Doe">Jane Doe</option>
<option value="George Washington"></option>
</select>
<input type="text" name="age" placeholder="Age" required>
<input type="text" name="race" placeholder="Select Race" required>
<select id="veteran_status" name="veteran_status" placeholder="Select Veteran Status" required>
<option value="Yes">Yes</option>
<option value="No">No</option>
</select>
</form>
When on change event for #contact_name is fired I need to update the fields with the values the db has.
How would you implement this? Thanks in advance.
Update: as requested here is my JQuery code, but I know my example doesn't use the same names.
<script type="text/javascript">
$(document).ready(function(){
$('#currency_select').on('change', function (e) {
$.ajax({
type: 'post',
url: 'getCurrentValues.php',
data: {currency: 'EUR'},
success: function () {
console.log('ajax was submitted');
}
});
});
});
</script>
Here is my understanding of how to do this:
First, detect event and pass data via ajax for the query to retrieve record. This is in the document ready function to ensure DOM is ready.
$("#contact_name).on("change", function(e){
$.ajax({
type: 'post',
url: 'editForm.php',
data: {name: this.value},
success: function () {
console.log('ajax was submitted');
}
});
};
editForm.php:
include 'includes/db.php';
$name = $_POST['contact_name'];
$sql = "SELECT * FROM contacts;";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
$lastValues = array();
while($row = mysqli_fetch_assoc($result)) {
$age = $row['age'];
}
<input type="text" name="age" value="<?php echo $age; ?>">
<?php
}
?>
your index:
<select id="contact_name" placeholder="Select Contact" required>
<option value="John Smith">John Smith</option>
<option value="Jane Doe">Jane Doe</option>
<option value="George Washington"></option>
</select>
<form action="editForm.php" id="form" method="POST" enctype="multipart/form-data">
<select name="contact_name" id="contact_form" placeholder="Select Contact" required>
<option value="John Smith">John Smith</option>
<option value="Jane Doe">Jane Doe</option>
<option value="George Washington"></option>
</select>
<input type="text" id="age" name="age" placeholder="Age" required>
<input type="text" id="race" name="race" placeholder="Select Race" required>
<select id="veteran_status" name="veteran_status" placeholder="Select Veteran Status" required>
<option value="Yes">Yes</option>
<option value="No">No</option>
</select>
</form>
$("#contact_name").on("change", function() {
var selected = $(this).val();
$("#form").load("formdata.php?contact="+selected); //normaly you do that with an id as value
OR
$.ajax({
type:"POST",
url:"formdata.php",
data: {user: selected},
dataType: "json",
success: function(response){
if(response.status == "success") {
$("#age").val(response.age);
$("#race").val(response.race);
$("#veteran_status").val(response.status);
} else {
alert("No data found for this user!");
}
});
});
and in your formdata.php file
//make your db-query
then either make the actual input fields which will be displayed if you use load
OR make something like if you use the ajax version
if($result) {
echo json_encode(array("status" => "success",age" => $result["age"], "race" => $result["race"], "status" => $result["status"]));
} else {
echo json_encode(array("status" => "failed"));
}
also you can delete the action, method and enctype in your form, as this will be set in the ajax function ;)
I would advice you to use the userid as the value in your select field, and you will also need to either also fill the contact_name IN the form OR make an hidden input field so that you can submit the form and know whos data this is..
just echo the $age variable in your editForm.php file and in the AJAX call success function alert the response. like so-
editForm.php
include 'includes/db.php';
$name = $_POST['contact_name'];
$sql = "SELECT * FROM contacts;";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
$lastValues = array();
while($row = mysqli_fetch_assoc($result)) {
echo $age = $row['age'];
}
}
?>
Ajax file
$("#contact_name).on("change", function(e){
$.ajax({
type: 'post',
url: 'editForm.php',
data: {name: this.value},
success: function (response) {
alert(response);
console.log(response);
}
});
};

Submit whole form when select option is changed without reload page

I need to submit the whole form when I change a select option and save it to mysql without reload page.
I have this code, but only POST the select option value and I need to POST the hidden values in the form too.
<form class="form-horizontal" method="POST" action='#' name="dateForm">
<input type='hidden' name='cond_acao' class='form-control' value="edit_seccao_formando">
<input type="hidden" name="id_formando" value="<?=$linha_formandos[id_formando];?>">
<select name="id_seccoes" class="form-control" onchange="return postSelection">
<option selected="selected" value="1">car</option>
<option value="2">boat</option>
<option value="3">plane</option>
</select>
<div id='response_seccoes'></div>
<script>
function postSelection(selectObject) {
var id_seccoes = window.dateForm.id_seccoes.value = selectObject.options[selectObject.selectedIndex].value;
var dataString = "id_seccoes=" + id_seccoes;
$.ajax({
type: "post",
url: "url.php",
data: dataString,
success: function(response) {
$('#response_seccoes').html("ok").fadeIn(100).delay(2000).fadeOut("slow");
//$("#list").html(response);
}
});
return false;
};
</script>
</form>
It's possible to serialize instead of especified the strings?
Can anyone help me please? Thank you
you are using Jquery, I suggest you monitor the select button for change using a jquery function like
$("id_seccoes").change(function(){
//call your post method here.
});
Replace your code with this.
<form class="form-horizontal" method="POST" action='#' name="dateForm">
<input type='hidden' name='cond_acao' class='form-control' value="edit_seccao_formando">
<input type="hidden" name="id_formando" value="<?=$linha_formandos[id_formando];?>">
<select name="id_seccoes" class="form-control">
<option value="1">car</option>
<option value="2">boat</option>
<option value="3">plane</option>
</select>
</form>
<div id='response_seccoes'></div>
<script>
$(document).ready(function(){
$('[name="id_seccoes"]').change(function(){
$.post('url.php',$('[name="dateForm"]').serialize(),function(response){
$('#response_seccoes').html("ok").fadeIn(100).delay(2000).fadeOut("slow");
});
});
});
</script>

retrieve value from database when onchange in dropdownbox by jquery

This is my html code
<select name="course" id="course" onchange="valuesOfAll(this.value)">
<option value=""> select </option>
<option value="1"> Diploma in Computing</option>
</select>
<input name="course_credits" id="course_credits" type="text" />
and my database table is like this
courseId courseName courseCredits
1 Diploma in Computing 5
So my request is, if i change the value in the 'select' the 'courseCredits' value should appear in the textbox. for this how can i write jquery code?
"Ajax with Jquery" is what your are looking for. It will work like this:
the user chooses an option from the select box
you submit via Javascript the chosen option to a PHP script
the php script fetches the data from the database
the php script returns the result as json encoded data
there is a callback function in your javascript code. This js code will manipulate the HTML in whatever way you want, e.g. "add the option to your select box"
There are tons of tutorials on how to do Ajax requests in detail, e.g. http://openenergymonitor.org/emon/node/107
Check out one of those tutorials - eventually you will want to update your question so that it becomes a bit more specific? :-)
it is good practice to seperate you html from scripts so i would like to change :
<select name="course" id="course" onchange="valuesOfAll(this.value)">
to
<select name="course" id="course" >
then my script will be following (hoping you add reference of latest jquery )
<script>
$(document).ready(function(){
//bind change event once DOM is ready
$('#course').change(function(){});
getResult($(this).val());
});
function getResult(selectedValue){
//call ajax method to get data from database
$.ajax({
type: "POST",
url: url,//this should be replace by your server side method
data: "{'value': '"+ selectedValue +"'}", //this is parameter name , make sure parameter name is sure as of your sever side method
contentType: "application/json; charset=utf-8",
dataType: "json",
async: false,
success: function (Result) {
alert(Result.d);
},
error: function (jqXHR, textStatus, errorThrown) {
alert(errorThrown);
}
});
}
</script>
use $.post.. (ajax or get)... i am using post here....go through the docs if you want to read more about jquery post..
javascript
function valuesofAll(val){
$.post('test.php',{data:val},function(result){
$('#course_credits').val(result.coursecredit)
}, "json"); //expect response as json..
}
test.php
$data=$_POST['data']; //this is the value posted
//make you query in db get result..
$courseCredits= $row; //db returned courseCreadit value
echo json_encode(array('coursecredit'=>$courseCreadits)); //send response as json
this will helps you .....
<script>
function valuesOfAll(value)
{
var base_url="http://.../../hello.php";
var ip=new Object();
ip.course=value;
var inputParam=JSON.stringify(ip);
var module="getcourseCredits"
$.ajax({
type: "POST",
url: base_url,
data: {coursevalue:inputParam,module :module},
dataType: "json",
success: function(msg)
{
$("#course_credits").val(msg.credit);
}
});
}
</script>
<body>
<select name="course" id="course" onchange="valuesOfAll(this.value)">
<option value=""> select </option>
<option value="1"> Diploma in Computing</option>
</select>
<input name="course_credits" id="course_credits" type="text" />
</body>
In your PHP file
<?php
if(isset($_POST['module']))
{
if($_POST['module']=='getcourseCredits')
{
$val=json_decode($_POST['coursevalue'],true);
$courseval=$val['course'];
// do the connectivity and query here and finally echo the result as json format that is the response to the ajax call
.
.
.
.
}
}
?>

How to access individual values of form data, passed from an HTML page with jQuery.ajax(), inside a PHP script?

I'm passing form data to a PHP script for processing via JS(jQuery.ajax()).
Problem is - I can't figure out a way to access individual form control values inside PHP( e.g. $_POST['zipcode'] ).
Instead I can only access the data with $_POST['form'], which is an entire form represented as one long string( e.g. string(89)"color=red&color=blue&zipcode=12345..." ).
How can I access individual values of form data inside PHP script passed from a HTML form via JS?
index.php(form)
<form id="myform">
<select name="color" id="color">
<option value="Red">Red</option>
<option value="Green">Green</option>
<option value="Blue">Blue</option>
</select>
<input type="text" id="zipcode" name="zipcode" />
<input type="submit" id="submit" name="submit" value="Submit" />
</form>
index.php(JS)
$('#myform').on('submit', function(e) {
e.preventDefault();
$.ajax({
type: 'POST',
dataType: 'html',
url : 'PHPscript.php',
data: {form : $('#myform').serialize()}
}).done(function(data) {
var myJSONresult = data;
alert(myJSONresult);
});
});
PHPscript
<?php
if(isset($_POST["form"])){
$form = $_POST["form"];
$myzipcode = $_POST['zipcode']; // won't work; will be null or empty
echo json_encode($form);
}
?>
EDIT: The zipcode field:
$("#zipcode").focus(function(){
if(this.value == "zipcode"){
$(this).val("");
}
}).blur(function(){
if(this.value == ""){
$(this).val("zipcode");
}
});
You need to use serializeArray() on the form data instead of serialize. That will submit as an array.
data: $('#myform').serializeArray()
HTML
<input type="hidden" name="action" value="submit" />
PHP
if(isset($_POST["action"]))
{
//code
}
Add dataType: 'json' to your ajax handler and further modify your code like this:
$.ajax({
type: 'POST',
dataType: 'json', // changed to json
url : 'PHPscript.php',
data: {form : $('#myform').serialize()},
success : function(data){ // added success handler
var myJSONresult = data;
alert(myJSONresult.yourFieldName);
}
});
set traditional to true like
$.ajax({
traditional:true,
//your rest of the ajax code
});
on the php end you are getting the value fine the problem is at the form serialization end

Usng JSON in codeigniter with form submit

Im trying to learn how to use JSON from within codeigniter. I'm trying to use a model to run the ajax code, but its not working.
I need to learn how to pass variables through to the model, and back out; or, if thats incorrect, I need to learn the correct process. The below is my code.
HTML
<div id="structures">
<h1>Build</h1>
<form name="buildForm" id="buildForm" method="POST">
<select name="buildID" class="buildClass">
<option value="0" selected="selected" data-skip="1">Build a Structure</option>
<option name='Town Center' value='1' data-icon='../img/structures/tc.png' data-html-text='Town Center<i>
500 minutes<br>50000 gold</i>'>Town Center</option>
<option name='Barracks' value='2' data-icon='../img/structures/barracks.png' data-html-text='Barracks<i>
25 minutes<br>1500 gold</i>'>Barracks</option>
<option name='Dragon Roost' value='3' data-icon='../img/structures/droost.png' data-html-text='Dragon Roost<i>
200 minutes<br>5000 gold</i>'>Dragon Roost</option>
<option name='Mage Hall' value='4' data-icon='../img/structures/mage.png' data-html-text='Mage Hall<i>
40 minutes<br>300 gold</i>'>Mage Hall</option>
<option name='Test Lab' value='6' data-icon='../img/structures/testlab.png' data-html-text='Test Lab<i>
1 minutes<br>10 gold</i>'>Test Lab</option>
</select>
<div id="buildSubmit">
<input id ="btnSubmit" class="button" type="submit" value="Submit"/>
</div>
</form>
</div>
Here is my ajax/js/json
I'm trying a very simple example. I'd like to post the value of the OPTION above (1-6) to the model function insert_build. I don't know if its doing it, as I can't really think of a good way to test it. However, I would assume if it returned anything, I would be alerted. I am not.
$(function(){
$(".button").click(function(e, value){
e.preventDefault();
$.ajax({
type: "POST",
url: "<?php $this->structure_model->insert_build() ?>", //the script to call to get data
str_id: value,
dataType: 'json', //data format
success: function(data) //on receive of reply
{
alert("success!");
}
});
});
});
The model code
public function insert_build()
{
$str_id = $this->input->post('str_id');
echo " TESTING $str_id";
}
Any help would be greatly appreciated. Why doesn't this work? I am still trying to understand the fundamentals of Codeigniter and JSON / JS.
Thanks
The url: you're setting in your javascript is the actual function you're wanting to call when the ajax is submitted. You'll need to have a controller in place that actually calls that function e.g.
url: "<?php echo base_url(); ?>/structure/insert/?str_id=" + $('#buildID').val(),
Also, setting dataType: to json means you're expecting what's returned from your function to be encoded as json.
Maybe change this:
url: "<?php echo $this->structure_model->insert_build() ?>"

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