With this code I can see empty data saving into my db once I click on the link button. But i need to the GET's data in my database. Any solution to this.
<a href="?id=1&pid=238874" id="day" onclick="this.form.submit();><button type="button" class="label label-danger" >Select</button></a>
<script>
$('#day').click(function(e)
{
e.preventDefault();
$.ajax({
type: 'post',
url: "index.php",
data: $("select.day").serialize(),
});
return false;
});
</script>
For PHP Code to save data
<?php
$a = $_GET['id'];
$b = $_GET['pid'];
// query
$sql = "INSERT INTO hos_patient(reg_id,pid) VALUES ('$a','$b')";
mysqli_query($db, $sql);
?>
Thanks
You specified using POST in your ajax function. But then you try to get the data by GET in your PHP-Script. I'd suggest you just use POST for this.
Alter your PHP-Script to use $_POST instead of $_GET
Try with this :
...
$.ajax({
type: 'post',
url: "index.php?id=1&pid=238874",
data: $("select.day").serialize(),
});
...
Use this in script file and in PHP file use below code
<script>
$('#day').click(function(e)
{
e.preventDefault();
$.ajax({
type: 'post',
url: "index.php?id=1&pid=238874",
data: $("select.day").serialize(),
});
return false;
});
</script>
<?php
$a = $_POST['id'];
$b = $_POST['pid'];
// query
$sql = "INSERT INTO hos_patient(reg_id,pid) VALUES ('$a','$b')";
mysqli_query($db, $sql);
?>
<button type="button" class="label label-danger" >Select</button>
<script>
$(function () {
$('.login_form_1').on('click', function (e) {
e.preventDefault();
$.ajax({
type: 'GET',
url: 'show.php?id=22&pid=33',
data: $('.login_form_1').serialize(),
success: function (data) {
$('div.logerrors').html(data);
}
});
});
});
</script>
For PHP output
<?php
$a = $_GET['id'];
$b = $_GET['pid'];
// query
$sql = "INSERT INTO hos_patient(reg_id,pid) VALUES ('$a','$b')";
mysqli_query($db, $sql);
?>
Related
I have a list of divs with unique IDs (they are inserted from my database). When I click on one of them I want to display content from my database in another div. For example, I have a div with class pizza. The query should look like this: SELECT * FROM product WHERE name = 'pizza'. So depending on what div you click you get different content. The code below doesn't work and is incomplete. I was trying to do some research myself, but I couldn't find anything useful.
//head
<script>
$(function () {
$('.product').on('click', function (e) {
e.preventDefault();
$.ajax({
type: "post",
url: 'php/recipe-container.php',
data: new FormData(this),
processData: false,
contentType: false,
success: function(response) {
$(".display_recipe").html(response);
},
error: function () {
}
});
});
});
</script>
//HTML
<div class="product" id="pizza">pizza</div>
<div class="product" id="lasagna">lasagna</div>
<div class="product" id="sushi">sushi</div>
<div class="display_recipe"></div>
// PHP (recipe-container.php)
<?php
function display_recipe(){
$con = mysqli_connect("localhost", "root", "", "cookbook");
$product = "'pizza'"; //just a placeholder
$sql = "SELECT * FROM product WHERE name = $product";
$res = mysqli_query($con,$sql);
while($row = mysqli_fetch_assoc($res)) {
$name = $row['name'];
$description = $row['description'];
$date = $row['date'];
echo $name;
echo "<br>";
echo $description;
echo "<br>";
echo $date;
echo "<br>";
}
mysqli_close($con);
}
display_recipe();
?>
Right now when I click the button nothing happens, even "pizza" placeholder doesn't work. Is there a simple way to do it?
JS file (AJAX code)
You can get the id attribute on click of the div with the class 'product' as coded below:
jQuery(function () {
jQuery('.product').on('click', function (e) {
var product = jQuery(this).attr('id');
$.ajax({
type: "post",
url: 'php/recipe-container.php',
data: {data:product},
processData: false,
contentType: false,
success: function(response) {
$(".display_recipe").html(response);
}
});
});
});
PHP file: get the posted data in this file use it in a query to fetch the result and return the result to the AJAX success handler as a response.
To fetch the data posted from the ajax in this php file you can use $_POST['data'] as stated below:
$product = $_POST['data'];
Use that variable in your sql query to fetch the result and then change the structure of your response as stated below:
//saving the html response in a variable named "response"
$response = $name.'<br>';
$response .= $description.'<br>';
$response .= $date.'<br>';
//echo response will send the response variable back to the AJAX success handler.
echo $response;
I am trying to make some code where a click of a SQL html button can display a different SQL table as a popup. I have already gotten the variable from the table to pass through using this:
echo "<td><a class='btn-floating btn-large waves-effect waves-light black' onclick='partSearch(".$product.")' value='display'><i class='material-icons'>search</i></a></td>";
The 'part search' code is as follows:
<script type="text/javascript">
function partSearch() {
$.ajax({
url: 'serv.php?id=<?php echo $product ?>',
type: 'GET',
success: function(result){
var obj = jQuery.parseJSON(result)
alert(obj)
}
})
}
</script>
Even though the variable is passed through to 'serv.php', I can't manage to get the sql data to be returned as a popup using alert. All I get is either nothing or [object, Object]. This is the SQL/php code:
<?php
include 'includes/dbh.inc.php';
$id = $_GET['id'];
$result = mysqli_query($conn,"SELECT * FROM pr WHERE product_ID='".$id."'");// test this
$rows = array();
while($r = mysqli_fetch_assoc($result)){
$rows[] = $r;
}
echo json_encode($rows);
?>
Any help is appriciated
<script type="text/javascript">
function partSearch() {
var product_id = "<?php echo $product; ?>";
$.ajax({
url: 'serv.php?id=product_id',
type: 'GET',
success: function(result){
alert(result);
}
})
}
</script>
I need help on how to post and retrieve multiple variables using Ajax Post. I actually needed to retrieve the posted variables for SQL query. See below the Ajax Code where i needed to include variable names selschool, selprogram, selsession to the post
<script>
$("#session").change(function()
{
$("#loding2").show();
var id=$(this).val();
var dataString = 'id='+ id;
var selschool=document.getElementById("selectedschool").val();
var selprogram=document.getElementById("selectedprogram").val();
var selsession=document.getElementById("selectedsession").val();
$("#semester").find('option').remove();
$("#class").find('option').remove();
document.getElementById("selectedclass").value= " ";
document.getElementById("selectedsemester").value= " ";
$.ajax
({
type: "POST",
url: "get_class.php",
data: dataString,
cache: false,
success: function(html)
{
$("#loding2").hide();
$("#class").html(html);
}
});
});
</script>
Also see below PHP script where i wanted to use the posted variable for the query;
<?php
include('dbconfig.php');
if($_POST['id'])
{
$id=$_POST['id'];
// Todo: I actually needed something like where session SELECT * FROM class where session_id=$id and program_id="selprogram" and school_id="selschool"
$stmt = $DB_con->prepare("SELECT * FROM class where session_id=$id ");
$stmt->execute(array(':id' => $id));
?><option selected="selected">Select Class :</option>
<?php while($row=$stmt->fetch(PDO::FETCH_ASSOC))
{
?>
<option value="<?php echo $row['class_id']; ?>"><?php echo $row['class_name']; ?></option>
<?php
}
}
?>
Let me explain the solution
consider the ajax call to demo.php
$.ajax({
url: 'demo.php',
type: 'post',
data: {
'name': 'abc',
'phone': '1234567899'
}, //data is in json form
success: function(res) {
console.log(JSON.parse(res)); //parsing because we will pass the data from demo.php in encoded form you will get it.
}
});
now in demo.php you will access data as $_POST['name'] and $_POST['phone']. lets pass the same to ajax call. will store it in array and will pass it.
<?php
$Arr = [];
$Arr[0] = $_POST['name'];
$Arr[1] = $_POST['phone'];
echo json_encode($Arr);
?>
like this, we can pass data to ajax and can pass the data from PHP file to request.
Hope that you got the result. Thank you.
I am trying to get data from one php page and pass it to another page using Ajax.
JS :
$.ajax({
url: "action.php",
success: function(data){
$.ajax({
url: "data.php?id=data"
}
});
action.php :
<?php
$test= 1;
?>
data.php :
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script type="" src="action.js"></script>
<?php
$id = $_GET['id'];
echo $id;
?>
First of all, you need to echo your data in action.php, and second, use data parameter of AJAX request to send data to data.php.
Here's the reference:
jQuery.ajax()
So the organization of pages should be like this:
JS :
$.ajax({
url: "action.php",
success: function(data){
$.ajax({
url: "data.php",
data: {id: data},
success: function(data){
// your code
// alert(data);
}
});
}
});
action.php :
<?php
$test = 1;
echo $test;
?>
data.php :
<?php
$id = $_GET['id'];
echo $id;
?>
Try to use $.get() method to get/send data :
$.get("action.php",{}, function(data){
//data here contain 1
$.get("data.php", {id: data}, function(id){
alert(id);
}
});
Just echo $test since just the data printed in page will return as responce to the query request.
action.php :
<?php
$test=1;
echo $test;
?>
Hope this helps.
For example,
Test
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script type="" src="action.js"></script>
action.js
$('.dataClass').click(function(){
var value=$(this).attr('data-value');
$.ajax({url:"Ajax_SomePage.php?value="+value,cache:false,success:function(result){
alert("success");
}});
});
Ajax_SomePage.php
<?php
$value = $_GET['value'];
echo $value;
?>
To get data as response in ajax call, you need to echo the result from your php page; action.php page.
echo $test = 1;
In your provided code
$.ajax({
url: "data.php?id=data"
} // closing bracket is missing
you are sending the string data as id to data.php page. Instead you have to append the result with the url using + symbol like shown in the below code.
$.ajax({
url: "action.php",
success: function(data){
$.ajax({
url: "data.php?id="+data
})
}
});
I just need some ideas how i get the value from the dropdown
<select id="ins" name="instructor" required>
<?php
//dropdown instructor name
echo "<option value=\"\">"."Select"."</option>";
$qry = "select * from instructor";
$result = mysqli_query($con,$qry);
while ($row = mysqli_fetch_array($result)){
echo "<option value=".$row['insA_I'].">".$row['insname']."</option>";
}
?>
</select>
and put the value in this .php file in $_POST['instructor']
<?php
session_start();
require 'connection.php'; //connection
$qry = "select course from instructor where insA_I = '".$_POST['instructor']."'";
$result = mysqli_query($con,$qry);
$_SESSION['row'] = mysqli_fetch_array($result);
$data = $_SESSION['row']['course'];
echo $data;
?>
am still new in ajax
$(document).ready(function() {
$('#ins').change(function(){
$.ajax({
type: "POST",
url: "json_php.php",
data: '',
datatype: 'json',
success: function(result){
alert(result);
}
});
});
});
i already searched this in internet but didn`t get my answer after some modifcation
just need idea, techniques .etc.
Try this ajax code
$(document).ready(function() {
$('#ins').change(function(){
$.ajax({
type: "POST",
url: "json_php.php",
data: {instructor:$(this).val()},
success: function(result){
alert(result);
}
});
});
});
php code for query
$qry = "select course from instructor where insA_I = '".$_POST['instructor']."'";
$result = mysqli_query($con,$qry);
$fetch = mysqli_fetch_array($result);
$_SESSION['row']['course']=$fetch['course'];
$data = $_SESSION['row']['course'];
echo $data;
Since you're using jQuery, use jQuery:
data: { instructor: $('#ins').val() }
Also, please note that your SQL query is wide open to SQL injection. This is a good place to start reading about that.