So I am submitting data to the database. Each data sent contains an id that is auto incremented. With ajax or PHP (I am very much new to this, and trying to learn I'm sure it's ajax along with some php) I need to fetch the id of the data that was submitted.
The idea is, after the form is submitted, the user gets the link back to the submitted page. Example:
Quote was submitted! [link] Click to go to the link or go back.
The link will look like this: http://example.com/quote-192
I pretty much have everything else set, I just don't know how I'll get the id and add it to the link.
Here is the PHP that processes the form:
require('inc/connect.php');
$quote = $_POST['quote'];
$quotes = mysql_real_escape_string($quote);
//echo $quotes . "Added to database";
mysql_query("INSERT INTO entries (quote) VALUES('$quotes')")
or die(mysql_error());
Oh, and the data is being sent with ajax:
$(document).delegate("'#submit-quote'", "submit", function(){
var quoteVal = $(this).find('[name="quote"]').val();
$.post("add.php", $(this).serialize(), function() {
var like = $('.quote-wrap span iframe');
$('.inner').prepend('<div class="quote-wrap group">' + like + '<div class="quote"><p>' + quoteVal+ '</p></div></div>');
// console.log("success");
});
return false;
});
So how would I get the id for each quote and add it to the page after the form has been submitted?
In you php:
echo mysql_insert_id($result)
Then in your jquery ajax:
$.ajax({
type:'post',
url:'url.php',
data:querystring,
success:function(data){
var id = parseInt(data);
}
]);
this will return the inserted ID as an integer value that you can work with in javascript
Have the PHP print the ID as a response to the request:
mysql_query("INSERT INTO entries (quote) VALUES('$quotes')")
or die(mysql_error());
// Print the id of last insert as a response
echo mysql_insert_id();
jQuery, test code to alert what was echoed by the PHP as a test
// add data as a param to the function to have access to the PHP response
$.post("add.php", $(this).serialize(), function(data) {
alert(data);
});
With this php-function. You can call after inserting.
int mysql_insert_id ([ resource $Verbindungs-Kennung ] )
mysql_insert_id
Related
I have a page with several buttons whose values and names are retrieved from the database. I'm trying to run an insert query on any button clicked, my code so far:
<?php
$sqlGetIllness = "SELECT * FROM illnissesandconditions ";
$resultGetIllness = $conn->query($sqlGetIllness);
while ($rowGetIllness= mysqli_fetch_array($resultGetIllness)){
echo "<div class='col-md-3'style='margin-top:20px;'><button onclick='insert(".$rowGetIllness['illness'].");' class='button button1' style=' color:white;' value='".$rowGetIllness['illness']."'>".$rowGetIllness['illness']."</button></div>";
}
function insert($value) {
$value='';
$sqlGetId = "SELECT commonID from common group by commonID DESC LIMIT 1 ";
$resultGetId = $conn->query($sqlGetId);
$r=mysqli_fetch_array($resultGetId);
$id=$r['commonID'];
$sqlGetIllness = "INSERT INTO medicalrecords (CommonID,Medical_Condition) VALUES (".$id.",'".$value."')";
$resultGetIllness = $conn->query($sqlGetIllness);
}
The value passed to the function inside onclick is correct when I inspect it in the browser, however nothing happens. I have a database connection on already, what could be wrong? Is it possible to do it like that in php without refreshing the page? Or do I need to use a client side lang like AJAX? Please note that I've never worked in AJAX btw.
New EDIT:
<script>
$("button").click(function(e) {
e.preventDefault();
$.ajax({
type: "POST",
data: {
condition: $(this).val(), // < note use of 'this' here
},
success: function(result) {
alert('Condition Inserted!');
},
error: function(result) {
alert('error');
}
});
});
</script>
Solution:
I got it worked out, after writing the script, i retrieved the variable value on top of the page
if (isset($_POST['condition'])) {
$value=$_POST['condition']; }
inside $_SERVER['REQUEST_METHOD'] == 'POST' ) and now it inserts the value when ever any button is clicked, my next step is to give the clicked button a background color
Solution is in the post under Solution, was my first time trying ajax and it did work indeed, gave the button an id, and took its value ( any button clicked ) through this.val and sent via post, retrieved and used the value in a variable for the insert query.
What I have
An url like this: http://localhost:8888/website/?key=ABC
A MySQL table with many rows, one with a key called ABC. I have a button that I want users to click (upvote) the MySQL row corresponding to key=ABC.
In my scripts.js I have this (incomplete?) Ajax code:
function increment() {
$.ajax({
type: 'POST',
url: '../js/ajax.php',
success: function(data) {
alert("function saved: " + data);
}
});
}
And my ajax.php looks like this:
<?php
if (isset($_GET['key']) {
$rowKEYtest = $_GET['key'];
$sql2 = "UPDATE database SET Score = Score + 1 WHERE UniqueKey = '$rowKEYtest'";
$conn->query($sql2);
} else {
echo "lol";
}
?>
It's not updating. I have no idea what to do.
Please have a look in your code, There is 2 mistakes.
1. In your ajax call you are using type: 'POST', you should use GET.
2. You are nor parsing your 'key' param in ajax call.
Actually you are using POST method in ajax but in ajax.php trying to get the value with GET method. second thing you are not passing any param in ajax call.
Hope this will help.
So I got this form which lets the user enter up to 10 item IDs and then check for the status or apply changes right away. If the user clicks check status the item IDs will be sent to another PHP.
What is the easiest way to send the result back to the form and display it in the red area?
if ($_POST['action'] == 'Check Status/Shelf') {
$itemids = "$itemID1, $itemID2, $itemID3, $itemID4, $itemID5, $itemID6, $itemID7, $itemID8, $itemID9, $itemID10";
while ($row = mysql_fetch_array($all)) {
$hyllplacering = $row['Hyllplacering'] . "";
$all = mysql_query("SELECT Hyllplacering, itemID FROM booking WHERE itemID IN ('$itemids')");
}
}
If you don't want to use Ajax, then save the details into a $_SESSION[] and upon submission the form is posted and then the details can be populated into SESSION and displayed back out to the form upon reloading, but it's a bit of a fiddle for not much return.
And use MySqli or PDO.
For that you have to use ajax
Give id to Apply Changes & Check Status/Self button
<script type="text/javascript">
$(document).ready(function(){
$('#apply').click(function() {
var id1 = $('#id1').val();
var id2 = $('#id2').val();
//same for all your 10 field
var cstring = "id1="+id1+"&id2="+id2 // etc;
$.ajax({
type: "POST",
url: "your file url to do database opration & fetch detail back to here",
data: cstring,
cache: false,
success: function(result) {
$('.resultdata').html(result)
}
});
});
});
</script>
<div class="resultdata"></div>
I started using jquery and jquery-ui. I am facing a problem trying to submit a form that is inside a tab. Once i hit 'submit', the page refresh itself, the tabs are gone, and the url is changed to the tab's function url. the submit button is working, and i get the desired result. However, it is not on the same page as the tabs.
Does anyone have any idea on how to keep the page from refreshing?
example of my problem:
I have a page called 'index.php' with 3 different tabs. one of the tabs is called 'submit form' where there I have a form using POST method, it is taking its source from 'form.php'. once i hit the 'submit' button, the page refreshes, the url changes from 'www.example.com' to 'www.example.com/form.php', i get the result there, but the page is "plain", means that its not under a tab, just a normal page.
I hope I explained myself correctly.
Thanks!
EDIT:
here is my submission code:
$submit = $_POST['submit'];
$name= $_POST['name'];
if($submit){
echo 'submitted <br><br>';
echo "hello $name";
}
Capture the form submission in jQuery and then stop it with preventDefault();
$(function(){
$('#formID').on('submit', function(e){
e.preventDefault();
var formSrc = $(this).attr('action');
var formMethod = $(this).attr('method');
var formData = $(this).serialize();
$.ajax({
url: formSrc,
type: formMethod,
data: formData,
success: function(data){
//work with returned data from requested file
alert(data);
}
});
});
});
EDIT:
i should add that I chose to use on(); as opposed to the default submit(). This is because the form may or may not be available at DOM.ready.
In answer to your 2nd question: "what if I wanned to run this information returned through a php code in the same tab, how could I do that?"
Have your php script return values as JSON like:
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
echo json_encode($row);
Then you can have javascript do whatever you wish with the returned values, such as -
$.ajax({
url: 'sc_getData.php?id=' + ID,
type:'GET'
})
.done(function(myData) { // data = array returned from ajax
myData = JSON.parse(myData);
for (var i in tFields){
thisSourceField = sFields[i];
thisTargetField = tFields[i];
targetID = '#' + thisTargetField;
thisValue = myData[thisSourceField];
$(targetID).val( thisValue );
console.log(targetID + ': ' + thisValue);
}
})
I've setup arrays for Target and Source Fields. The source field arrays match the field names returned by the php script while the target field arrays will match the form field id's to be filled with the returned values.
The attached picture shows the results page of the search engine that I'm building. For each return result, the user may click on the result (i.e. "Food Science") and it will expand out accordion-style to reveal information about that particular result.
I want to log each time the user clicks on a result (for learning/intelligence purposes) and store it in a database table that I have created which stores the session ID, the query, the position of the result, and the order in which the user clicked the item.
Using JQuery, I already have a function that will pull the title of the result that was clicked, and I have it set where I want to log the click, but I don't know how to do it since JQuery is client side and PHP is server side.
How can I use the JQuery to trigger a PHP function so that I can query the database to insert the click logs into my table?
Below is the JQuery function.
$(document).ready(function() {
$('.accordionButton').click(function(e) {
if($(this).next().is(':hidden') == true) {
$(this).addClass('on');
$(this).next().slideDown('normal');
$(this).next().slideDown(test_accordion);
// SEND CLICK ACTION TO LOG INTO THE DATABASE
alert($(this).find('h3:last').text()); // displays the title of the result that was just clicked
}
else {
$(this).removeClass('on');
$(this).next().slideUp('normal');
$(this).next().slideUp(test_accordion);
}
});
}
You can do something like this (untested):
Define a javascript variable to track the order of the clicks, outside your click function:
var order = 0;
Add this into your click function, at the bottom:
order++;
var sessionID = $("input[name='sessionID']").val(); // assuming you have sessionID as the value of a hidden input
var query = $("#query").text(); // if 'query' is the id of your searchbox
var pos = $(this).index() + 1; // might have to modify this to get correct index
$.post("logClick.php", {sessionID:sessionID, query:query, pos:pos, order:order});
In your php script called "logClick.php" (in the same directory):
<?php
// GET AJAX POSTED DATA
$str_sessionID = empty($_POST["sessionID"]) ? '' ; $_POST["sessionID"];
$str_query = empty($_POST["query"]) ? '' ; $_POST["query"];
$int_pos = empty($_POST["pos"]) ? 1 ; (int)$_POST["pos"];
$int_order = empty($_POST["order"]) ? 1 ; (int)$_POST["order"];
// CONNECT TO DATABASE
if ($str_sessionID && $str_query) {
require_once "dbconnect.php"; // include the commands used to connect to your database. Should define a variable $con as the mysql connection
// INSERT INTO MYSQL DATABASE TABLE CALLED 'click_logs'
$sql_query = "INSERT INTO click_logs (sessionID, query, pos, order) VALUES ('$str_sessionID', '$str_query', $int_pos, $int_order)";
$res = mysql_query($sql_query, $con);
if (!$res) die('Could not connect: ' . mysql_error());
else echo "Click was logged.";
}
else echo "No data found to log!";
?>
You can add a callback function as a third parameter for the $.post() ajax method if you want to see if errors occured in the script:
$.post("logClick.php", {sessionID:sessionID, query:query, pos:pos, order:order},
function(result) {
$('#result').html(result); // display script output into a div with id='result'
// or just alert(result);
})
);
EDIT: If you need the value of the order variable to persist between page loads because you paginated your results, then you can pas the value of this variable between pages using either GET or POST. You can then save the value in a hidden input and easily read it with jQuery. (Or you could also use cookies).
Example (put this in every results page):
<?php
$order = empty($_POST["order"]) ? $_POST["order"] : "0";
$html="<form id='form_session' action='' name='form_session' method='POST'>
<input type='hidden' name='order' value='$order'>
</form>\n";
echo $html;
?>
In your jQuery, just change var order = 0; to
var order = $("input[name='order']").val();
Then, when a user clicks on a page link, prevent the default link action, set the order value and the form action, and then submit the form using javascript/jQuery:
$("a.next_page").click(function(event) {
event.preventDefault();
var url = $(this).attr("href");
$("input[name='order']").val(order);
$("#form_session").attr('action', url).submit();
});
All the 'next' and 'previous' pagination links must be given the same class (namely 'next_page' (in this example).
EDIT: If your pagination is as follows:
<div class='pagination'>
<ul><li><a href='page1.url'>1</a></li>
<li><a href='page2.url'>2</a></li>
</ul>
</div>
then just change this:
$("div.pagination a").click(function(event) {
etc.
This one is pretty easy, you need a PHP-Script to handle AJAX requests which are sent from your Search page.
In your search page you'll need to add an .ajax to create an AJAX request to your Script.
Everything you need to know about AJAX can be found here: http://api.jquery.com/jQuery.ajax/
In your PHP-Script you'll handle the Database action, use GET or POST data to give the script an ID over Ajax.
Use Ajax. Write a simple php-script that writes clickes to the database. I don't know how you log the clicks in the database exactly, but you can send the clicked item unique identifier to a php script with ajax, for example via POST variables.
A little example, on click:
$.post(
'count_click.php',
{ id: "someid" },
function(data) {
// data = everything the php-script prints out
});
Php:
if (isset($_POST['id'])) {
// add a click in the database with this id
}
Send a request to a PHP page using jQuery AJAX. See here for more info (it is really simple):
http://api.jquery.com/jQuery.ajax/
In this particular case, as you do not need to return anything, it may be better to just use the POST or GET methods in jQuery:
http://api.jquery.com/jQuery.post/
http://api.jquery.com/jQuery.get/
Something like:
$.ajax({
type: "POST",
url: "some.php",
data: "name=John&location=Boston"
success: function(data){
alert('done');
});