Develop Reference

Ajax Search of MySql table to create href link from result

I currently have an Ajax search function that works and returns results.
However i am unable to get the search result to become a href link so that it adds the 'payroll_number' the search finds to the url to become candidatecard.php?pr=1388 for example. Where 1388 is the payroll number
<a href="candidatecard.php?pr='.$Result['payroll_number'].'">
My web page code:
<h2>Payroll <span>Number Search</span></h2>
<!-- Search box. -->
<input type="text" id="search" placeholder="Payroll Number Search" />
<b>Eg: </b><i>5100, 6237, 7259. </i>
<!-- Suggestions will be displayed in below div. -->
<?php
echo "<a href='candidatecard.php?pr=$pn'>Search $pn</a>";
?>
<div id="display"></div>
My js code
//Getting value from "ajax.php".
function fill(Value) {
//Assigning value to "search" div in "search.php" file.
$('#search').val(Value);
//Hiding "display" div in "search.php" file.
$('#display').hide();
}
$(document).ready(function() {
//On pressing a key on "Search box" in "search.php" file. This function will be called.
$("#search").keyup(function() {
//Assigning search box value to javascript variable payroll_numberd as "payroll_number".
var payroll_number = $('#search').val();
//Validating, if "payroll_number" is empty.
if (payroll_number == "") {
//Assigning empty value to "display" div in "search.php" file.
$("#display").html("");
}
//If payroll_number is not empty.
else {
//AJAX is called.
$.ajax({
//AJAX type is "Post".
type: "POST",
//Data will be sent to "ajax.php".
url: "ajax.php",
//Data, that will be sent to "ajax.php".
data: {
//Assigning value of "payroll_number" into "search" variable.
search: payroll_number
},
//If result found, this funtion will be called.
success: function(html) {
//Assigning result to "display" div in "search.php" file.
$("#display").html(html).show();
}
});
}
});
});
My Ajax code:
<?php
//Including Database configuration file.
include "Include/db.php";
//Getting value of "search" variable from "script.js".
if (isset($_POST['search'])) {
//Search box value assigning to $payroll_number variable.
$payroll_number = $_POST['search'];
//Search query.
$Query = "SELECT payroll_number FROM candidatedetails WHERE payroll_number LIKE
'%$payroll_number%' LIMIT 10";
//Query execution
$ExecQuery = MySQLi_query($con, $Query);
//Creating unordered list to display result.
echo '
<ul>
';
//Fetching result from database.
while ($Result = MySQLi_fetch_array($ExecQuery)) {
?>
<!-- Creating unordered list items.
Calling javascript function named as "fill" found in "script.js" file.
By passing fetched result as parameter. -->
<li onclick='fill("<?php echo $Result['payroll_number']; ?>")'>
<a>
<!-- Assigning searched result in "Search box" in "search.php" file. -->
<?php echo $Result['payroll_number'];?>
</li></a>
<!-- Assign searchbox result to pn variable.-->
<?php$pn = $Result['payroll_number'];?>
<!-- Closing parenthesis. -->
<?php
}}
?>
</ul>

Returning a value from server to html for manipulation

I want to manipulate the value that it is stored in this variable $result[]. Specifically i want to return that value from php file to my html file. What should i do? Can you give me some reference code when i want to return things from server side to client side for further manipulation?
My php file:
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<?php
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("127.0.0.1", "root", "", "mysql3");
// Check connection
if($link === false) {
die("ERROR: Could not connect. " . mysqli_connect_error());
}
$user_id =$_POST['user_id'];
$book_id =$_POST['book_id'];
$game_id =$_POST['game_id'];
$site_id =$_POST['site_id'];
//Attempt insert query execution
$query = "SELECT site_id FROM components WHERE user_id='$user_id' && book_id='$book_id' && game_id='$game_id' ORDER BY site_id DESC LIMIT 1";
$res = mysqli_query($link,$query);
$result = array();
$res = mysqli_query($link,$query) or die(mysqli_error($link));
while($row = mysqli_fetch_assoc($res)){
$result[]=$row['site_id'];
}
// Close connection
mysqli_close($link);
?>
My html file:
<html>
<head>
<script src="https://code.jquery.com/jquery-3.3.1.min.js" integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8=" crossorigin="anonymous"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<script src="https://unpkg.com/sweetalert/dist/sweetalert.min.js"></script>
<script>
function load3() {
var flag1 = true;
do{
var selection = window.prompt("Give the User Id:", "Type a number!");
if ( /^[0-9]+$/.test(selection)) {
flag1=false;
}
}while(flag1!=false);
$("#user_id").val(selection)
//$("#user_id").val(prompt("Give the User Id:"))
var flag2 = true;
do{
var selection2 = window.prompt("Give the Book Id:", "Type a number!");
if ( /^[0-9]+$/.test(selection2)) {
flag2=false;
}
}while(flag2!=false);
$("#book_id").val(selection2)
//$("#book_id").val(prompt("Give the Book Id:"))
var flag3= true;
do{
var selection3 = window.prompt("Give the Game Id:", "Type a number!");
if ( /^[0-9]+$/.test(selection3)) {
flag3=false;
}
}while(flag3!=false);
$("#game_id").val(selection3)
//$("#game_id").val(prompt("Give the Game Id:"))
}
var fieldNameElement = document.getElementById('outPut');
if (fieldNameElement == 4)
{
window.alert("bingo!");
}
</script>
</head>
<body>
<form name="LoadGame" id="LoadGame" method="post" action="http://127.0.0.1/PHP/mine1.php" enctype="multipart/form-data">
<input type="submit" value="Load" id="load" onclick="load3()" class="button12" />
<input type="hidden" name="user_id" id="user_id">
<input type="hidden" name="book_id" id="book_id">
<input type="hidden" name="game_id" id="game_id">
<input type="hidden" name="site_id" id="site_id">
</form>
</body>
</html>

Note that this answer is making use of jQuery notation, so you will need to include a jQuery library in your application in order to make this example work:
<script src="/js/jquery.min.js" type="text/javascript"></script>
Since you have your HTML and PHP in separate files, you can use AJAX to output your HTML in an element you so desire on your HTML page.
Example of jQuery AJAX:
<script>
function submitMyForm() {
$.ajax({
type: 'POST',
url: '/your_page.php',
data: $('#yourFormId').serialize(),
success: function (html) {
//do something on success?
$('#outPut').html(html);
var bingoValue=4;
if( $('#outPut').text().indexOf(''+bingoValue) > 0){
alert('bingo!');
}
else {
alert('No!');
}
}
});
}
</script>
Note that I encapsulated the AJAX function in another function that you can choose to call onclick on a button for example.
<button id="mySubmitButton" onclick="submitMyForm();">Submit form!</button>
Step-by-step:
What we do in our AJAX function, is that we declare our data type, just like you would do with a form element. In your PHP file I noticed that you used the POST method, so that's what I incorporated in the AJAX function as well.
Next we declare our URL, which is where the data will be sent. This is the same page that your current form is sending the data to, which is your PHP page.
We then the declare our data. Now, there are different ways of doing this. I assume you are using a form currently to POST your data to your PHP page, so I thought we might as well make use of that form now that you have it anyways. What we do is that we basically serialize the data inside your form as our POST values, just like you do on a normal form submit. Another way of doing it, would be to declare individual variables as your POST variables.
Example of individual variables:
$.ajax({
type: 'POST',
url: '/your_page.php',
data: {
myVariable : data,
anotherVariable : moreData
//etc.
},
success: function (html) {
//do something on success?
$('#outPut').html(html);
}
});
A literal example of a variable to parse: myVariable : $('input#bookId').val().
The operator : in our AJAX function is basically an = in this case, so we set our variable to be equal to whatever we want. In the literal example myVariable will contain the value of an input field with the id bookId. You can do targeting by any selector you want, and you can look that up. I just used this as an example.
In the success function of the AJAX function, you can basically do something upon success. This is where you could insert the HTML that you wish to output from your PHP page into another element (a div for example). In my AJAX example, I am outputting the html from the PHP page into an element that contains the id outPut.
We also write a condition in our success function (based off comments to this answer), where we check for a specific substring value in the div element. This substring value is defined through the variable bingoValue. In my example I set that to be equal to 4, so whenever "4" exists inside the div element, it enters the condition.
Example:
<div id="outPut"></div>
If you make use of this example, then whatever HTML you structure in your PHP file, making use of the PHP values in your PHP file, will be inserted into the div element.
<?php
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("127.0.0.1", "root", "", "mysql3");
// Check connection
if($link === false) {
die("ERROR: Could not connect. " . mysqli_connect_error());
}
$user_id =$_POST['user_id'];
$book_id =$_POST['book_id'];
$game_id =$_POST['game_id'];
$site_id =$_POST['site_id'];
//Attempt insert query execution
$query = "SELECT site_id FROM components WHERE user_id='$user_id' && book_id='$book_id' && game_id='$game_id' ORDER BY site_id DESC LIMIT 1";
$res = mysqli_query($link,$query);
$result = array();
$res = mysqli_query($link,$query) or die(mysqli_error($link));
while($row = mysqli_fetch_assoc($res)){
$result=$row['site_id'];
echo $result.' ';
}
// Close connection
mysqli_close($link);
?>
Your form also no longer needs an action defined as all of that is now taken care of by the AJAX function.
So change:
<form name="LoadGame" id="LoadGame" method="post" action="http://127.0.0.1/PHP/mine1.php" enctype="multipart/form-data">
to:
<form name="LoadGame" id="LoadGame" method="post" enctype="multipart/form-data">
And make sure that your button: <button id="mySubmitButton" onclick="submitMyForm();">Submit form!</button> is outside of your form tag, as buttons without a defined type attribute will have type="submit" by default inside a form tag.
If you need anything elaborated, let me know. :)

First of all: remove the script tag from your php.
Secondly: Why are you executing the sql statement two times?
To your question:
You have to send a request to your PHP script via AJAX: (Place this inside <script> tags and make sure to include jquery correctly)
$(() => {
$('form').on('submit', () => {
event.preventDefault()
$.ajax({
type: 'POST',
url: '<your-php-file>', // Modify to your requirements
dataType: 'json',
data: $('form').serialize() // Modify to your requirements
}).done(function(response){
console.log(response)
}).fail(function(){
console.log('ERROR')
})
})
})
Your PHP-Script which needs to return JSON:
$query = "SELECT site_id FROM components WHERE user_id='$user_id' && book_id='$book_id' && game_id='$game_id' ORDER BY site_id DESC LIMIT 1";
// Execute Query
$res = mysqli_query($link,$query) or die(mysqli_error($link));
// Get Rows
while($row = mysqli_fetch_assoc($res)){
$result[] = $row['site_id'];
}
// Return JSON to AJAX
echo json_encode($result);
Take a look at your developer console.
Haven't tested it.

Changing value of a PHP string based on a condition check with AJAX

Basically, on my HTML display page, mainHTML.php I include a PHP file called phpFunctions.php with some of the PHP functions I made and call these functions. One of them in particular gets information from a SQL database and makes a table from it.
On mainHTML.php, I have an ajax script that gets the value from a select box that is on same page, on change.
<script>
$(document).ready(function() {
$('select[name="selectBox"]').change(function(){
var value = $(this).val();
$.ajax({
type: 'POST',
url: 'ajaxScript.php', //if bug change to ajaxScript.php
data: {valueChange: value },
dataType: 'html'
}).done(function(response){
$('.response-holder').html(response);
});
});
});
</script>
My ajaxScript looks like so. At the moment, all I can get it to do is echo something based on the value of the select box.
<?php
$status = $_POST['valueChange'];
if($status == 'height'){
echo "ORDER BY height";
}
if($status == 'weight'){
echo "ORDER BY weight";
}
?>
All of this works fine but then I come into a problem. I don't know how I can append the SQL string present in a function in phpFunctions.php so that it may be modified based on the ajaxScript.php condition.
Thanks in advance.

Passing a php variable to another php file using ajax

I have some divs with class content-full which are created according to the data stored in a MySQL table. For example, if I have 3 rows there will be 3 divs.
When I click on any of the divs then the data associated with the ID for that div should be displayed inside div content-full. However, the content associated with the last div appears when I click on any of the divs because I am not passing any kind of variable ID that can be used to specify the clicked div.
<script type="text/javascript">
$(document).ready(function() {
$(".content-short").click(function() { //
$.ajax({
type: "GET",
url: "content.php", //
dataType: "text",
success: function(response){
$(".content-full").html(response);
}
});
});
});
</script>
content.php
<?php
include "mysql.php";
$query= "SELECT Content FROM blog limit 1";
$result=mysql_query($query,$db);
while($row=mysql_fetch_array($result))
{
echo $row['Content'];
}
mysql_close($db);
?>
Is there an alternate way to do this or something that I need to correct for it to work?
Thanks

First off, you will have to pass actual data to the your PHP script. Let's first deal with the JavaScript end. I am going to assume that your HTML code is printed using php and you are able to create something like the following: <div class=".content-short" data-id="2" />:
$(".content-short").click(function() {
// get the ID
var id = $(this).attr('data-id');
$.ajax({
type: "post",
url: "content.php",
data: 'id=' + id
dataType: "text",
success: function(response){
$(".content-full").html(response);
}
});
});
Next up is reading the value you passed in using the script:
<?php
include "mysql.php";
// Use the $_POST variable to get all your passed variables.
$id = isset($_POST["id"]) ? $_POST["id"] : 0;
// I am using intval to force the value to a number to prevent injection.
// You should **really** consider using PDO.
$id = intval($id);
$query = "SELECT Content FROM blog WHERE id = " . $id . " limit 1";
$result = mysql_query($query, $db);
while($row=mysql_fetch_array($result)){
echo $row['Content'];
}
mysql_close($db);
?>
There you go, I have modified your query to allow fetching by id. There are two major caveats:
First: You should not use the mysql_ functions anymore and they are deprecated and will be removed from PHP soon if it has not happened yet!, secondly: I cannot guarantee that the query works, of course, I have no idea what you database structure is.
The last problem is: what to do when the result is empty? Well, usually AJAX calls send and respond with JSON objects, so maybe its best to do that, replace this line:
echo $row['Content'];
with this:
echo json_encode(array('success' => $row['Content']));
Now, in your success function in JS, you should try to check if there a success message. First of, change dataType to json or remove that line entirely. Now replace this success function:
success: function(response){
$(".content-full").html(response);
}
into this:
success: function(response){
if(response.success) $(".content-full").html(response);
else alert('No results');
}
Here the deprecation 'notice' for the mysql_ functions: http://php.net/manual/en/changelog.mysql.php

You can look at passing a parameter to your script, generally like an id by which you can search in the db.
The easiest way to achieve this is by get parameters on the url
url: "content.php?param=2",
Then in php:
<?php $param = $_GET['param'];
...

Code returns the latest value before last refresh instead of the latest value inserted?

I have a column of buttons in a table, declared like this:
(file index.php)
echo '';
Then this script reads the data in the row of the button clicked and posts it to another php file:
<!-- scripts that gets the lecturer chosen to SHOW functionality-->
<script>
$(document).ready(function(){
$(".show-button").click(function() {
var $row = $(this).closest("tr"); // Find the row
var names = $row.find(".name").text(); // Find the name
var surname = $row.find(".surname").text(); // Find the surname
$.ajax({ type: "POST", url: "show_lecturer.php", data: { x: names, y: surname} })
});
});
</script>
That file (show_lecturer.php) stores the data read in a table (keep_track) in the database:
(file show_lecturer.php)
<?php
ob_start(); //eliminates buffer collisions
require_once('connect_db.php');
$name = $_POST['x'];
$surname = $_POST['y'];
$result = pg_query(connect(), "INSERT INTO keep_track VALUES ('$name', '$surname')");
?>
Then I create an empty dialogbox with jquery, to populate it with the data taken from the database:
(file index.php)
<!-- The following script generates the empty dialog box -->
<script src="/js/jquery.min.js"></script>
<link rel="stylesheet" href="/css/jquery-ui.css">
<script src="/js/jquery-ui.min.js"></script>
<script>
$(function() {
//show lecturer dialog
$("#show_dialog").dialog({autoOpen: false});
$(".show-button").on("click", function() {$("#show_dialog").dialog("open");});
});
</script>
Then these data are taken from the table keep_track and echoed in the above dialog:
(file index.php)
$name; $surname;
require_once('connect_db.php');
$firstname = pg_query(connect(), "SELECT name FROM keep_track");
while($row = pg_fetch_array($firstname)){ $name = $row['path']." ".$row['name']; }
$lastname = pg_query(connect(), "SELECT surname FROM keep_track");
while($row = pg_fetch_array($lastname)){ $surname = $row['path']." ".$row['name']; }
echo '<div id="show_dialog" class="ui-dialog-content ui-widget-content">';
echo $name."".$surname;
echo '</div>';
?>
So when I click the button of row x, a dialogbox opens with the data from the row x.
The only thing that is not working correctly is this:
The moment I click button x, it opens a dialog but displays a value, but not that of row x. However, when i see the database, the row x is stored there. The value in the checkbox is that of the button clicked before the latest refresh on the page. Its as if there is some mistake in my chain of calls or something (that I cant figure out, thats why Im asking).
To illustrate the data I get:
(Initially the table keep_track is empty)
Press button 1 -> row 1 stored, dialogbox has no content
Press button 2 -> row 2 stored, dialogbox has no content
Press button 3 -> row 3 stored, dialogbox has no content
Refresh page manually
Press button 4 -> row 4 stored, dialogbox has content from row 3
Press button 5 -> row 5 stored, dialogbox has content from row 3
Refresh page manually
Press button 6 -> row 6 stored, dialogbox has content from row 6
Press button 7 -> row 7 stored, dialogbox has content from row 3

I suggest you return your data from the POST via JSON. And please be aware that an AJAX Call is asynchronous. So you won't know when the reply is coming.
So you need to process your results using the ajax Success callback function.
</script>
$(document).ready(function(){
$(".show-button").click(function() {
var $row = $(this).closest("tr"); // Find the row
var names = $row.find(".name").text(); // Find the name
var surname = $row.find(".surname").text(); // Find the surname
do_post_and_show_info(names, surname);
});
});
function do_post_and_show_info(names, surname){
request= $.ajax({
type: "post",
cache: false,
url: "show_lecturer.php",
data: { x: names, y: surname} ,
dataType: "json",
});
request.done(function(json){
if (json.status =="ok"){
// DO YOUR THING!
Alert(json.data.names + " " + json.data.surnames);
}
else {
alert("Error! " + json.error + " : " + json.remarks);
}
});
request.fail(function(jqXHR, textStatus) {
alert( "Request failed: " + textStatus + ":" + jqXHR.responseJSON);
});
}//do_post_and_show_info
</script>
I usually return a datastructure like this in PHP (so in your show_lecturer.php)
<?
// get your data before this in the variable $data
// put your status "OK" or "ERROR" in $status
// put some error info in $extraInfo
// of course some processing is involved, but here's a simple example
require_once('connect_db.php');
$name = $_POST['x'];
$surname = $_POST['y'];
$result = pg_query(connect(), "INSERT INTO keep_track VALUES ('$name', '$surname')");
// obviously you need to do some error checking, but here's the happy flow
$status = "OK";
$error = "";
$data['names'] = $name;
$data['surnames'] = $surname;
echo json_encode(array(
"status" => $status,
"error" => $error,
"remark" => $extraInfo,
"data" => $data
));
?>
Please be aware this is an example that I have created here in the editor and not in a real working setup. SO please try to understand it instead of copy-pasting it and giving it a run.

I wrote the content of the dialog (div) in another file and used
$("#div").load("content.php", {x:parameter_1, y:parameter_2, ......});
instead of
$.ajax({ type: "POST", url: "show_lecturer.php", data: { x: names, y: surname} })
This did the trick.
Now the div is initially invisible and empty, but once the button is clicked, it requests the content.php page to load. Since I'm passing the search parameters when I request the content, I get the data that I wanted.
The problem from before was that when the page loaded, the div was created with the data (even though I hadn't clicked any button). Therefore, when I 'd click a button, it would show me the div with the content from the last page load (last refresh).
There were also other minor changes I had to do to make it work, but this is the main idea.