Develop Reference

Dropdown post live search ,add onclick redirect to post (php page)

Hi,
i am coding a homepage to learn php and javascript. I decided to use a livesearch using jQuery and php.
It is working well ,but i wonder how i can integrate to the found titles an onclick function that will redirect to the viewpost.php so it opens the clicked title and opens the post.
My HTML search part on index page:
<!-- Search Widget -->
<div class="card my-4">
<div class="card bg-success">
<h5 class="card-header">Search</h5>
<div class="card-body">
<div class="search-box">
<input type="text" autocomplete="off" placeholder="Search country..." />
<div class="result"></div>
</div>
</div>
</div>
</div>
jQuery part for livesearch that redirect to php page(backend-search.php)
<script type="text/javascript">
$(document).ready(function(){
$('.search-box input[type="text"]').on("keyup input", function(){
/* Get input value on change */
var inputVal = $(this).val();
var resultDropdown = $(this).siblings(".result");
if(inputVal.length){
$.get("backend-search.php", {term: inputVal}).done(function(data){
// Display the returned data in browser
resultDropdown.html(data);
});
} else{
resultDropdown.empty();
}
});
// Set search input value on click of result item
$(document).on("click", ".result p", function(){
$(this).parents(".search-box").find('input[type="text"]').val($(this).text());
$(this).parent(".result").empty();
});
});
</script>
PHP backend-search.php
<?php
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
require_once "pdoconfig.php";
// Attempt search query execution
try{
if(isset($_REQUEST["term"])){
// create prepared statement
$sql = "SELECT * FROM articles WHERE title LIKE :term";
$stmt = $db->prepare($sql);
$term = $_REQUEST["term"] . '%';
// bind parameters to statement
$stmt->bindParam(":term", $term);
// execute the prepared statement
$stmt->execute();
if($stmt->rowCount() > 0){
while($row = $stmt->fetch()){
echo "<p>" . $row["title"] . "</p>";
}
} else{
echo "<p>No matches found</p>";
}
}
} catch(PDOException $e){
die("ERROR: Could not able to execute $sql. " . $e->getMessage());
}
// Close statement
unset($stmt);
// Close connection
unset($db);
?>
That is my table structure called articles:
id title content categorie_id pubdate views short_details
And finally my viewpost.php
<?php
$stmt = $db->prepare('SELECT id, title, text, pubdate FROM articles WHERE id = :id');
$stmt->execute(array(':id' => $_GET['id']));
$row = $stmt->fetch();
//if post does not exists redirect user.
if($row['id'] == ''){
header('Location: ./');
exit;
}
echo "<br>";
echo "<div class='card mb-4'>" . "<div class='card-body'>";
echo "<h2 class='card-title'>";
echo $row['title'] . "</h2>";
echo "<div class='card-footer text-muted'>";
echo $row['pubdate'];
echo "</h2>";
echo "<p class='card-text'>";
echo $row['text'];
echo "</p>";
echo '</div>';
?>
Do i need to get the articles id with the jQuery and somehow post it onclick to viewpost.php ?
I do appreciate all help ..

You Need To Change This PHP "backend-search.php" File :
This Code To
if($stmt->rowCount() > 0)
{
while($row = $stmt->fetch())
{
echo "<p>" . $row["title"] . "</p>";
}
}
else
{
echo "<p>No matches found</p>";
}
This Code
if($stmt->rowCount() > 0)
{
while($row = $stmt->fetch())
{
echo "<p>". $row["title"] . "</p>";
}
}
else
{
echo "<p>No matches found</p>";
}

how should i put data fetched from ajax call in hidden div box

i am working on a project and come across a module.
page1
user have to search from search bar which will take him to page 2.
page2
On page 2 all fetched results will get displayed to user in div's. Each result has a checkbox associated with it.
when i click on add to compare check box ,ajax call is executed and fetched selected result should appear in hidden div.
my problem is it is only shows first result in hidden div and not working with another result.
My code of page 2
<script type="text/javascript">
$(document).ready(function()
{
var check = $('#compare').val();
$("#compare").change(function() {
if(this.checked) {
$.ajax({
type: 'POST',
url: 'compare.php',
dataType : 'JSON',
data:{value : check},
success: function(data)
{
console.log(data);
$('#compare_box').html(data);
}
});
$("#compare_box").show();
}
else
{
$("#compare_box").hide();
}
});
});
</script>
</head>
<body>
<?php
$query = $_GET['search_bar'];
$query = "call fetch_data('$query')"or die(mysqli_error($conn));
$result = mysqli_query($conn,$query);
while($row = mysqli_fetch_array($result))
{
$id = $row['course_id'];
$title = $row['course_title'];
$description = $row['course_description'];
$course_url = $row['course_url'];
$video_url = $row['course_video_url'];
$fee = $row['course_fee'];
$duration = $row['course_duration'];
$start_date = $row['course_start_date'];
$university = $row['university_name'];
$course_provider = $row['course_provider_name'];
$instructor = $row['instructor_name'];
$_SESSION['result'][$id] = Array('id'=> $id,'course_title' => $title,'course_description'=> $description,'course_url' => $course_url,'video_url' => $video_url,'fee' => $fee,'course_duration'=>$duration,'start_date'=>$start_date,'university' => $university,'course_provider'=>$course_provider,'instructor'=>$instructor);
?>
<div id='compare_box'>
</div>
<div class="col-md-3 photo-grid " style="float:left">
<div class="well well-sm">
<a href="final.php?id=<?php echo $id;?>&name=<?php echo $title;?>" target="_blank">
<h4><small><?php echo $title; ?></small></h4>
</a>
<br>
<input type ='checkbox' name="compare" id="compare" value="<?php echo $id;?>">add to compare
</div>
</div>
<?php
}
?>
page3 compare.php
<?php
session_start();
include 'includes/dbconfig.php';
$check = $_POST['value'];
$sql = "SELECT * from course_info_table where course_id = '$check' " or die(mysqli_error($conn));
$result = mysqli_query($conn,$sql);
$index = 0;
while($row = mysqli_fetch_array($result))
{
$title = $row['course_title'];
?>
<?php
}
echo json_encode($title);
?>

You can change
<input type ='checkbox' name="compare" id="compare" value="<?php echo $id;?>">
to
<input type ='checkbox' name="compare" class="compare" value="<?php echo $id;?>">
^you can only have one unique 'id' value in your html doc, which means your first id="compare" will work fine and others with id="compare" will be ignored by the DOM tree
Reference:
http://www.w3schools.com/tags/att_global_id.asp

Im trying to populate a dropdowm menu with a database column in php

I am trying to make a dropdown menu from a database. I am returning nothing right now. For the menu, I need to choose a object then run a SQL query on it. This query will populate a table and will be dynamic.
Here is the code, please help.
<html>
<?php
include('header.php');
?>
<h1>Chemicals Search</h1>
<br/ >
<br/ >
</head>
<h1>
<center>Chemical Search</center>
</h1>
<form action="chemicals.php" method="post">
<label>Search By Product:</label>
<?php
//making the chemical array
$con = mysqli_connect("...");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$p = mysqli_query($con, "SELECT product_name FROM product");
$products = mysqli_fetch_array($p);
echo '<select name="product">';
foreach ($product as $key => $value) {
echo "<option value=\"$key\"> $value</option>\n";
}
echo '</select>';
echo "<br>";
?>
</form>

Your Query only selects product name make a little changes to it
$p = mysqli_query($con,"SELECT product_id,product_name FROM product");
echo'<select name="product">';
if(mysqli_num_rows($p) > 0)
{
while($products =mysqli_fetch_array($p))
{
echo"<option value='".$products['product_id']."'>".$products['product_name']."</option>\n";
}
}
echo '</select>';
echo"<br>";
I am sure this will works, works for me everytime

Get mysql column of a clicked item

I have searched endlessly for an answer but have found none. I am trying to get the id of a clicked item. The item that I am clicking is from mysql database and has been displayed through a for loop. When the item is clicked I am taken to another page. In this page I want to utilize the id from the clicked item to get other information from that row in mysql database; this much I can do. The problem is getting the id from the clicked item and sending it.
This is the most recent way that I tried:
First I displayed the items from mysql.
<?php
$query = "SELECT `video` FROM `challenge_name` ORDER BY `id`";
$result = mysql_query($query);
if($result = mysql_query($query))
{
for($i = 0; $i < mysql_num_rows($result); $i++)
{
$id = $i;
$code = "<div id=\"challenge_preview\"><h7 class=\"challenge_preview_item\"
id=\"challenge_preview_name\"></h7><a href=\"challengeprofile.php\">
<video
class=\"challenge_preview_item\" id=\"challenge_video\"
src=\"".mysql_result($result, $i)."\"></video></a></div>";
echo $code;
}
}
else
{
die('Couldn\'t connect'. mysql_error());
}
?>
Than I put the id in a hidden form so that I could attempt to POST it to the other page:
<form action="<?php echo $current_file; ?>" method="POST">
<input type="hidden" name="id" value="14">
</form>
On the script.php file that is included in both pages, I put
$(#challenge_video).click(function(){ <?php $id = $_POST['id']; ?> ;});
And on the page that the id is being posted to I put
<?php
include 'script.php';
echo getChallengeData('name', 'id', $id)
?>
Please help, Thank you
These are the edits
<div id='challenge_previews'>
<?php
$query = "SELECT `video` FROM `challenge_name` ORDER BY `id`";
$result = mysql_query($query);
if($result = mysql_query($query))
{
for($i = 0; $i < mysql_num_rows($result); $i++)
{
$id = $i;
echo "<div id=\"challenge_preview\"><video class=\"challenge_preview_item challenge_video\" src=\"".mysql_result($result, $i)."\"></video></div>";
}
}
else
{
die('Couldn\'t connect'. mysql_error());
}
?>
<form action="<?php echo $current_file; ?>" method="GET">
<input type="hidden" name="id" value="14">
</form>
</div>
This is the code for page 2
<h1 id="challenge_profile_name">
<?php
$id = substr(base64_decode($_GET['id']),6);
echo getChallengeData('name', 'id', $id);
?>
</h1>
This is the code for the getChallengeData method in the core.inc.php
function getChallengeData($field1, $field2, $field3)
{
$query = "SELECT `$field1` FROM `challenge_name` WHERE `$field2` = '$field3'";
if($query_run = mysql_query($query))
{
if($query_result = mysql_result($query_run, 0, $field1))
{
return $query_result;
}
}
}
That error that I'm getting has to do with the implementation of the getChallengeData method on page 2. The error says
Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 10 in C:\xampp\htdocs\ChallengeNetworkWebsite\core.inc.php on line 42

how to fetch data from sql using form $_Post id in where clause

I am using a form with javascript which is used to add n numbers of rows dynamical and post data to mysql.
now i want to post more information to mysql using where clause (form data) in sql statement.
This is my code to submit and post data.
<script src="jquery.min.js"></script>
<script type="text/javascript">
$(function() {
var addDiv = $('#addinput');
var i = $('#addinput p').size() + 1;
$('#addNew').live('click', function() {
$('<p><select name="stockid[]' + i +'" onchange="showUser(this.value)"> <?php echo $item; ?></select> <select name="desc[]' + i +'" id="txtHint"> <?php echo $description; ?></ </select>Remove </p>').appendTo(addDiv);
i++;
return false;
});
$('#remNew').live('click', function() {
if( i > 2 ) {
$(this).parents('p').remove();
i--;
}
return false;
});
});
</script>
<body>
<?php if (!isset($_POST['submit_val'])) { ?>
<h1>Add your Hobbies</h1>
<form method="post" action="">
<div id="container">
<p id="addNew"><span>Add New</span></p>
<div id="addinput">
<input type="submit" name="submit_val" value="Submit" />
</form>
<?php } ?>
<?php
?>
<?php
if (isset($_POST['submit_val']))
{
$stockid = $_POST["stockid"];
$desc = $_POST["desc"];
foreach($stockid as $a => $B)
{
$query = mysql_query("INSERT INTO 0_stock_master (stock_id,description) VALUES ('$stockid[$a]','$desc[$a]')", $connection );
}
echo "<i><h2><strong>" . count($_POST['stockid']) . "</strong> Hobbies Added</h2></i>";
}
?>
its working fine now when am trying to use a select statement and post data to mysql its not working
here is code
<?php
$con=mysqli_connect("localhost","root","","inventory");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM 0_stock_master where id = '".$$_POST['stockid']."'");
while($row = mysqli_fetch_array($result))
{
echo $row['price'];
}
mysqli_close($con);
?>
then i modify the post code of above file like this
<?php
if (isset($_POST['submit_val']))
{
$stockid = $_POST["stockid"];
$desc = $_POST["desc"];
$price = $row['price'];
foreach($stockid as $a => $B)
{
$query = mysql_query("INSERT INTO 0_stock_master (stock_id,description,price) VALUES ('$stockid[$a]','$desc[$a]','$price[$a]')", $connection);
}
echo "<i><h2><strong>" . count($_POST['stockid']) . "</strong> Hobbies Added</h2></i>";
}
?>
but nothing is inserted in to database in price column

Change your code to store the price value in a new variable:-
<?php
$con=mysqli_connect("localhost","root","","inventory");
$price = array(); //declare
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM 0_stock_master where id = '".$_POST['stockid']."'");
while($row = mysqli_fetch_array($result))
{
echo $row['price'];
$price = $row['price']; //initiate
}
mysqli_close($con);
?>
<?php
if (isset($_POST['submit_val']))
{
$stockid = $_POST["stockid"];
$desc = $_POST["desc"];
$query = mysql_query("INSERT INTO 0_stock_master (stock_id,description,price) VALUES ('$stockid','$desc','$price')", $connection);
}
?>
Your $row['price'] variable will only exist within the while loop so you have to store it in something that is present beforehand and use that variable instead.
Assuming that both code snippets are in the same file, that is. Take a look over the code and see the changes on line 3 and line 27.
Also, as the other guys have said remove the double $$ and just use one on this line:-
$result = mysqli_query($con,"SELECT * FROM 0_stock_master where id = '".$$_POST['stockid']."'");
Hope this is of some help to you :)

As said by aconrad in comments, replacing $$_POST by $_POST would probably solve your problem.
But I suggest you to change mysqli_query() to mysqli_prepare (and to change all mysql_* by the equivalent mysqli_* function)
I suggest you to transform all into mysqli_ and use prepared statements instead of direct query like this :
Change this:
<?php
$result = mysqli_query($con,"SELECT * FROM 0_stock_master where id = '".$$_POST['stockid']."'");
while($row = mysqli_fetch_array($result))
to this:
<?php
$stmt = mysqli_prepare($con,"SELECT price FROM 0_stock_master where id = ?");
mysqli_stmt_bind_param($stmt, 'i', $_POST['stockid']);
$result = mysqli_stmt_execute($stmt);
if (!$result)
echo 'Mysql error : '.mysqli_stmt_error($stmt);
mysqli_stmt_bind_result($stmt, $price); // values will
mysqli_stmt_fetch($stmt); // this call send the result in $price
mysqli_stmt_close($stmt);
Change this:
<?php
$query = mysql_query("INSERT INTO 0_stock_master (stock_id,description,price) VALUES ('$stockid[$a]','$desc[$a]','$price[$a]')", $connection );
to this :
<?php
$stmt = mysqli_prepare($connection, "INSERT INTO 0_stock_master (stock_id,description,price) VALUES (?, ?, ?)");
// I assume stock_id must be int, desc must be string, and price must be float
mysqli_stmt_bind_param($stmt, 'isf', $stockid[$a],$desc[$a],$price[$a]);
$query = mysqli_stmt_execute($stmt);
$affected_rows = mysqli_stmt_affected_rows($stmt);
EDIT :
Some documentation:
MySQLi
mysqli_prepare (sql queries more protected from sql injection)
mysqli_stmt_bind_param
mysqli_stmt_execute
mysqli_stmt_bind_result
mysqli_stmt_fetch