I've done some research and followed a tutorial exactly, with minor changes, but I can't seem to get this to work. No dropdown appears. Can you look at the code and tell me what is not happening?
<div class="ui-widget">
<label for="skills">Skills: </label>
<input id="skills">
</div>
<script>
$(function() {
$( "#skills" ).autocomplete({
width:448,
delimiter: /(,|;)\s*/,
lookup: <?php echo "'autocomplete(#skills)'" ?>
});
});
</script>
<?php
$servername = 'localhost';
$username = 'xx';
$password = 'xx';
$con = mysql_connect($servername,$username,$password);
mysql_select_db('c9', $con);
function autocomplete($inputName){
$searchTerm = $_GET[$inputName];
$query = mysql_query("SELECT * FROM clients WHERE fname LIKE '%".$searchTerm."%' ORDER BY fname ASC");
while ($row = mysql_fetch_array(MYSQL_ASSOC)) {
$data[] = $row['fname'];
}
//return json data
echo json_encode($data);
}
?>
Related
I need to define my website var from mySQL, but I don't know how to get the data to the var. This is what I have so far.
I'm able to get the data in JSON with this:
$json_array = array();
while($row = mysqli_fetch_assoc($result))
{
$json_array[] = $row;
}
echo json_encode($json_array);
?>
I'm stuck in this part.
<?php
$connect = mysqli_connect("localhost", "user", "", "pricesdb");
$sql = "SELECT * FROM precios";
$result = mysqli_query($connect, $sql);
while($row = mysqli_fetch_array($result)) ?>
<script type="text/javascript">
var websiteVars = {
priceusd: <?php echo ''.$row['priceusd'].''?>,
pricebs: <?php echo ''.$row['pricebs'].''?>
};
</script>
You're redefining the variable websiteVars each time through the loop.
You should keep the original loop that creates the array $json_array, and then encode the entire thing into the JavaScript variable;
var websiteVars = <?php echo json_encode($json_array); ?>;
Thank you very much, I was able to solve the problem this way.
<?php
$db = mysqli_connect("localhost", "root", "", "db");
$sql = "SELECT * FROM precios";
$result = mysqli_query($db, $sql);
while($row=mysqli_fetch_array($result, MYSQLI_ASSOC)){
?>
<script type="text/javascript">
var websiteVars = {priceUsd: <?php echo ''.$row['priceUsd'].''?>, priceBS: <?php echo ''.$row['priceBS'].''?>};
<?php
}
?>
</script>
I want to retrieve a row value from database.
Since we need serverside to access database we have to use PHP to retrive the value but I am not sure how to use this php in a .js file.
This is what I am trying but it does not work.
The reason is simple.. we can not write PHP inside a javascript file.
file.js:
<?php
$con = mysqli_connect($server, $db_user, $db_pwd, $db_name);
$username = $_SESSION['username'];
$sql = "SELECT * FROM Holders WHERE username='$username'";
$data = mysqli_query($con, $sql);
$row = mysqli_fetch_assoc($data);
$address = $row['address'];
?>
$( document ).ready(function() {
var addressFrmPHP = "<?= $address; ?>";
var url = "https://api.examlesite.io/api?module=account&action=bal&add=000000000&address="+addressFrmPHP+"&tag=latest&apikey=APIKEY";
});
How can I assign the php $address variable to javascript addressFrmPHP variable?
You have to create two files as shown below
getdata.php
<?php
$con = mysqli_connect($server, $db_user, $db_pwd, $db_name);
$username = $_SESSION['username'];
$sql = "SELECT * FROM Holders WHERE username='$username'";
$data = mysqli_query($con, $sql);
$row = mysqli_fetch_assoc($data);
$address = $row['address'];
echo $address;
?>
file.js
$( document ).ready(function() {
/* AJAX for get data from php page */
$.post('getdata.php', function(address) {
var addressFrmPHP = address;
});
var url = "https://api.examlesite.io/api?module=account&action=bal&add=000000000&address="+addressFrmPHP+"&tag=latest&apikey=APIKEY";
});
you have to do like this
test.php
<?php
$con = mysqli_connect($server, $db_user, $db_pwd, $db_name);
$username = $_SESSION['username'];
$sql = "SELECT * FROM Holders WHERE username='$username'";
$data = mysqli_query($con, $sql);
$row = mysqli_fetch_assoc($data);
$address = $row['address'];
echo '<script> var address= '.$address.';</scrpit>';
?>
<script type='text/javascript' src='js/file.js'></script>
file.js
$( document ).ready(function() {
var addressFrmPHP = address;
var url = "https://api.examlesite.io/api?module=account&action=bal&add=000000000&address="+addressFrmPHP+"&tag=latest&apikey=APIKEY";
});
Create One file Getdata.php and write the below code:
<?php
$con = mysqli_connect($server, $db_user, $db_pwd, $db_name);
$username = 'nirav4491';
$sql = "SELECT * FROM Holders WHERE username='$username'";
$data = mysqli_query($con, $sql);
$row = mysqli_fetch_assoc($data);
$address = $row['address'];
?>
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script type="application/javascript">
$(document).ready(function() {
var addressFrmPHP = '<?php echo $address; ?>';
var url = "https://api.examlesite.io/api?module=account&action=bal&add=000000000&address="+addressFrmPHP+"&tag=latest&apikey=APIKEY";
alert(url);
});
</script>
</head>
</html>
Used a form to create a php search for a MySQL database in a header.php file.
Attempting to use simplePagination.js with php. I am able to correctly calculate the number of results and display the appropriate amount of page links. However, search.php is not limiting the number of items on the page, and all of the pagination links lead to a blank page.
<form action="search.php" method="POST">
<input type="text" name="search" placeholder="search site">
<button type="submit" name="submit-search"><img src="../assets/search icon-05.png"></button>
</form>
search.php code:
<?php
include 'header.php';
?>
<section class="searchPage">
<div class="searchResults">
<?php
if (isset($_POST['submit-search'])){
$searchTerm = trim( (string) $_POST['search'] );
if (isset( $searchTerm[0] )) {
$search = mysqli_real_escape_string($conn, $_POST['search']);
$sql = "SELECT * FROM articles WHERE title LIKE '%$search%' OR abstract LIKE '%$search%' OR keywords LIKE '%$search%'";
$result = mysqli_query($conn, $sql);
$queryResult = mysqli_num_rows($result);
$limit = 10;
$numberOfPages = ceil($queryResult/$limit);
if ($queryResult > 0){
echo $queryResult . " results found";
while ($row = mysqli_fetch_assoc($result)){
echo "<div class='articleItem'>
<h2>".$row['title']."</h2>
<p>".$row['abstract']."</p>
<a href=".$row['link']." target='_blank'>".$row['link']."</a>
</div>";
}
$pageLinks = "<nav><ul class='pagination'>";
for ($i=1; $i<=$numberOfPages; $i++) {
$pageLinks .= "<li><a href='search.php?page=".$i."'>".$i."</a></li>";
};
echo $pageLinks . "</ul></nav>";
}
else {
echo "There are no results matching your search.";
}
}
}
?>
</div>
</section>
<script type="text/javascript">
$(document).ready(function(){
$('.pagination').pagination({
items: <?php echo $queryResult;?>,
itemsOnPage: <?php echo $limit;?>,
currentPage : <?php echo $page;?>,
hrefTextPrefix : 'search.php?page='
});
});
</script>
You don't need to create the page links on your own, because this is what the plugin does through JavaScript events. So you can replace the ul with a div element. This is the reason why you get a blank page.
echo "<nav><div class='pagination'></div></nav>";
In the following is what I added to make it work:
$(document).ready(function(){
var pageParts = $(".articleItem");
pageParts.slice(<?php echo $limit;?>).hide();
$('.pagination').pagination({
items: <?php echo $queryResult;?>,
itemsOnPage: <?php echo $limit;?>,
onPageClick: function(pageNum) {
var start = <?php echo $limit;?> * (pageNum - 1);
var end = start + <?php echo $limit;?>;
pageParts.hide().slice(start, end).show();
}
});
});
I have a search box where search is done through database. In the code I have, the search is done in one input box and the dynamic search output is shown in a text area below it.
What I want is a search like Google, where when the user stars typing, it should show similar items from the db table.
For example, if I have two organizations named "Dummy 1" and "Dummy 2" and the user types in "du", the search bar should show the 2 results and user should be able to select one.
The code I have is:
<form action="newbrand.php" method="post">
<br>
Brand Name: <input type="text" name="bname" /><br><br>
Search for an Organization: <input type="text" name="search" onkeyup="searchq()" id="output"><
Selected Organization:<textarea id="output"></textarea>
</form>
The js is like this:
<script type="text/javascript">
function searchq(){
var searchTxt = $("input[name='search']").val();
$.post("search.php", {searchVal: searchTxt},function(output){
$("#output").html(output);
}
}
</script>
This is the search.php file:
<?php
include 'db_connect.php';
$link = mysqli_connect($host, $username, $password, $db);
if(!link){
echo "DB Connection error";
}
$output = '' ;
$output2 = '' ;
if (isset($_POST['searchVal'])){
$searchq = $_POST['searchVal'];
//$searchq = preg_replace("#[^0-9a-z]#i","",$searchq);
$query = mysqli_query($link, "SELECT * FROM `organisations_info` WHERE `Organisation_Name` LIKE '%$searchq%'")or die("Could not search!");
$count = mysqli_num_rows($query);
if($count == 0){
$output = '<div>No results!</div>';
}else{
while($row = mysqli_fetch_array($query)){
$orgname = $row['Organisation_Name'];
$orgid = $row['Organisation_Id'];
$subs = $row['Subscription_Type'];
//$output = echo "<option value='".$orgname."'>" . $orgname . "</option>";
$output = $orgname;
$output2 = $orgid;
$output3 = $subs;
//$output = '<div>'.$orgname.'</div>';
}
}
}
echo ($output);
?>
How can I achieve that?
In the JS code...
<script type="text/javascript">
function searchq(){
var searchTxt = $("input[name='search']").val();
$.post("search.php", {searchVal: searchTxt},function(output){
$("#output").html(output);
}
}
</script>
you have given the id(#output) of a input type element to display(or to return) the HTML statements and the js script also not closed properly (syntax error).So the valid statement will be...
<form action="newbrand.php" method="post">
<br>
Brand Name: <input type="text" name="bname" /><br><br>
Search for an Organization: <input type="text" name="search" onkeyup="searchq()" id="output"><
Selected Organization:<textarea id="output"></textarea>
</form>
<br>
<div id="mydiv"></div>
<script type="text/javascript">
function searchq(){
var searchTxt = $("input[name='search']").val();
$.post("search.php", {searchVal: searchTxt},function(output){
$("#mydiv").html(output);
});
}
</script>
Just change your query :
$query = mysqli_query($link, "SELECT * FROM `organisations_info` WHERE `Organisation_Name` LIKE '%".$searchq."%' ")or die("Could not search!");
And the query will work fine :)
Then output the response in HTML in your search.php (manage the css accordingly) :
<?php
include 'db_connect.php';
$link = mysqli_connect($host, $username, $password, $db);
if(!link){
echo "DB Connection error";
}
$output = '' ;
$output2 = '' ;
if (isset($_POST['searchVal'])){
$searchq = $_POST['searchVal'];
//$searchq = preg_replace("#[^0-9a-z]#i","",$searchq);
$query = mysqli_query($link, "SELECT * FROM `organisations_info` WHERE `Organisation_Name` LIKE '%".$searchq."%' ")or die("Could not search!");
$count = mysqli_num_rows($query);
if($count == 0){
$output = '<div>No results!</div>';
}else{
while($row = mysqli_fetch_array($query)){
$orgname = $row['Organisation_Name'];
$orgid = $row['Organisation_Id'];
$subs = $row['Subscription_Type'];
?>
<div><?php echo $orgname; ?></div>';
<div><?php echo $orgid ; ?></div>';
<div><?php echo $subs ; ?></div>';
<?php
} // while
} // else
} // main if
?>
I hope this is what you required !!
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