I tried this tutorial to send/retrieve data from my mysql database, using ajax and php.
This is the ajax part:
<script>
function showUser(str) {
if (str=="") {
document.getElementById("txtHint").innerHTML="";
return;
}
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
} else { // code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function() {
if (xmlhttp.readyState==4 && xmlhttp.status==200) {
document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","suchen_ma.php?id="+str,true);
xmlhttp.send();
}
</script>
The .php file looks like this:
<?php
//Connection Details
$username = 'root';
$password = '';
$hostname = 'localhost';
$databasename = 'plzdb';
$id = ($_GET['id']);
//Connection-string
$con = mysqli_connect($hostname,$username,$password,$databasename);
//SQL Query
$sql = "SELECT per_vorname,per_nachname from plz_person
WHERE per_id = $id";
$result = mysqli_query($con,$sql);
echo "<table border='0'>
<tr>
<th></th>
<th></th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['per_vorname'] . "</td>";
echo "<td>" . $row['per_nachname'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
The data is formatted within the php file, but I want to send the "raw" data back to the html and display it inside a list and format it with css. Unfortunately I have no idea how to do this
Just give back the data as JSON:
$sql = "SELECT per_vorname,per_nachname from plz_person
WHERE per_id = " . mysqli_real_escape_string($con, $id);
$result = mysqli_query($con, $sql);
while ($row = mysqli_fetch_array($result)) {
$persons[] = array(
"vorname" => $row['per_vorname'],
"per_nachname" => $row['per_nachname']
);
}
echo json_encode($persons);
and then create a loop in javascript, to build the HTML.
NOTE: Avoid sql injection by escaping your string.
Related
On the body of the document, lets call it "form.php" we have the following:
On the head we have a JavaScript code:
<script>
function showUser(str) {
if (str == "") {
document.getElementById("txtHint").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("txtHint").innerHTML = this.responseText;
}
};
xmlhttp.open("GET", "getchauffeur.php?q=" + str, true);
xmlhttp.send();
}
}
</script>
We query to database and populate a dropdown. We switch content using (showUser):
<div>
<?
$result = $mysqli -> query("select id, nomchauffeur from chauffeurs");
echo "<select name='id' onchange='showUser(this.value)'>";
while ($row = $result -> fetch_assoc()) {
unset($id, $name);
$id = $row['id'];
$name = $row['nomchauffeur'];
echo '<option value="'.$id.
'">'.$name.
'</option>';
}
?>
Here we are still in body. We put the content of AJAX into div.
<div id="txtHint"><b>chauffeur info will be listed here...</b> </div>
</div>
Here is our script that populate form fields with the content of AJAX request:
<script>
var table = document.getElementById('table');
for (var i = 1; i < table.rows.length; i++) {
table.rows[i].onclick = function() {
//rIndex = this.rowIndex;
document.getElementById("nomchauffeur").value = this.cells[0].innerHTML;
document.getElementById("prenomchauffeur").value = this.cells[1].innerHTML;
document.getElementById("agechauffeur").value = this.cells[2].innerHTML;
document.getElementById("cinchauffeur").value = this.cells[3].innerHTML;
};
}
</script>
Now here is our getchauffeur.php:
<?php
$q = intval($_GET['q']);
$con = mysqli_connect('localhost','root','','nouveau');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"ajax");
$sql="SELECT * FROM chauffeurs WHERE id = '".$q."'";
$result = mysqli_query($con,$sql);
echo "<table>
<tr>
<th>nom</th>
<th>prenom</th>
<th>age</th>
<th>adresse</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['nomchauffeur'] . "</td>";
echo "<td>" . $row['prenomchauffeur'] . "</td>";
echo "<td>" . $row['agechauffeur'] . "</td>";
echo "<td>" . $row['adressechauffeur'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
The Problem: Everything works fine if the table is on the same page. But here the AJAX request constraints us to put the table in other php page(chauffeur.php).
What we need is populating the form fields automatically by clicking on the row displayed from dropdown Change actions. It appears that the row inserted into table inside 'chauffeur.php' is not printed on the html DOM. When we click on page view source, it displays only:
<div id="txtHint"><b>chauffeur info will be listed here...</b> </div>
And not the content of the following fields:
nomchauffeur prenomchauffeur agechauffeur adressechauffeur
How could we grab the content of row and fill automatically the form and where is it?
This whole javascript AJAX should be after your HTML div #txtHint.
if (this.readyState == 4 && this.status == 200) {
document.getElementById("txtHint").innerHTML = this.responseText;
}
Also you can return only row instead of whole table. On main page create table as id 'txtHint' and insert that row in response.
I want the output of "cor_resp_template" where is nome and select that values for in next page do some operations like(sum, multiplication,...). I try the first time use JavaScript, but i don´t know if the code is correct. I try to analysis step by step and seems ok for that operation.
In next picture like you see the empty values. Don´t show the table with the values for after submit that values for next page.
The output of database of the table template where is all the values:
That is the output of that page:
http://localhost/dissertacao/conceptualization13.php?menuId=1
After we have the code where we use the JavaScript for get the values in next page where is the PHP query.
<script>
function showUser(str) {
if (str == "") {
document.getElementById("$option").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("$option").innerHTML = this.responseText;
}
};
xmlhttp.open("GET","template1.php?option="+str,true);
xmlhttp.send();
}
}
</script>
</head>
<body>
<form>
<?php
echo "<select name='template' class='form-control' onchange='showUser(this.value)'>";
mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("teste") or die(mysql_error()); // selecciona a database do server escolhido
echo "<center>";
// Get the county names from database - no duplicates
$query = "SELECT DISTINCT nome FROM template";
{
// execute the query, $result will hold all of the Counties in an array
$resultado = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($resultado))
$option .="<option>" . $row['nome'] . "</option>";
echo $option;
}
echo "</select>";
echo "<br>";
?>
<input type="submit" value="Submit">
</form>
<br>
<div id="valores"><b>Os valores para este template são:</b></div>
template1.php
The following code is the PHP QUERY where is called for give the output of the table with the values where is nome in the SQL table.
<?php
$option = (isset($_POST['nome']) ? $_POST['nome'] : '');
mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("teste") or die(mysql_error()); // selecciona a database do server escolhido
$query="SELECT * FROM template WHERE 'nome' = '".$option."'";
$resultado = mysql_query($query) or die(mysql_error());
echo "<table>
<tr>
<th>Valores</th>
</tr>";
while($row = mysql_fetch_array($resultado)) {
var_dump ($resultado);
echo "<tr>";
echo "<td>" . $row['corp_resp_template'] . "</td>";
echo "</tr>";
}
echo "</table>";
?>**
I would like to display one column of data, [pin], based on the [plan] and [order_id] values. plan=9 and order_id=0. Would like to load data without reloading page, using ajax.
Here is my HTML/Script:
<script>
function showPins(str) {
if (str == "") {
document.getElementById("txtHint").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
}
}
xmlhttp.open("GET","getPins.php?q="+str,true);
xmlhttp.send();
}
}
</script>
HTML:
<div align="center">
<h3>View PIN's</h3>
<form>
<select name="users" onchange="showPins(this.value)">
<option value="">Select Plan Type:</option>
<option value="1">Plan1</option>
<option value="2">Plan2</option>
<option value="3">Plan3</option>
</select>
</form>
<br/>
<div id="txtHint"></div>
</div>
This is my PHP file (getPins.php):
<?php
$q = intval($_GET['q']);
$con = mysqli_connect('myHost','myUsername','myPw','my_db');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"my_db");
$sql="SELECT * FROM attPins WHERE (order_id=0, plan=9 and id = '".$q."')";
$result = mysqli_query($con,$sql);
echo "<table>
<tr>
<th>PIN's</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['pin'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
This is based off the tutorial shown here: http://www.w3schools.com/php/php_ajax_database.asp
Trying to make it work for showing the correct pins for plan type chosen.
your query would be wrong read manual where
$sql="SELECT * FROM attPins WHERE (order_id=0, plan=9 and id = '".$q."')";
It would be
WHERE (order_id=0 and plan=9 and id = '".$q."')
Or
WHERE (order_id=0 OR plan=9 and id = '".$q."')
according to your requirment
Let's say I have this piece of code:
$templatesf = $DB->query('SELECT * FROM templates WHERE category="something"');
Is there a way to change that "something" using javascript/AJAX? For example:
function changesomething(selse){
if (selse == '1'){
something = 'this'
}else{
something = 'that'
}
}
The following will be your input page from where you'll call your ajax handler to process something.
<html>
<head>
<script language="javascript" type="text/javascript">
//Browser Support Code
function ajaxFunction(){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
}
catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
}
catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
}
catch (e){
// Something went wrong
alert("Your browser broke!");
return false;
}
}
}
// Create a function that will receive data
// sent from the server and will update
// div section in the same page.
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
var ajaxDisplay = document.getElementById('ajaxDiv');
ajaxDisplay.innerHTML = ajaxRequest.responseText;
}
}
// Now get the value from user and pass it to
// server script.
var somedata = document.getElementById('something').value;
var queryString = "?something=" + somedata ;
ajaxRequest.open("GET", "ajax-example.php" +
queryString, true);
ajaxRequest.send(null);
}
</script>
</head>
<body>
<form name='myForm'>
Something: <input type='text' id='something' /> <br />
<input type='button' onclick='ajaxFunction()'
value='Query MySQL'/>
</form>
<div id='ajaxDiv'>Your result will display here</div>
</body>
</html>
Below is the php code to handle your request and process your something
<?php
$dbhost = "localhost";
$dbuser = "dbusername";
$dbpass = "dbpassword";
$dbname = "dbname";
//Connect to MySQL Server
mysql_connect($dbhost, $dbuser, $dbpass);
//Select Database
mysql_select_db($dbname) or die(mysql_error());
// Retrieve data from Query String
$something = $_GET['something'];
// Escape User Input to help prevent SQL Injection
$something = mysql_real_escape_string($something);
//build query
$query = "SELECT * FROM ajax_example WHERE something = '$something'";
$qry_result = mysql_query($query) or die(mysql_error());
//Build Result String
$display_string = "<table>";
$display_string .= "<tr>";
$display_string .= "<th>Name</th>";
$display_string .= "<th>something</th>";
$display_string .= "<th>other</th>";
$display_string .= "<th>blahblah</th>";
$display_string .= "</tr>";
// Insert a new row in the table for each person returned
while($row = mysql_fetch_array($qry_result)){
$display_string .= "<tr>";
$display_string .= "<td>$row[name]</td>";
$display_string .= "<td>$row[something]</td>";
$display_string .= "<td>$row[other]</td>";
$display_string .= "<td>$row[blahblah]</td>";
$display_string .= "</tr>";
}
echo "Query: " . $query . "<br />";
$display_string .= "</table>";
echo $display_string;
?>
You'll need to place a variable instead of that SOMETHING, and then pass an ajax variable to that method.
You'll catch the variable with a $_POST and place it instead of the variable you had previously.
You could also use a default(in case nothing was passed, or wrong type of POST was passed).
This is my table subjects in lecturer database
No subject credit_hour
1 (111) AAA 4
2 (222) BBB 3
3 (222) CCC 4
4 (333) DDD 3
This is what I have done using ajax
This is my testing1.php
<?php
$conn = mysql_connect('localhost','root','password');
mysql_select_db('lecturer');
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
?>
<html>
<head>
<script>
function showUser(str)
{
if (str==="")
{
document.getElementById("txtHint").innerHTML="";
return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState===4 && xmlhttp.status===200)
{
document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","testing3.php?q="+str,true);
xmlhttp.send();
}
</script>
</head>
<body>
<form>
<select name="subjects" onchange="showUser(this.value)">
<option value="">Select a subject:</option>
<?php $result= mysql_query('SELECT * FROM subjects'); ?>
<?php while($row= mysql_fetch_assoc($result)) {
$list=array($row['subject'],$row['credit_hour']);
?>
<option>
<?php echo htmlspecialchars($row['subject'] ); ?>
<?php echo htmlspecialchars($row['credit_hour'] ); ?>
</option>
<?php } ?>
</select>
</form>
<br>
<div id="txtHint"><b>subject info will be listed here.</b></div>
</body>
</html>
This is my testing3.php
<?php
$q = intval($_GET['q']);
$con = mysqli_connect('localhost','root','password','lecturer');
if (!$con)
{
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"lecturer");
$sql="SELECT * FROM subjects WHERE No = '".$q."'";
$result = mysqli_query($con,$sql);
echo "<table border='1'>
<tr>
<th>Subject</th>
<th>Credit_hour</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['subject'] . "</td>";
echo "<td>" . $row['credit_hour'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
The problem is that when I select (111)AAA 4
it should be appearing a table which is located in testing3.php
but it did't fetch any data in the table
Thank you
After your first select query you should put the subject, hour_credit pair in an array, then on onchange event of the first option(your subject list) you should check which index is selected then set the second options selected index.