I'm trying to get a dynamically loaded content from a web page. Specifically the options loaded to a select. So if I do:
$options = $html->find('select[class=theSelectClass]')[0]->find('option');
foreach($options as $option){
echo $option->text().'<br>';
}
This works as expected and my output is:
Select an option
Why? Because the other options are loaded with JS after the page loads. So my question is how can I get this dynamically loaded options inside the select?
This is my attempt using JS Ajax and another PHP page:
in my php that includes the simple_html_dom:
$html->load_file($base);
$var = '<script>
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
this.responseText;
}
};
xhttp.open("GET", "http://localhost/crawler/ajax.php?param=HelloWorld", true);
xhttp.send();
</script>';
$e = $html->find("body", 0);
$e->outertext = $e->makeup() . $e->innertext . $var . '</body>';
and my ajax.php file:
file_put_contents ( 'ajax.txt' , $_GET['param']);
I was trying to see if I could send an Ajax call from the html loaded file, but I feel far from being able to do it. So how can I make this happen?
Thank you
It might be easier for you to first use a headless browser to render the page then pass that to simple html dom. You could do this with CasperJS/PhantomJS or another tool that renders the page with javascript.
`
require("vendor/autoload.php");
use Sunra\PhpSimple\HtmlDomParser;
use Browser\Casper;
$casper = new Casper();
// forward options to phantomJS
// for example to ignore ssl errors
$casper->setOptions(array(
'ignore-ssl-errors' => 'yes'
));
$casper->start('https://www.reddit.com');
$casper->wait(5000);
$output = $casper->getOutput();
$casper->run();
$html = $casper->getHtml();
$dom = HtmlDomParser::str_get_html( $html );
$elems = $dom->find("a");
foreach($elems as $e){
print_r($e->href);
}
?>`
Related
Is it possible to get data php with Ajax without display them ? Simply stock data in JS variable?
I need this data to manipulate dates but no show it.
When I tried to simply return data without echo, etc. Data ajax in JS is empty
Ps : sorry my English is bad
try it this way
File *.php
<?php
$var_1 = null;
$var_2 = null;
/** ... */
$response = new stdClass;
$response->var_1 = $var_1;
$response->var_2 = $var_2;
echo json_encode($response);
?>
File *.html or *.js
<script>
var state = {};
$.ajax({
url: 'getData.php',
type: 'post',
dataType: 'json',
success: function (response) {
console.warn(response);
state = response;
}
});
</script>
Assuming you are trying to pass data from a PHP file to HTML/JS where it happens that your PHP file is also included in the HTML that's why it's displaying the echo (if I understood correctly!)
Using AJAX PHP example from w3school.
HTML sample file:
<?php include "PHP_SAMPLE_FILE.php" ?>
<header>
<meta name="temp_files" content="<?= htmlspecialchars($jsonData) ?>">
<!-- The rest of HTML content -->
JS sample file:
if (str.length == 0) {
// do something if there was nothing entered
} else {
var xmlhttp = new XMLHttpRequest();
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
if (this.responseText.includes('{')){
result = JSON.parse(this.responseText);
// do something if response is JSON
} else {
// do something if response is null
}
}
}
xmlhttp.open("GET", "PHP_SAMPLE_FILE.php?q="+str, true);
xmlhttp.send();
}
PHP sample file:
$q = $_REQUEST["q"] ?? $_POST["q"] ?? "";
$sql = "GET SOMETHING FROM DATABASE";
$result = mysqli_query($con, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
$json[] = $row;
}
}
$jsonData = json_encode($json ?? null);
if($q != ""){
echo $jsonData;
}
What happens exactly is that once the page loads initially it won't display the output of the PHP query as we have surrounded the echo with an if statement that requires to have query value (q) to search and it shouldn't be empty (""). Of course, assuming that once the page is loaded the data is shared with the client-side through defined PHP variables using various approaches, using a meta tag in the header for instance.
Once the data is received from the PHP file through echo, we use the JSON.parse function to parse it as in this scenario JS receives it as a string.
Hope that helped :)!
I have a HTML page JavaScript which send a GET request data to PHP file to return all datas saved in the database . PHP replies with a HTML-table - that works fine!
But: When if i click a button (which calls the same JavaScript function) to update my table in order to display the new data, i get the same result (and i have definitely new data on table).
If I call the PHP manually via the browser it'll show me the new results immediately and at this moment it is also working with JavaScript (but only once).
Here is a part of my code.
HTML/JS:
<button onclick="GetData()"></button>
<div id="test"></div>
<script>
function GetData(){
var xhttp = new XMLHttpRequest();
document.getElementById("test").innerHTML = "";
xhttp.onreadystatechange = function(){
if (xhttp.readyState == 4 && xhttp.status == 200){
document.getElementById("test").innerHTML = xhttp.responseText;
}
};
xhttp.open("GET", "../GetData.php", true);
xhttp.send();
}
</script>
PHP:
//DB details
$dbHost = 'localhost';
$dbUsername = 'lalalala';
$dbPassword = 'lalalalal';
$dbName = 'lalalala';
//Create connection and select DB
$db = new mysqli($dbHost, $dbUsername, $dbPassword, $dbName) or die ("UUUUPS");
$sql = "select name, beschreibung, image, video from data";
$result = $db->query($sql);
if ($result->num_rows > 0) {
$return = '<table class ="table table-hover"><thead><tr><th scope="col">Name</th><th scope="col">Beschreibung</th><th scope="col">Bilddatei</th><th scope="col">Video-Url</th></tr></thead><tbody>';
// output data of each row
while($row = $result->fetch_assoc()) {
$return .= "<tr><td>".$row["name"]."</td><td>".$row["beschreibung"]."</td><td><img src=data:image/png;base64,".base64_encode($row["image"])."/></td><td>".$row["video"]."</tr>";
}
$return .= "</tbody></table>";
$db->close();
echo $return;
} else {
echo "0 results";
}
Thank you for your help!
It seems your browser is caching your result, that's why you see data.
You can test it like this:
var random = Math.floor(Math.random() * 100);
xhttp.open("GET", "../GetData.php?"+random, true);
If this helps, look into expire headers in your PHP script. Also, the way you're doing queries in quite outdated. It's a very PHP4 way. Have a look here: http://php.net/manual/en/book.mysqli.php
I guess you probably know this, but just in case. Have you had a look in your browsers inspector, when testing you html page? especially the network tab within that inspector. There you can see the actual response from the server and you can see if it is served from cache or fetched (you can even disable cache there), maybe this helps.
Kind regard,
Mark
I'm having problems with passing php variables to javascript.
It does pass through the variable that is declared at the top, but I don't know how to call the function to get the new version of variable after the IF statement is done.
$info = "A message";
if (true){
$info = 'Message to be passed';
}
The script that is used to pass the php variable to javascript file:
<script type='text/javascript'>
var info = "<?php echo $info; ?>";
</script>
I was wondering what could I do to fix this problem?
The simple way (this requires both files to be PHP files):
<?php
require_once "your_php_file_here.php"; // Change to your PHP file here
?>
<script type='text/javascript'>
var info = "<?php echo $info; ?>";
alert(info);
</script>
This will only allow you to get the value on page load. You need to reload the page if you want it to get a new value.
The (in my opinion) better way (the file can be HTML) using Ajax:
<script type='text/javascript'>
var info;
var xhr = new XMLHttpRequest();
xhr.open('GET', 'your_php_file_here.php'); // Change to your PHP file here
xhr.onload = function() {
if (xhr.status === 200) {
info = xhr.responseText;
alert(info);
} else {
alert('Request failed: ' + xhr.status);
}
};
xhr.send();
</script>
This can be put in a function and called as many times as you want. It can get the new value without the need to reload the page.
For this to work, you need to change your PHP code to:
$info = "A message";
if (true){
$info = 'Message to be passed';
}
echo $info;
I did not add support for IE6 and below because I think it's about time we stop supporting browsers that lost support by their developers many years ago.
I have a simple AJAX function being called when a user clicks a button that sends the text of a HTML textarea and alerts the response from the backend:
send_button.onclick = function ()
{
var ajax = new XMLHttpRequest();
var text = text_input.value;
ajax.onreadystatechange = function ()
{
if (ajax.readyState == 4 && ajax.status == 200) alert(ajax.responseText);
};
ajax.open("POST", "write.php", true);
ajax.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
ajax.send("text=" + text);
};
as well as a PHP script on the backend which is supposed to write the received text to a file and echo the file's content:
<?php
$filename = "preview/preview.html";
$text = (isset($_POST["text"]) ? $_POST["text"] : "");
try
{
$fh = fopen($filename, "w+");
if (!$fh) throw new Exception("File open error");
fwrite($fh, $text);
$filetext = (filesize($filename) > 0 ? fread($fh, filesize($filename)) : "");
echo $filetext;
fclose($fh);
}
catch (Exception $e)
{
header("Location: error.php");
}
?>
But every time the response is empty. I tried echoing a hardcoded string instead of fread()and it worked, I also tried echoing filesize($filename)which worked perfectly fine as well.
The POST data sent by the AJAX function gets through as well, and the fwrite($fh, $text) function does exactly what it is supposed to.
What am I doing wrong?
You didn't rewind your file:
you open your file for writing
you write out some text - file pointer is at the END of the file
you try to read some text from the file, but the pointer is at the END of the file
no data is read, so you output an empty string
Why not use something more like this:
file_put_contents('preview/preview.html', $_POST['text'], FILE_APPEND);
readfile('preview/preview.html');
The "can't read file" is all fine and dandy, but all of the open/write/read business is redundant and can be reduced to the above two lines of code.
You could use $filetext = file_get_contents($filename); instead. I think you moved the file pointer to the end after writing in it, so you only see the end-of-file character.
You can use file_get_contents($file_name)
ok I have edited this to another couple of questions I've asked on a similar issue, but I really am in a rush so thought I'd start a new one, sorry if it bothers anyone.
first I have a php script on test.php on the apache server
<?php
//create connection
$con = mysqli_connect("localhost", "user", "password", "dbname");
//check connection
if (mysqli_connect_errno()){
echo "failed to connect to MySQL: " . mysqli_connect_error();
}
$grab = mysqli_query($con, "SELECT * FROM table");
$row = mysqli_fetch_array($grab);
$name = $row["name"];
$color = $row["color"];
$price = $row["price"];
$n1 = $name[0];
$c1 = $color[0];
$p1 = $price[0];
?>
Then I've got this ajax script set to fire onload of page a webpage written in html. so the load() function is onload of the page in the body tag. This script is in the head.
function load(){
var xmlhttp = new XMLHttpRequest();
xmlhttp.open("GET", "test.php", true);
xmlhttp.send();
xmlhttp.onreadystatecahnge = function(){
if(xmlhttp.readyState == 4 && xmlhttp.status == 200){
document.getElementById("itemNameLink1").innerHTML = "<?php echo $n1;?>;
}
}
}
ok so what I want is the $n1 variable in the php script to be used in the javascript ajax code. Where the script is, but I'm not sure where or how to make use of the variable, I've tried a few things. All that happens right now is the innerHTML of itemNameLink1 just disappears.
I'm quite new so any advise would be appreciated, thanks.
The response (this is what you echo in php) returned from request you can get by responseText attribute of XMLHttpRequest object.
So first your JS code should be:
function load(){
var xmlhttp = new XMLHttpRequest();
xmlhttp.open("GET", "test.php", true);
xmlhttp.send();
xmlhttp.onreadystatecahnge = function(){
if(xmlhttp.readyState == 4 && xmlhttp.status == 200){
document.getElementById("itemNameLink1").innerHTML = xmlhttp.responseText;
}
}
}
now in php echo $n1 variable:
....
$grab = mysqli_query($con, "SELECT * FROM table");
$row = mysqli_fetch_array($grab);
$name = $row["name"];
$color = $row["color"];
$price = $row["price"];
$n1 = $name[0];
$c1 = $color[0];
$p1 = $price[0];
// echo it to be returned to the request
echo $n1;
Update to use JSON for multiple variables
so if we do this:
$name = $row["name"];
$color = $row["color"];
$price = $row["price"];
$response = array
(
'name' => $name,
'color' => $color,
'price' => $price
);
echo json_encode($response);
Then in javascript we can parse it again to have data object containing 3 variables.
var data = JSON.parse(xmlhttp.responseText);
//for debugging you can log it to console to see the result
console.log(data);
document.getElementById("itemNameLink1").innerHTML = data.name; // or xmlhttp.responseText to see the response as text
Fetching all the rows:
$row = mysqli_fetch_array($grab); // this will fetch the data only once
you need to cycle through the result-set got from database: also better for performance to use assoc instead of array
$names = $color = $price = array();
while($row = mysqli_fetch_assoc($grab))
{
$names[] = $row['name'];
$color[] = $row['color'];
$price[] = $row['price'];
}
$response = array
(
'names' => $names,
'color' => $color,
'price' => $price
);
You can dynamically generate a javascript document with php that contains server side variables declared as javascript variables, and then link this in the head of your document, and then include this into your document head whenever server side variables are needed. This will also allow you to dynamically update the variable values upon page generation, so for example if you had a nonce or something that needs to change on each page load, the correct value can be passed upon each page load. to do this, you need to do a few things. First, create a php script and declare the correct headers for it to be interpreted as a script:
jsVars.php:
<?php
//declare javascript doc type
header("Content-type: text/javascript; charset=utf-8");
//tell the request not to cache this file so updated variables will not be incorrect if they change
header('Cache-Control: no-cache, no-store, must-revalidate'); // HTTP 1.1.
header('Pragma: no-cache'); // HTTP 1.0.
header('Expires: 0'); // Proxies.
//create the javascript object
?>
var account = {
email: <?= $n1; ?>,
//if you need other account information, you can also add those into the object here
username: <?= /*some username variable here for example */ ?>
}
You can repeat this for any other information you need to pass to javascript on page load, and then reference your data using the namespaced javascript object (using object namespacing will prevent collisions with other script variables that may not have been anticipated.) wherever it is needed as follows:
<script type="text/javascript>
//put this wherever you need to reference the email in your javascript, or reference it directly with account.email
var email = account.email;
</script>
You can also put a conditional statement into the head of your document so it will only load on pages where it is needed (or if any permission checks or other criteria pass as well). If you load this before your other scripting files, it will be available in all of them, provided you are using it in a higher scope than your request.
<head>
<?php
//set the $require_user_info to true before page render when you require this info in your javascript so it only loads on pages where it is needed.
if($require_user_info == TRUE): ?>
<script type="text/javascript" href="http://example.com/path-to-your-script/jsVars.php" />
<?php endif; ?>
<script type="text/javascript" href="your-other-script-files-that-normally-load" />
</head>
You can also do this for any other scripts that have to load under specific criteria from the server.
You should define the PHP variable. And use that variable in your javascript:
<?php
$n1 = "asd";
?>
<html>
<head></head>
<body>
<div id="itemNameLink1"></div>
<script>
function load()
{
var xmlhttp = new XMLHttpRequest();
xmlhttp.open('GET', '/test.php', true);
xmlhttp.send(null);
//Note you used `onreadystatecahnge` instead of `onreadystatechange`
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("itemNameLink1").innerHTML = '<?=$n1?>';
}
}
}
load();
</script>
</body>
</html>