This question already has answers here:
Creating a PHP header/footer
(5 answers)
Closed 4 years ago.
I have a html file which is just my footer. How can I use this in another file? Do I need to use php?
I'm looking for something like this:
<section>
<p>site content</p>
</section>
<?php footer html from a different file goes here ?>
Will this include the styling and javascript that is referenced in the head of the file i'm getting the html from?
Thanks!
Now, when you visit index.html, you should be able to click the link tags.
IN HTML
<html>
<head>
<title></title>
<script
src="https://code.jquery.com/jquery-3.3.1.js"
crossorigin="anonymous">
</script>
<script>
$(function(){
$("#header").load("header.html");
$("#footer").load("footer.html");
});
</script>
</head>
<body>
<div id="header"></div>
<!--Remaining section-->
<div id="footer"></div>
</body>
</html>
IN PHP
<html>
<head>
<title></title>
</head>
<body>
<?php include('header.php'); ?>
<!--Remaining section-->
<?php include('footer.php'); ?>
</body>
</html>
If you are using PHP you can require you can find more about it here.
If you don't want to use PHP you will have to use AJAX.
Example :
$.ajax({
url: "yourHTMLFile.html",
dataType: "html",
success: function(data) {
$("body").html(data);
}
});
Related
I am getting the issue
Warning: Use of undefined constant otherInput - assumed 'otherInput'
(this will throw an Error in a future version of PHP) Warning: A
non-numeric value encountered
I know there is some issue with the quote.
What I am doing is, I have to click on the anchor tag called Click me one and sending the data-id in the script. I am getting the id value in the script but I am getting the error on $(".showme'+otherInput+'").show();
I am using WordPress and I have to use the below code in the function.php so I have to use my script inside the PHP tag.
This is the screenshot of the code
Here is the code
<!DOCTYPE html>
<html>
<head>
<title></title>
<style type="text/css">
.showme{display: none;}
</style>
</head>
<body>
<div class="clickme" data-id="1">Click me one</div>
<div class="showme showme1">this is example</div>
<?php
echo '<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
<script type="text/javascript">
$(".clickme").click(function(){
var otherInput=$(this).data("id");
$(".showme'+otherInput+'").show();
});.
</script>';
?>
</body>
</html>
It looks like everything in your php block is all JQuery/Javascript.
Q: What do you even need the PHP block and the "echo" for?
Current:
<?php
echo '<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
<script type="text/javascript">
$(".clickme").click(function(){
var otherInput=$(this).data("id");
$(".showme'+otherInput+'").show();
});.
</script>';
?>
Suggested change:
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
<script type="text/javascript">
$(".clickme").click(function(){
var otherInput=$(this).data("id");
$(".showme"+otherInput).show();
});.
</script>'
Just follow the regular js standard, and if you willing to implement your script inside the php code:
Replace your code with these changes:
<!DOCTYPE html>
<html>
<head>
<title></title>
<style type="text/css">
.showme{display: none;}
</style>
</head>
<body>
<div class="clickme" data-id="1">Click me one</div>
<div class="showme showme1">this is example</div>
<?php
echo '<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
<script type="text/javascript">
$(".clickme").click(function(){
var otherInput=$(this).data("id");
$(".showme"+otherInput).show();
//here otherInput js variable is just used as regular js variable inside the script, so no need to extra commas to call that js variable, just use it like you do in js files, and will work fine.
});
</script>';
?>
</body>
</html>
Hiii everyone,
I just want to show one div if the database value is equal to some value.I tried like this
<?php echo $show['birth_certificate_available']=='Available') { ?>
<script type="text/javascript">
$(".birth").show();
</script>
<?php } ?>
But the div is not showing.I dont know why its not working.Kindly guide me to correct this issue.
Please modify the code accordingly :
<?php if($show['birth_certificate_available']=='Available') { // put a condition?>
<script type="text/javascript">
$(document).ready(function(){ // jQuery ready function
$(".birth").show();
});
</script>
<?php } ?>
Wrap it in ready() function as follows,
Assuming you have already imported the relevant jquery lib in your code.
<?php echo $show['birth_certificate_available']=='Available') { ?>
<script type="text/javascript">
$(document).ready(function(){
$(".birth").show();
});
</script>
<?php } ?>
I missed JQuery reference document so that its not worked.Now I added script document its working.Thanks all who helped me to fix this.
Solution
echo outputs data to the page. To write a condition, you would like to use if statements.
<?php if ($show['birth_certificate_available'] === 'Available'): ?>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
<script>
$(document).ready(function() {
$('.birth').show();
});
</script>
<?php endif; ?>
Note: if jQuery is already linked to your document, remove the <script src="..."></script> line. Otherwise, I suggest you to move the <script src="..."></script> line right before the </body> HTML closing tag to speedup HTML elements loading:
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8" />
<title></title>
</head>
<body>
<div class="birth" hidden="hidden">Foobar</div>
<?php if ($show['birth_certificate_available'] === 'Available'): ?>
<script>
$(document).ready(function() {
$('.birth').show();
});
</script>
<?php endif; ?>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
</body>
</html>
Note: I use the === equal assertion to force type testing. Whatever, be careful: testing is case-sensitive, whether it is == or ===
The div element must have the birth class, must exist and must be hidden in order to successfully show() it
Debug
$show variable
If this is not working, and in order to debug the $show variable, you may want to use var_dump() as so:
var_dump($show);
and ensure it is an array and has the birth_certificate_available index defined.
Javascript errors
Use the browser console to retrieve javascript errors (such as inexisting div elements...)
Probably I'm misunderstanding something about Fancybox v1 but have the following minor issue. First I'm calling fancybox via clicking a link:
$(document).ready(function() {
$(".href").click(function() {
$.fancybox.open({
href : $(this).attr("data-id"),
type : 'iframe'
});
});
});
This loads a simple PHP/HTML page with JS, to be populated into the Fancybox window:
<!DOCTYPE html><html><head><title></title></head>
<body>
<?php
# this is the fancybox content
echo "<script>var dcwrtext = 'random_text'; document.write(dcwrtext);</script> Some other content to show on fancybox.";
?>
</body></html>
The content appears without a problem, but the nuance is, the document.write value is echoed after the closing </script> tag: and this makes the content visible on the page, not the document.write() function. You can see this in the source code:
<body>
<script>var dcwrtext = 'random_text'; document.write(dcwrtext);</script>**random_text** Some other content to show on fancybox.
</body>
The problem seems to be with fancybox, if I use the document.write() on a standard HTML (not fancyboxed) page, as expected, it will correctly show 'random_text' on the page, inside <script></script> only. This is the source code:
<body>
<script>var dcwrtext = 'random_text'; document.write(dcwrtext);</script> Some other content to show on fancybox.
</body>
What I'd need is if I open the fancbox window, 'random_text' shouldn't appear literally after the closing </script> tag, but would be displayed only by the document.write() function.
This is how I tested:
main page:
<html>
<head>
<title></title>
<meta name="" content="">
<link rel="stylesheet" type="text/css" href="fancybox/source/jquery.fancybox.css?v=2.1.2" media="screen" />
</head>
<body>
<div data-id="fancy.php" class="href">open fancy</div>
<script src="http://code.jquery.com/jquery-1.9.0.js"></script>
<script src="http://code.jquery.com/jquery-migrate-1.2.1.js"></script>
<script type="text/javascript" src="fancybox/source/jquery.fancybox.js?v=2.1.3"></script>
<script>
$(document).ready(function() {
$(".fancybox").fancybox();
$(".href").click(function() {
$.fancybox.open({
href : $(this).attr("data-id"),
type : 'iframe'
});
});
});
</script>
<div id="theid" class="fancybox"></div>
</body>
</html>
and the page being opened in fancybox: fancy.php
<html>
<head>
<title></title>
</head>
<body>
<?php
$randomtext = "random_text";
echo "<script>var dcwrtext = '$randomtext'; document.write(dcwrtext);</script>Some other content to show on fancybox";
?>
</body>
</html>
[I had to include jquery-migrate-1.2.1.js because of an error [ Uncaught TypeError: Cannot read property 'msie' of undefined] when including newer versions of jquery with fancybox]
I'm trying to pass the data-value of the < li > I clicked to PHP in the same page. I'm using AJAX to pass the data to PHP, to be used in querying. I think the ajax request is working, but the php code inside the if is not running.
The codes:
offices.php
<?php
include 'connect.php';
$page = 'offices';
include 'header.php';
if (isset($_POST['postdata'])){
$filter1 = $_POST['postdata'];
echo $filter1;
}
?>
<!DOCTYPE html>
<html>
<head>
<script src="js/jquery.min.js"></script>
</head>
<body>
<div class="filter1">
<ul>
<li class="fteacher" data-value="teacher">Professor</li>
<li class="foffice" data-value="office">Gabinete</li>
</ul>
</div>
<script src="js/offices.js"></script>
</body>
</html>
offices.js
$('.filter1 li').click(function(){
$.ajax({
type: 'POST',
url: 'offices.php',
data: {'postdata': $(this).attr('data-value')},
success: function(msg){
alert('Success');
}
});
});
I tried the code you provided, but had to ommit the two includes at the beginning of your PHP script, and it worked. Testing was done using file_put_contents("test.txt", "test"); in the if-clause.
You may have a look at what the PHP script returns by using alert(msg). Note that everything after the PHP part is returned, too. It might also help to use error_reporting(E_ALL) in the PHP to search for problems in the included scripts.
Please try this. Hope you are only expecting the output $filter1
<?php
if (isset($_POST['postdata'])){
$filter1 = $_POST['postdata'];
echo $filter1;
}else {
include 'connect.php';
$page = 'offices';
include 'header.php';
?>
<!DOCTYPE html>
<html>
<head>
<script src="js/jquery.min.js"></script>
</head>
<body>
<div class="filter1">
<ul>
<li class="fteacher" data-value="teacher">Professor</li>
<li class="foffice" data-value="office">Gabinete</li>
</ul>
</div>
<script src="js/offices.js"></script>
</body>
</html>
<?php
}
?>
I thought this would be simple enough, but I can't get it to work.
I have two files. main.html and data.php
The two files are in the same folder on a server.
I want to get a string from the PHP file and use it in jQuery. In the example below, I want the browser to create a pop-up, with the text "xxxx". I get no pop-up.
data.php
<?php
$var = "xxxx";
echo json_encode($var);
?>
main.html
<!DOCTYPE html>
<html>
<head>
<script type="text/javascript" src="scripts/jquery-1.11.1.js"></script>
<script>
$(document).ready(function() {
var testString = <?php $var = json_decode(file_get_contents('data.php'), true); echo $var; ?>;
alert(testString);
});
</script>
</head>
<body>
<h1>Test</h1>
</body>
</html>
Any takers?
<!DOCTYPE html>
<html>
<head>
<script>
$(document).ready(function() {
$.ajax({
url:"data.php",
success:function(result){
alert(result);
}
});
});
</script>
</head>
<body>
<h1>Test</h1>
</body>
</html>
The first problem with this is that on many servers .html is not parsed for PHP
The second problem is that the file_get_contents will actually display the full contents of your .php file (including the <?php) which is likely not what you're looking for.
Instead I would use an AJAX request such as http://api.jquery.com/jquery.get/
For this you will need to include jQuery in your <head>
You should get the php data before html initiates, like below
<?php
$var = json_decode(file_get_contents('data.php'), true);
?>
<!DOCTYPE html>
<html>
<head>
<script>
$(document).ready(function() {
var testString = <?php echo $var; ?>;
alert(testString);
});
</script>
</head>
<body>
<h1>Test</h1>
</body>
</html>