Reputation: 49
I have four text files and I want to display it on the browser using the buttons I want to create four buttons when click on the 1st button first text should open. Please help me out from this I'm new to HTML and PHP.
Upvotes: 2
Views: 9879
Reputation: 26288
It seems that you dont need to necessarily read the file on the triggered action of clicking a button. If you are already reading the file, load it into a hidden HTML element, like a <div>
, then use Javascript to show that on the click of the button.
PHP:
$myFile = "file.txt";
$fh = fopen($myFile, 'r');
$theData = fread($fh, filesize($myFile));
fclose($fh);
echo '<div id="hidden_content" style="display:none">'.$theData.'</div>';
HTML Link:
<a href="#" onclick="show_content()">Show it!</a>
Javascript:
function show_content(){
document.getElementByID('hidden_content').style.display = 'block';
}
Or you can read the data dynamically using the ajax and show it on the page.
Upvotes: 1
Reputation: 2147
To do this you need to read about php file functions.
Refer this link - http://php.net/manual/en/function.file.php
Try below code:
<h2>Click</h2>
<form action="" method="post">
<button name="click" class="click">Click me!</button>
</form>
<?php
if(isset($_POST['click']))
{
$filename = "test.txt";
$before_editing = file_get_contents($filename);
?>
<!--this is for display in table-->
<table border="1"><tr><td><?php echo "Content of the file " . $filename . " before editing: " . $before_editing . "<br>"; ?></td></tr></table>
<!-- this is for display in single-->
<?php
echo "Content of the file " . $filename . " before editing: " . $before_editing . "<br>";
file_put_contents($filename, "test test test test test");
$after_editing = file_get_contents($filename);
echo "Content of the file " . $filename . " after editing: " . $after_editing . "<br>";
}
?>
You can replace your file name to "test.txt" as i am just giving example to you.
Upvotes: 4