printing textbox value on page

[myselfamit47]

<form method="post" action="">

<input type="text" name="username">

<input type="button" name="sbutton" value="click button ">

<?php

if(isset($_POST['sbutton']))
{
        $value=$_POST['username'];
        echo "$value";
}
?>

</form>

i want to print the value of textbox on that same page ..but it is not working..

plz help me...

i have done this also

<?php

        $value=$_POST['username'];
        echo "$value";

?>

and

this also

<?php

        $value=$_GET['username'];
        echo "$value";

?>

but it is not working..........plz help me

Edited August 15, 2017 by cyberRobot
added [code][/code] tags

[Barand]

Instead of

type="button"

use

type="submit"

so that your form is submitted.

[myselfamit47]

thank u ...

In visual studio ,,when we create .net projects we have a feature of debugging the code, in which we execute the code line by line and can check for errors,,like this in php have we debugging feature?? if yes then in which IDE.??

[Barand]

I use Nusphere's PHPEd IDE, It has an excellent debugging facility, But any decent IDE should provide one,

[ginerjm]

Try putting all your html together and don't put php in the middle.  Plus if you want the value to show up actually IN the text field use a value attribute on that tag.  Basic HTML stuff.

[myselfamit47]

i have written this code for uploading a file ...

 

<form method="post" action="" enctype="multipart/form-data">
<input type="file" name="upload">
<input type="submit" name="fileupload" value="upload the file">

</form>


<?php
$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES['upload']['name']);
$uploadOk = 1;
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
// Check if image file is a actual image or fake image
if(isset($_POST['submit'])) {
    $check = getimagesize($_FILES['upload']['tmp_name']);
   if($check !== false) {
        echo "File is an image - " . $check['mime'] . ".";
        $uploadOk = 1;
    } else {
        echo "File is not an image.";
        $uploadOk = 0;
    }
}
?>

 

but it is not working.....plz help me....

[Jacques1]

Why checking for submit button names is a bad idea.

[ginerjm]

Your code is looking for something named "submit" in your $_POST array.  There isn't anything with that name.  Look at your code again.