I have the following code sample upload3.php:
<html>
<head>
<title>PHP Form Upload</title>
</head>
<body>
<form method='post' action='upload3.php' enctype='multipart/form-data'>
Select a File:
<input type='file' name='filename' size='10' />
<input type='submit' value='Upload' />
</form>
<?php
if (isset($_POST['submit']))
{
echo "isset submit";
}
else
{
echo "NOT isset submit";
}
?>
</body>
</html>
The code always returns "NOT isset submit".
Why does this happen? Because the same script upload3.php calls itself?
You do not have your submit button named:
Change
<input type='submit' value='Upload' />
To:
<input type='submit' value='Upload' name="submit"/>
Two things:
You'll want to try array_key_exists instead of isset when using arrays. PHP can have some hinky behavior when using isset on an array element.
http://www.php.net/manual/en/function.array-key-exists.php
if (array_key_exists('submit', $_POST)) { }
Second, you need a name attribute on your button ( "name='submit'" )
Because you don't have any form element whose name property is submit.
Try to use var_dump($_POST) to see the keys that are defined.
Notice that files are an exception; they're not included in $_POST; they're stored in the filesystem and they're metadata (location, name, etc) is in the $_FILES superglobal.
Try looking at the REQUEST_METHOD and see if it's POST. It's a little bit nicer.
<input type='submit' value='Upload' />
should be
<input type='submit' value='Upload' name='subname'/>
and that subname should be in $_POST[' ']
it will look like
if (isset($_POST['subname']))
{
echo "isset submit";
}
else
{
echo "NOT isset submit";
}
Related
<form method='get' action='y.php'>
<div>
<input type='text' id='txtName' name='txtName'/>
<input type='submit' value='submit' id='submit'/>
</div>
</form>
<?php
if ($_SERVER['REQUEST_METHOD'] == 'GET')
{
if (isset($_GET['btnSave'])) {
$name=isset(($_GET['txtName'])?isset($_GET['txtName']:'');
//then Logic of insert goes here
}
}
?>
so before moving to y.php the record must be saved.
but I cant get the $name value, as action given to y.php.
How can I get $name which contain value in text box.
if you change the action to this (same/current) page record is going to database without any flaw or error.
try using post method instead and change your code accordingly, try this:
<form method='post' action=''>
<div>
<input type='text' id='txtName' name='txtName'/>
<input type='submit' value='submit' id='submit' name='submit'/>
</div>
</form>
<?php
if (isset($_POST['submit'])) {
$name=$_POST['txtName'];
//then Logic of insert goes here
//redirect to y.php with name value
echo "<script>window.open('y.php?user=$name','_self')</script>";
}
?>
Then use $nme = $_GET['user']; to get the value of $name in y.php
Try this code,
<form method='get' action=''>
<div>
<input type='text' id='txtName' name='txtName'/>
<input type='submit' value='submit' id='submit'/>
</div>
</form>
<?php
if ($_SERVER['REQUEST_METHOD'] == 'GET')
{
if (isset($_GET['txtName'])) {
$name=$_GET['txtName'];
//then Logic of insert goes here
}
}
?>
I suggest you use method post. You can code like this
<form method='post' action=''>
<div>
<input type='text' id='txtName' name='txtName'/>
<input type='submit' value='submit' id='submit'/>
</div>
</form>
<?php
if(isset($_POST['txtName'])){
$name =$_POST['txtName'];
echo $name;
}
?>
you can use the session if you want make the value use in another file.
I hope that can solve your problem
I am trying to make a like button for posts on my website.
PHP for like query (d_db_update is
function d_db_update($string) {
return mysql_query($string);
}
)
if($_GET['like']) {
$like = d_db_update("UPDATE posts set post_rating = post_rating+1 WHERE post_id = {$_GET['like']}");
}
Button
<form action='{$_SERVER['PHP_SELF']}&like={$posts_row['post_id']}' method='get'>
<p align='left'>{$posts_row['post_rating']}
<input type='submit' name='like' value='Like' /></p>
</form>
What can I do to fix it/make it work?
Use below form with a hidden input it solve your problem.
<form action='{$_SERVER['PHP_SELF']}' method='get'>
<p align='left'>{$posts_row['post_rating']}
<input type='hidden' name='like' value='{$posts_row['post_id']}' />
<input type='submit' value='Like' /></p>
</form>
You are using your form action wrong.. if you are using get method than there is not need to use the form..
try this..
<a href='yourpage.php?like=<?php echo $post_id ?>'>Like</a>
your submit button name and like variable which you have used in action url are the same , and you used get method in method of form.So, you need to change the submit button name.
or
you can do it without using form only on button click try below code
<input type='button' name='like' value='Like' onclick="location.href='yourpage.php?like=<?php echo $post_id ?>'" />
Change your code to this
You can not write PHP variables using {}. You need to echo them out.
<form action='' method='get'>
<p align='left'><?php echo $posts_row['post_rating'] ?>
<input type='hidden' name='like' value='<?php echo $posts_row["post_id"] ?>' />
<input type='submit' value='Like' /></p>
</form>
Edit--
You were not returning the post id correctly, I made the changes, also there is no need to provide any action as it will be self only.
I can't really use PHP and from what I've seen in tutorials this should work, but it doesn't:
<html>
<head></head>
<body>
<form>
<input type='text' name="name" value='myName'>
</form>
<p>
<?php
$name = $_POST['name'];
echo $name
?>
</p>
</body>
</html>
Is there a reason I can't get the value of name?
Sorry for asking such a simple question...
here is the fiddle http://jsfiddle.net/DCmu5/1/, so, please try what you said and send it to me only when it works before answering
PHP is a server-side language, so you'll need to submit the form in order to access its variables.
Since you didn't specify a method, GET is assumed, so you'll need the $_GET super-global:
echo $_GET['name'];
It's probably better though, to use $_POST (since it'll avoid the values being passed in the URL directly. As such, you can add method attribute to your <form> element as follows:
<form method="post">
<input type='text' name="name" value="myName" />
<input type="submit" name="go" value="Submit" />
</form>
Notice I also added a submit button for good measure, but this isn't required. Simply hitting return inside the textbox would submit the form.
Well, you have to POST your form to get the value of $_POST variables. Add this to your <form>
<form action="yourpagename.php" method="post">
<input type='text' name="name" value='myName'>
<button type="submit">Submit</button>
</form>
Click button and whatever is typed in your field will show up.
<html>
<body>
<form method="post" action="1.php">
Name: <input type="text" name="fname">
<input type="submit">
</form>
<?php
if ($_SERVER["REQUEST_METHOD"] == "POST") {
// collect value of input field
$name = $_POST['fname'];
if (empty($name)) {
echo "Name is empty";
} else {
echo $name;
}
}
?>
</body>
</html>
try this
<?php
if(isset($_REQUEST['name'])){
$name = $_REQUEST['name'];
echo $name;
}
?>
On my website i got a form looks like this on index.html:
<form id="demo" action='submit.php' method='post' enctype='text/plain'>
link: <input type='text' name='web'></br>
<input type='submit' value='submit'>
</form>
And then in the submit.php i got:
<?php
$fp=fopen('sub.txt','a');
fwrite($fp,addslashes($web));
fclose($fp);
header('location: thanks.html');
exit();
?>
But when i press submit the result in sub.php is beeing
/n
But it should be
example.com/n
What is wrong in the php code.
I want the link submited in the form printed in the file sub.txt and then the person be redirected to a thank you page.
You missing value attribute as well as you need $_POST to get the value inside sub.php
<form id="demo" action='submit.php' method='post' enctype='text/plain'>
link: <input type='text' name='web' value="" /></br>
<input type='submit' value='submit' />
</form>
<?php
$fp=fopen('sub.txt','a');
fwrite($fp,addslashes($_POST['web']). "\r\n");
fclose($fp);
header('location: thanks.html');
exit();
?>
On my form I have 3 submit,I want to know the best way to handle these buttons because the default action is the same form:
<form method="POST" action="file_where_form_is.php">
Name:<input type='text' name='name'>
<input type='submit' name='add' value='Add Names' ONCLICK='window.location.href="file_where_form_is.php">
<input type='submit'name='Preview'value='Preview'ONCLICK='window.location.href="file_where_form_is.php">
<input type='submit' name='submit'value='Submit' ONCLICK='window.location.href="send.php">
</form>
This not seems to work completely,just the 2 first work and for the last submit does nothing(with name='submit')
Any suggestions?
I finally found in header part add something like this:
if(isset($_POST['submit'])){
header("location: somefile.php");
}
elseif(isset($_POST['Preview'])){
header("location: anotherfile.php");
}
elseif(isset($_POST['add'])){
header("location: anotheronefile.php");
}