<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 $name;
}
?>
you can use the session if you want make the value use in another file.
I hope that can solve your problem
Related
so I have a form. The form consists of 10 lines by default. It goes like this:
<form method="post" action="actionhere">
<?php
for($i=0; $i<10;$i++) {
?>
<div class='clone_me'>
<span>Line <?php echo $i;?></span>
<input type='checkbox' name='ck_<?php echo $i;?>'/>
<input type='text' name='tx_<?php echo $i;?>'/>
</div>
<?php } ?>
<input type='submit' name='submit' value='Submit'/>
</form>
So inside the form, we will have 10 rows of checkbox+textbox.
What I'm trying to make is, I want to place a button to add new row (the checkbox+textbox). Now, problem is, I need the $i value (since it's form the for loop). Is that possible that when we click the add row button, the value of $i that we set inside for loop be incremented by 1 on each click? I know we can clone the div using jquery, but how about the $i value?
I think you are doing it in wrong way you do not need $i value inside name attribute you have to use array for it for example
<form method="post" action="test.php">
<div class='clone_me'>
<span>Line 1</span>
<input type='checkbox' name='ck[]'/><!--this field should menditory-->
<input type='text' name='tx[]'/>
<span>Line 2</span>
<input type='checkbox' name='ck[]'/><!--this field should menditory-->
<input type='text' name='tx[]'/>
</div>
<input type='submit' name='submit' value='Submit'/>
</form>
Now implement this code in actionhere.php
<?php
$cks = $_POST['ck'];
$txs = $_POST['tx'];
foreach($cks as $key => $ck) {
echo $ck."<br>";
echo $txs[$key]."<br>";
}
?>
Well, basically no. Your PHP script is already over when the html has been generated. So you can't rely anymore on PHP. But you don't need to make it explicitly appear in your html.
You should count the rows using jquery :
var i = $('form').find('.clone_me').length
and then add a new row using javascript again :
$('form .clone_me:last').clone().insertAfter('form .clone_me:last');
<input type='hidden' name='counter' id='counter'/>
<input type='checkbox' name='chk' id='chk'/>
<?php
$counter=$_POST['counter'];
for($i=0;$i<=$counter;$i++)
{
$chk=$_POST['chk'.$i];
// Your Insert Code Here
}
?>
may be this can be but have some limit
if you have no problem with page reload the you can do it is:-
by this way your page will reload and the value in textbox and checkbox will gone:---:)
every time new page generate and send by server to client browser
<?php
if(isset($_POST['submit'])){
$ends = $_POST['ttl_rows'];
}else{
$ends = 10;
}
?>
<form method="post" action="#">
<?php
for($i=1 ; $i<$ends;$i++) {
?>
<div class='clone_me'>
<span>Line <?php echo $i;?></span>
<input type='checkbox' name='ck_<?php echo $i;?>'/>
<input type='text' name='tx_<?php echo $i;?>'/>
</div>
<?php } ?>
<input type='hidden' name='ttl_rows' value='<?php echo ($i+1); ?>'/>
<input type='submit' name='submit' value='Submit'/>
</form>
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 have a little problem with my code and I don't know what is wrong in my code... So I have a two form where the first load is the "view.php" then the second is "checkbox_building.php".. My problem is when I removed the button of delete in "view.php" the dropdown list can proceed to "checkbox_building.php". but when I placed it back, its not working (its not proceeding to "checkbox_building.php") I'm kinda confuse about this.
Here's my code for "view.php"
<fieldset width= "200px">
<form name='form' method='post' action=''>
Select Network: <select name="netName" onChange="this.form.action='checkbox_building.php'; this.form.submit()">
<option value="" >- Select -</option>
<?php
include 'connect.php';
$q = mysql_query("select fldNetname from tblnetwork");
while ($row1 = mysql_fetch_array($q))
{
echo "<option value='".$row1[fldNetname]."'>".$row1[fldNetname]."</option>";
$net = $row1[fldNetname];
}
?>
</select>
<input type='submit' name='submit' value='Delete Building/s' onClick="this.form.action='delete_building.php'; this.form.submit()">
</form>
</fieldset>
thanks.!
just as a guess, just from your code:
correct the flaw, then
change
<input type='submit' name='submit'
to
<input type='button' name='dosubmit'
and
<form name='form' method='post' action=''>
to
<form name='<someUsefullName>' method='post' action=''>
and all references to this as well, since "form" is a reserved word in IE and this.form.action may lead to errors there.
just remove name='submit' and try
<input type='submit' name='submit' value='Delete Building/s' onClick="this.form.action='delete_building.php'; this.form.submit()">
so it will be
<input type='submit' value='Delete Building/s' onClick="this.form.action='delete_building.php'; this.form.submit()">
its because this.form.submit is default submit function, but in your code its an input element :)
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();
?>
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";
}