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 $posts_row["post_id"] ?>' />
<input type='submit' value='Like' /></p>
</form>
Edit--
You were not returning the post id correctly, I made the changes, also there is no need to provide any action as it will be self only.
Related
<form method='get' action='y.php'>
<div>
<input type='text' id='txtName' name='txtName'/>
<input type='submit' value='submit' id='submit'/>
</div>
</form>
<?php
if ($_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 $name;
}
?>
you can use the session if you want make the value use in another file.
I hope that can solve your problem
I have a page that prints out rows of information. In each row is a notes box:
<?php
<td style='font-size:12px;width:300px;'>
{$row['Notes']} <br /><center><br />
<form action=\"http://********/functions/notes.php\" id='formNotesform' method='post'>
<input type='hidden' id='ID' name='ID' value='{$row['ID']}' />
<textarea placeholder=\"Add more notes here...\" name=\"notes\" rows=\"5\" cols=\"30\"'></textarea><br />
<input type='submit' name='formNotes' id='formNotes' value='Add to Notes' />
</form>
</center></td>
?>
Then there's also another button on the page in each row.
<form action=\"http://********/functions/archive.php\" method='post' id='formArchiveform' onclick=\"return checkArchive()\">
<input type='hidden' id='ID' name='ID' value='{$row['ID']}' />
<input type='submit' name='formArchive' id='formArchive' value='Archive' style=\"height:25px; width:80px\"/>
</form>
What I need to happen is that when someone clicks the "Add to Notes" button it does its job but when someone clicks the "archive" button it checks to see if notes is empty and if not then it submits that as well (kind of like a failsafe).
Ideally I'd like to just pick up the Notes data and post it to the archived.php file the form is going to anyways since that would cause minimal disruption to the code base but I can't get it to work.
I understand this isn't really a sensible choice. It just has to be done. Thanks for the help!
If I had reputation I would ask first because what I understood is that you want to now what button has been pressed to then do another things.
If that's it do this:
if($_SERVER['REQUEST_METHOD'] == 'POST')
{
if(isset($_POST['formNotes']))
{
//...
}
elseif(isset($_POST['formArchive']))
{
//...
}
}
I have a HTML form that acts as a "confirm deletion?" page and my "Back" button is playing up.
$string = "foo.php?id=" . $_POST['fooid'] . "&id2=" . $_POST['barid'];
<!-- ^^ this ends up being "foo.php?id=1&id2=2" -->
<form action='<?php echo $string; ?>'>
<input type='submit' value='Back'>
</form>
the problem is, when the button is pressed it links to foo.php without any of the $_GET data in my string, even though the string contains "id=1&id2=2"
P.S I changed the code above so people could better understand it, here is the raw code:
<?php
$string = "xrays.php?id=" . $_POST['visitid'] . "&id2=" . $_POST['patientid'];
?>
<form action='delete.php' method='post'>
<input type='hidden' name='xrayid' value='<?php $_POST['xrayid']?>'>
<input type='submit' name='submit' value='Confirm Delete?'>
</form>
<form action='<?php echo $string; ?>'>
<input type='submit' value='Back'>
</form>
Maybe it's not the best answer but I would like do something like that:
$string = "foo.php?id=" . $_POST['fooid'] . "&id2=" . $_POST['barid'];
<!-- ^^ this ends up being "foo.php?id=1&id2=2" -->
<form action='foo.php'>
<input type="hidden" name="fooid" value ="<php echo $_POST['fooid']; ?>" />
<input type="hidden" name="barid" value ="<php echo $_POST['barid']; ?>" />
<input type='submit' value='Back'>
</form>
This should work properly.
EDIT: change $_GET na $_POST
You need to put the get variable in the form hidden inputs, like so:
<form action='foo.php'>
<input type="hidden" name="id" value="1" />
<input type="hidden" name="id2" value="2" />
<input type='submit' value='Back'>
</form>
Or you could use a link:
Back
Let's start with form submission in general. W3C: Form Submission
Next, let's review $_GET and $_POST. PHP Manual: $_GET | PHP Manual: $_POST
In summary, inside of your <form> tag, use either method="get" or method="post". Only one of the superglobal arrays will be populated by successful controls, based upon your method of sending the data. I believe the query string must result from a GET request url (which may be the default), not just a plain string slapped into the action="" attribute. I could be wrong about the query string, but you have another problem. You are using two forms on one page. Presently, I think only one form's controls can be submitted successful at a time.
<form action='delete.php' method='post'> <!-- FORM 1 -->
<input type='hidden' name='xrayid' value='<?php $_POST['xrayid']?>'>
<input type='submit' name='submit' value='Confirm Delete?'>
</form>
<form action='<?php echo $string; ?>'> <!-- FORM 2, add a method="" attribute -->
<input type='submit' value='Back'>
</form>
Upon adding a method="get" to form two, it should become clear that a composite $_POST + $_GET request is not possible in the two form approach, and that you need to start with making a single, monolithic form instead of two modular ones. Using the type="hidden" attribute of an <input /> tag, inside of one form, as in #machineaddict's answer, will help. However, what will really help is if you explicitly use all the correct attributes of each tag so that you can spot errors like this in the future.
In a situation like this, it is helpful to know that the $_SERVER['QUERY_STRING'] element would hold the complete query string if your web server received one.
PHP Manual: $_SERVER
so I have a form. The form consists of 10 lines by default. It goes like this:
<form method="post" action="actionhere">
<?php
for($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>
This is a simple demonstration of my code :
if(isset($_POST['login'])) {
do this
}
<form method='POST' action="index.php">
Username : <input type='text' name='username'/>
Password : <input type='password' name='password' />
<input type='image' src='images/login.png' name='login' value='login'/>
</form>
The problem is that , when i use type='image' as a submit button, in Firefox and IE nothing happends. But if i use type='submit' everything works fine..
Well i dont want to display the button as a submit one, but as an image, so what im doing wrong ?
Simply change your if to look for any other field in the POST. Image type button wont send that value.
Change:
if(isset($_POST['login'])) {
To
if(isset($_POST['username'])) {
If you want to check login, you need to check for login_x or login_y values instead of login, which is not created when button is an image.
<?php
if(isset($_POST['login_x'])) {
echo "Here I m";
}
?>
<form method='POST' action="index.php">
Username : <input type='text' name='username'/>
Password : <input type='password' name='password' />
<input type='image' src='images/login.png' name='login' value='login'/>
</form>
Sorry, I explained in a comment the type=image creating _x and _y, but I forgot the params login does not exists in that case