There I have three buttons and when I click them I want to get there value using PHP here's my code
<form action="" method="POST">
<button name="apple" value="apple">apple</button>
<button name="windows" value="windows">windows</button>
<button name="linux" value="linux">linux</button>
</form>
Thank you
You can try this:
//for apple button or change to windows/linux if you want to get the other value
if (isset($_POST['apple'])) {
echo $_POST['apple'];
}
Related
i am new to php and while i am practing i came across a problem. actually,i have two files index1.php and index2.php. in index1.php i have a link with a unique id as
<a href="index2.php?companyid=<?php echo $row('company_id');?>>details</a>
i have got this value in index2.php as
if(isset($_GET['companyid'])){
$companyid = $_GET['companyid'];
}
now i have a search form in the index2.php as
<form method="POST" action="index2.php">
<input type="text" name="search">
<button type="submit" name="submit">submit</button>
</form>
now on button click i want the search results be displayed in the same page as
'index2.php?companyid=$companyid'
but some how if i try to use $_POST['submit']; in the same page it takes me to index2.php and instead of index2.php?companyid=$companyid and also it throws error as undefined index of $companyid if i don't use $_POST['submit']; and echo $companyid; it gives value and works fine. all i want is that to use $companyid' value inside ``$_POST['submit']; as and display the result in the same url as before
if(isset($_POST['submit']){
$companyid //throws an error index of company id
}
any help will be appreciated
First off, it looks like you are not using the company id in the form itself, so it will not be submitted as part of the the POST. You could possibly use:
<form method="POST" action="index2.php">
<?php if (isset($companyid)): ?>
<input type="hidden" name="companyid" value="<?= $companyid; ?>">
<?php endif; ?>
<input type="text" name="search">
<button type="submit" name="submit">submit</button>
</form>
But you would probably also need to change your logic to:
if(isset($_POST['companyid'])){
$companyid = $_POST['companyid'];
}else if(isset($_GET['companyid'])){
$companyid = $_GET['companyid'];
}
As Josh pointed out in the comments, PHP is not able to remember your previous GET request but this can easily be solved by altering the action attribute of the form element. By doing this you can pass on the previous data. This would look a little something like this:
<form method="POST" action="index2.php?companyid=<?php echo $companyid;?>">
<input type="text" name="search">
<button type="submit" name="submit">submit</button>
</form>
This way you will be redirected to index2.php with the URL parameters present and you will be able to retrieve both search and companyid using $_POST and $_GET or use $_REQUEST for both.
I want to show a button in my page only if a certain condition is met.
Also i want to run a query (DELETE QUERY) when i press that button.
<?php if (isset($_POST['finduser_btn']) && $noerr) :
echo "<div class='green'>
<button type='submit' class='btn' name='scoreDel'>Delete scores</button>
</div>
endif ?>
I use $noerr as FLAG to display or not the button if another button is pressed (the other button is not shown in code)
Well, how do i use scoreDel button to run a query like:
DELETE FROM scores
WHERE name = '$username$;
I think i have some issue with " and ' into the PHP echoing html tags but i'm not sure... I hope in some help, i'm getting mad.
Thanks in advance
You need a form in order to submit your action.
echo '<form action="mypage.php" method="POST"><div class="green">
<button type="submit" class="btn" name="scoreDel">Delete scores</button>
</div></form>';
Try the following:
<?php if (isset($_POST['finduser_btn']) && $noerr) : ?>
<div class='green'>
<form method="post">
<input type="text" name="finduser">
<button type='submit' class='btn' name='scoreDel'>Delete scores</button>
</form>
</div>
<?php endif ?>
Use a form to submit the button tag. Also, write html outside of PHP code if possible.
I want to create a checkbox and check whether checkbox is checked or not in another PHP page so I created form like :
<form action="heartbeat.php" method="post">
<input type="checkbox" name="keepme">
<input type="submit" value="log in">
</form>
and then I tried to extract this input on heartbeat.php like:
<?php
/*
* A PHP file for laying down a heartbeat JavaScript call.
*/
if(!isset($_POST["keepme"])){
$auto_logout = 10;
}
else{
$auto_logout = 1000;
}
?>
but it never gets $_POST["keepme"] value. Any idea??
If you don't pass value to checkbox tag, the POST array will be empty. You need to do something like that
<input type="checkbox" name="keepme" value='1'>
Make Sure you have submit button
If(isset($_POST['submit-btn'])){
echo $_POST['keepme'];
}
As the title says This is the code that I tried with. The forms must appear one by one because information from previous forms determine how the next ones will look.
$(document).ready(function(){
$('#first_form').submit(function(){
$('#first_form').fadeOut('fast');
$('#second_form').fadeIn('fast');
});
});
<form action="new_patch.php" method="POST" id="first_form">
Title: <input type="text" name="patch" placeholder="Patch 4.20">
<br/>
Number of Champions: <input type="number" name="champ_number" min="1" max="99">
<br/>
<input type="submit" value="submit">
</form>
<form action="new_patch.php" method="POST" id="second_form" style="display: none;" >
<input type="text" value="text">
<input type="submit" value="submit">
<?php
$champ_number = null;
if(isset($_POST['champ_number']))
{
$champ_number = $_POST['champ_number'];
for($champ_number;$champ_number>0;$champ_number--)
{
echo "<br/>Champion ".$champ_number."<input type=\"number\" name=".$champ_number." min=\"1\" max=\"99\">";
}
}
?>
</form>
You're mixing client-side and server-side form code. Submitting the form will reload the page entirely, so from the looks of your code it will fade in the new form when the old form is submitted, but then reload the page so the old form will show again anyway.
You could either:
Let the PHP determine how the next form appears based on the submission of the first form, e.g. if (isset($_POST["First_form_submit"]) { Show second form... }
Probably better and more user-friendly: make the second form appear below once the user has filled in the relevant inputs on the first form before they've submitted
you can use:
$('#first_form').submit(function(){
$('#first_form').fadeOut(function() {
$('#second_form').fadeIn('fast');
});
return false;
});
From the jQuery documentation the syntax is fadeIn( [duration ] [, complete ] ) it accepts a duration and a onComplete callback that you can use to execute the next action when the first is completed.
I did this once too, just add a submit class to the button and make it like this:
<input type="submit" value="submit" class="submit">
Change script to a click function.
$(document).ready(function(event){
event.preventDefault();
$('.submit').click(function(){
$('#first_form').fadeOut(400);
$('#second_form').fadeIn(400);
});
});
PS, also you need to prevent submit default...otherwise it will just submit the form, see this JSfiddle
I have two submit buttons in a form which has different values, one to voteup a post and another to votedown a post, how do i submit the value of the input type="submit" buttons?
Without using javascript, i'm making my script flexible to browsers with javascript disabled.
<?php
if (isset($_POST)) {
var_dump($_POST);
}
echo '
<form action="" method="post">
<input type="submit" name="vote_up" value="up">
<input type="submit" name="vote_down" value="down">
</form>';
?>
The value of the clicked button will always be submitted with the form. This is how you can disambiguate between different submit buttons on the same form.
Further clarification can be found at: http://www.javascript-coder.com/html-form/html-form-submit.phtml
Just include it into the form you are submitting and do something like this for disambiguate the buttons:
<input class="submitButton" type="submit" name="up" id="up">
Then in your code you can do something like:
if (isset($_POST["up"])){
// do something
}
you can also use
print_r($_POST);
in place of
var_dump($_POST);