<input type="button" value="Delete" onclick=show_confirm(http://mydomain.com/delete.php)>
I have this button and this JavaScript function, but they don't work.
When I click the button, nothing happens.
What is wrong with my code?
function show_confirm(url){
var r = confirm("Do you want to delete this?");
if (r == true){
window.navigate(url);
}
else{
alert("You pressed Cancel.");
}
}
Your html is not well formed.
<input type="button" value="Delete" onclick=show_confirm(http://mydomain.com/delete.php)>
should be
<input type="button" value="Delete" onclick="javascript:show_confirm('http://mydomain.com/delete.php');" />
The javascript: is only necessary if you also have VBScript on the page, but it's a good habit to be in.
Pass the url inside double quotes.
<input type="button" value="Delete" onclick=show_confirm("http://mydomain.com/delete.php")>
Working sample : http://jsfiddle.net/93QWJ/
<input type="button" value="Delete" onclick="show_confirm('http://mydomain.com/delete.php')">
no quotes around onclick function
like this : "show_confirm(url)"
Related
I have a simple form in HTML that contains two buttons. Button 1 which action in the form tag submits it to another php page e.g. button1-action.php which submits data to a third party API and Button 2 which I want to submit to the same page if it is clicked without going to button1-action.php.
In its simplest method the form is as follows:
<?php
echo '<form name="form123" id="form123" action="button1-action.php" method="POST">';
echo '<input type="text" name="first_name" id="first_name></input>';
echo '<button name="button1" id="button1" value="button1">Button 1</button>';
echo '<button name="button2" id="button2" value="button2">Button 2</button>';
echo '</form>';
?>
This is what I tried so far
$action = null;
if (isset($_POST['button1'])) {
$action = 'button1-action.php';
} elseif (isset($_POST['button2'])) {
$action = $_SERVER["PHP_SELF"];
}
echo '<form name="form123" id="form123" action="' . $action . '" method="POST">';
echo '<input type="text" name="first_name" id="first_name></input>';
echo '<button name=" button1" id="button1" value="button1">Button 1</button>';
echo '<button name="button2" id="button2" value="button2">Button 2</button>';
echo '</form>';
However, it doesn't seem to be working. I tried to look for solutions but I haven't been successful.
I'm interested in any solution, but I would prefer solving it using PHP and not JavaScript.
The Issue might be that you forgot to close the <form> tag with </form> and you should use the <input> for buttons aswell with type="submit" .
If this still doesn't resolve your issue then maybe you should try this :
On the same page.
<?PHP
//// place this on top
if($_POST["button1"]) {
// add code to send data to Third Party API
}
if($_POST["button2"]) {
// will show data here
} ?>
////////
<?php
echo '<form name="form123" id="form123" action="/">';
echo '<input type="text" name="first_name" id="first_name></input>';
echo '<input type="submit" name="button1" id="button1" value="button1" >';
echo '<input type="submit" name="button2" id="button2" value="button2" >';
echo '</form>';
?>
I hope this answers your question đ
This is the solution for you in html
<form name="form123" id="form123" method = "post">
<input type="text" name="first_name" id="first_name"></input>
<button name=" button1" id="button1" value="button1" formaction="button1-action.php" >Button 1</button>
<button name="button2" id="button2" value="button2" >Button 2</button>
</form>
button 1 will submit the form to button1-action.php and button 2 will submit the form to same page.
Documentation: https://developer.mozilla.org/en-US/docs/Web/HTML/Element/button#attr-formaction
You appear to be using the value submitted from the form to set the action of the form. This makes no sense - by the time you read the submitted values, the action has already happened. So your code would just set the action for next time the form is submitted. Not useful.
At the heart of this there seems to be a conceptual / design issue. A more sensible approach (but not the only one) would be to simply post the form to the same place every time, and then use if statements to decide what to do next.
e.g.
if (isset($_POST['button1'])) {
require_once "button1-action.php";
} elseif (isset($_POST['button2'])) {
//do whatever it is you want to do in ths script
}
else {
?>
<form name="form123" id="form123" method="POST">';
<input type="text" name="first_name" id="first_name></input>
<button name=" button1" id="button1" value="button1">Button 1</button>
<button name="button2" id="button2" value="button2">Button 2</button>
</form>
<?php
}
To improve a bit more on that, instead of using a bare require to include the code from another script, we could encapsulate the code from button1-action.php into a function which we can call, instead of a script with global scope. This makes the code more re-usable, maintainable, testable, less likely to cause scope conflicts, etc.
e.g.
if (isset($_POST['button1'])) {
callTheApi($_POST["first_name"]);
} elseif (isset($_POST['button2'])) {
doSomethingElse($_POST["first_name"]);
}
else {
?>
<form name="form123" id="form123" method="POST">';
<input type="text" name="first_name" id="first_name></input>
<button name=" button1" id="button1" value="button1">Button 1</button>
<button name="button2" id="button2" value="button2">Button 2</button>
</form>
<?php
}
(Even better if you then encapsulate that function in a class containing closely related functionality, but let's just get as far as a funtion for now.)
Alternatively, Virender Kumar's answer would also be reasonable - simply setting the form action of each button directly.
First of all your form is not structured properly.
index.php
<form name="form123" id="form123" action="button1-action.php">
<input type="text" name="first_name" id="first_name"></input>
<button name=" button1" id="button1" value="button1">Button 1</button>
<button name="button2" id="button2" value="button2">Button 2</button>
</form>
button1-action.php
if (isset($_GET['button1'])) {
echo 'button1 submitted'; // Send data to the third party API
} else if (isset($_GET['button2'])) {
echo 'button1 submitted'; // Submit on the same page
}
Edit: ignore my solution; Virender Kumarâs solution here is correct, elegant and doesnât need JS.
Original answer:
The issue is not with your buttons, but with the fact that a form can only post to a single endpoint (the action attribute). You will have to handle what happens with the form data from there. If you truly want your form to be posted to a different endpoint in the client based on what button the user clicks, you canât do it without JavaScript.
If you can live with JS, this could work:
<body>
<!-- your form here -->
<script>
const form = document.forms[0]; // assuming your form is the first form on the page, or the only one
document.querySelectorAll('button').forEach(button => {
button.addEventListener('click', event => {
if (event.target.name === 'button1') {
form.action = 'button1-action.php';
} else if (event.target.name === 'button2') {
form.action = 'other-destination';
}
});
});
</script>
</body>
I have a markup like this
<button type="button" class="btn btn-success pull-right add-customer" name="add-new-customer">Add New Cusomer</button>
so when the add new customer will be clicked it should show some message. like this
<?php
if(isset($_POST['add-new-customer'])) {
echo 'Set';
}
?>
but it is not doing isset for the button. So can somone tell me how to solve this using php. I have not used any kind of form. I want to simply check the button is set and show some message. Any help will be really appreciable. Thanks
This is because if you click a button within a form it doesnt actually submit the form.
Try using
<input type="submit" value="Submit" name="add-new-customer" />
EDIT:
Having just seen your comment, you must wrap the elements in a
<form>
Button elements aren't linked to anything in HTML. PHP won't detect if you click on a button.
You have to use a form or an AJAX query in order to populate the $_POST variable.
<form action="" method="post">
<input type="submit" value="Submit" name="add-new-customer" />
</form>
button type must be submit like type="submit" not type="button"
<button type="submit" class="btn btn-success pull-right add-customer" name="add-new-customer">Add New Cusomer</button>
also use form tag
This is my problem - when I hit the submit button it is supposed to execute some code. Unfortunately all it does is refresh the page at a weird url using the submit's ID and value.
File is called Q2C.php.
This is the code:
<?php
if(isset($_POST['submit'])) {
$arraylength = count($selectedQuotes);
for($i = 0; $i < $arraylength; $i = $i + 1) {
$supplierQuoteID = $selectedQuotes[$i];
$updateSQL = "UPDATE supplierquotes SET `Selected` = 1 WHERE `SupplierQuoteID` = '$supplierQuoteID'";
$updateQuery = mysql_query($updateSQL);
}
}
?>
<div class="buttons">
<input type="submit" name="submit" id="submit" value="Create Q2C" />
<input type="submit" name="cancel" id="cancel" value="Cancel" />
</div>
It redirects me to this url: (path)/Q2C.php?submit=Create+Q2C
How can I stop it from redirecting me here, and instead just perform what is in the if isset statement?
You need to put your buttons into a form.
<form method="post">
<input type="submit" name="submit" id="submit" value="Create Q2C" />
<input type="submit" name="cancel" id="cancel" value="Cancel" />
</form>
You should consider changing your second submit button to a cancel button.
I don't see a <form> in your HTML, but you need to be using POST, not GET
<form method=post>
You have not submitted your form completely in the above code, but it is clear from the redirected url that you are using GET as post method of the form. And in your isset check you are using $_POST. Change the form method to post then it will work fine.
You need to specify method. By default its get
<div class="buttons">
<form action="Q2C.php" method="post">
<input type="submit" name="submit" id="submit" value="Create Q2C" />
<input type="submit" name="cancel" id="cancel" value="Cancel" />
</form>
</div>
You should probably use ajax to submit a form if you don't want to refresh the page. Here is the same code you can refer to:
$.ajax({
type: "GET",
url: "abc.php",
data: "name="+form.name.value,
success: function(rs)
{
// do success stuffs here
}
});
Hello am not sure if i get your problem correct its like You want Submit a form to Q2C.php with out a form attribute called "Method" if u dont set the method at all, the form will use the GET method as default..... but from your code it seems the form is using get to submit to Q2C.php.... and you are saying (isset($_POST['submit'])) which is wrong because the Submit button was never set.. what u need to do is change set the attribute method of the form to "post" method and try running the code again... your form should look like this...
.........
I only just started learning JS like 5-10 minutes ago, I was told by someone to try and create a basic validation feature, however it doesn't seem to work as wanted. It checks if field is empty, that part works. But checking if it's got something in it and carrying on running the code doesn't.
my form:
echo '<form action="index.php?action=getHashedText" method="post" name="formHash">
<br/><textarea name="text" rows="4" cols="50" placeholder="Add your text/pharse/word which you want hashing here." autofocus></textarea><br/>
<button type="button" name="button" onclick="return validate();">Hash</button>';
function validate():
<script>
function validate() {
with (window.document.formHash) {
if (formHash.text.value === "") {
alert('Field is empty!');
return false;
} else {
return true;
}
}
}
</script>
The problem is you are using a button that is not submitting the form anyway (regardless of your js), change this dom:
<button type="button" name="button" onclick="return validate();">Hash</button>
To this: Fiddle
<input type="submit" name="button" value="Hash" onclick="return validate();" />
Or you can just add type="submit" on your <button> (HT #RocketHazmat): Fiddle
<button type="submit" name="button" onclick="return validate();">Hash</button>
Or you can just remove the type on your <button> all together as the default type is submit (HT #FabrĂcioMattĂ©): Fiddle
<button name="button" onclick="return validate();">Hash</button>
Also, slightly off topic but I would get in the habit of avoiding putting javascript onclicks directly on your elements. You can create listeners instead: addEventListener
I have a doubt on the following code. My function is not called when the save button is clicked .
This is the following code for save function,
if(isset($_POST['Save'])) // If the submit button was clicked
{
echo "hgfd";
$post['ProductSegmentCode'] = $_POST['ProductSegmentCode'];
$post['ProductSegment'] = $_POST['ProductSegment'];
$post['ProductGroup'] = $_POST['productgroup'];
// This will make sure its displayed
if(!empty($_POST['ProductSegment'])&&!empty($_POST['ProductSegmentCode'])&&!empty($_POST['productgroup']))
{
echo "SAVE";
$news->addNews($post);
?>
<script type="text/javascript">
alert("Created Sucessfully..!!");
</script>
<?
}
else
{
?>
<script type="text/javascript">
alert("Enter Mandatory Fields");
</script>
<?
}
}
following is the button format in html,
<div style="width:70px; height:32px; float:left; margin-top:16px; margin-left:4px;">
<input name="Save" type="button" class="button" value="Save">
</div>
Your button is type="button"; to get the form to submit, it needs to be type="submit". Try updating it with this and it should work (also pending you form has action="post", or no action specified; the default is post):
<input name="Save" type="submit" class="button" value="Save" onclick="Save" />
Also, you're using onclick="Save" in your button. This indicates you have a corresponding JavaScript function named Save() - though, per your code examples you do not show one. I'm assuming that this is in error and can safely be removed (the value="Save" can also be removed as you only need to check isset($_POST['Save']) and not it's actual value). All changes in-place should give you:
<input name="Save" type="submit" class="button" />
If you do, in fact, have a JavaScript function named Save(), please post its code and I can revise.
use form for sending data and use type submit
<form action="" method="post">
<input name="Save" type="submit" class="button" value="Save">
</form>
and if you want to use this
<input name="Save" type="button" class="button" value="Save" onclick="Save()">
create Save() function in javascript and use ajax call for sending data.
It looks like you should change
<input name="Save" type="button" class="button" value="Save" onclick="Save">
to a summit button.
<input name="Save" type="submit" class="button" value="Save">