I'm very interested in if it is possible to connect javascript variables to php. I know that in php we can write javascript code, but on the contrary we could not. To express my aim better , lets bring example like that:
<form name="some" action="<?php $_SERVER['php_self']; ?>" method="post">
<input type="submit" name="but" value="Action">
</form>
My question is how to make after pressing submit button to confirm (alert) with javascript and if it is confirmed do something (with php) and if isn't cancel (php operation).
You can do two things to pass javascript variable to php:
You can pass it as a hidden input field and submit it using POST
<input id="myHidden" name="myHidden" type="hidden"/>
Assign javascript variable to hidden input something like
var myVariable;
document.getElementById("myHidden").value = myVariable;
You can pass it as a query string with in your URL
As for confirming you can use javascript confirm, which will post on OK and not post/cancel on CANCEL
Something like:
<input type="submit" onclick="return confirm('Are you sure you want to submit?')"/>
<form name="some" action="<?php $_SERVER['php_self']; ?>" method="POST"
onsubmit="document.getElementById("response")
= confirm('some question') ? 'yes' : 'no';
return true;">
<input type="hidden" name="response" id="response" value="">
<input type="submit" name="but" value="Action">
</form>
BTW action="<?php $_SERVER['php_self']; ?>" opens up your page to XSS attacks.
Related
I'll try and explain this as best as I can. Basically, I'm using a form to receive a comment. Upon hitting submit, the action creates a link similar to this: http://localhost:8080/camagru/comment.php?comment=test&post=Post
I have a variable with the image name in it that I want to pass as well, so something like this: http://localhost:8080/camagru/comment.php?img=test.png&comment=test&post=Post
I've tried using <form action="<?php echo commentpost.php?img=$img?>"> But everytime the submit button is pressed, it erases the img variable from POST and only puts in the new variables from the form.
Any suggestions?
add new hidden field in form tag like that
<form action="commentpost.php" method="post">
<input type="hidden" value="<?php echo $img ?>" name="img" />
<input type="submit" value="Save" name="IsSubmit" />
</form>
Now you can able to use $_POST['img']
The img variable is in GET.
If you want it in POST, try <input type="hidden" name="img" value="test.png">
use quotes in your case:
<form action="<?php echo "commentpost.php?img=$img"; ?>">
the best practice is to insert hidden element into your form:
<input name="img1" type="hidden" value="test.png" />
<input name="img2" type="hidden" value="test2.png" />
first of all, greetings and excuse my English...
I explain my problem:
I have some JS functions that generate some variables, I would like to send these variables via the POST method to a php file to be consulted in a database...
I have read that the best way to do this is sending the variables to the respective values about inputs within a form, Below I show the variables and the form:
Javascript variables:
On click in "Buscar"
$('#Buscar').on('click', function () {
document.getElementById("tipo").value=TipoDeInmuebleDATA.selectedData.value;
document.getElementById("operacion").value=TipoDeOperacionDATA.selectedData.value;
document.getElementById("habitaciones").value=HabitacionesDATA.selectedData.value;
document.getElementById("MetrosCuadrados").value=MetrosCuadrados;
document.getElementById("banos").value=BanosDATA.selectedData.value;
document.getElementById("precio").value=Precio;
});
The Form:
<form name="FormBuscar" method="post" action="consulta.php">
<input id="tipo" name="tipo" type="hidden" />
<input id="operacion" name="operacion" type="hidden" />
<input id="MetrosCuadrados" name="MetrosCuadrados" type="hidden" />
<input id="habitaciones" name="habitaciones" type="hidden" />
<input id="banos" name="banos" type="hidden" />
<input id="precio" name="precio" type="hidden" />
<input type="submit" name="Buscar" class="BotonBuscar">
</form>
I suspect that something is wrong, in the sense that I think the variables are not being sent to consultation
In consultation, if I do the following:
$tipo=$_POST['tipo'];
echo $tipo;
No result :-(
Greetings, I will await your answers in!
The event is wrong. You want to change the values right before submitting. So also add id="buscar-form" to the form tag. Then you can change the jQuery to this:
$('#buscar-form').on('submit', function () {
$('#tipo').val(TipoDeInmuebleDATA.selectedData.value);
$('#operacion').val(TipoDeInmuebleDATA.selectedData.value);
...
});
Forget about the #buscar button. When you click the button, the form is going to submit. So that's the event captured above. Careful with your typing, there are a lot of typos!
I have read the answer to this question, to execute PHP scripts with the click of a button. But what if I have a "nested button", like this :
<?php
if(!empty($_POST['act'])) {
echo "Ready to rock!";
$someVar = "Rock n Roll";
if(!empty($_POST['act2'])) {
echo $someVar;
} else {
?>
<form method="POST" action="">
<input type="hidden" name="act2" value="run">
<input type="submit" value="Rock It!">
</form>
<?php
}
} else {
?>
<form method="POST" action="">
<input type="hidden" name="act" value="run">
<input type="submit" value="Show It!">
</form>
<?php } ?>
I heard my problem can be solved with jQuery, but I no idea.
anyone please.
To execute a script on the server you use the action property of your form:
<form method="POST" action="myscript.php">
When clicking a input type="submit" the browser will go to to action of the form surrounding the input type="submit"
Nesting is not a issue, as the browser always will look for the 'surrounding' form.
Problem is in second form, so it will never calls in this code, because it fails in first $_POST variable IF statement, because in second form you do not POST variable "act". so you need to add it
<form method="POST" action="">
<input type="hidden" name="act" value="run">
<input type="hidden" name="act2" value="run">
<input type="submit" value="Rock It!">
</form>
with this form you should see echo $someVar;
p.s. if form action property is emtpy, by default it submits form to the same php script
Just like #DTukans said here, you need the hidden field. If you would post the second form, the value of act will be lost if you are not having a hidden field with the value of act from the first form.
In php you can also check which submit button you submitted by giving the input[type="submit"] a name, such as <input type="submit" name="form2">, then you could check if you submitted that form by:
if (isset($_POST['form2'])) {}, but this is not the case here.
Use the hidden input and you will be good to go.
How can I convert this $_GET method to $_POST method?
$linkword .= "\n$alpha[$c] ";
You cannot pass $_POST data in a url, you must use cURL (PHP), AJAX (javascript), or a similar tool to build full HTTP requests.
ALTERNATE SOLUTION
You could however build a small form that submits, but you would have to use a submit button control for the "link" and use some hidden form inputs. You can Re-style the button anyway you wish with CSS.
<form action="link/url.php" method="post">
<input type="hidden" name="letters" value="value of letters" />
<input type="hidden" name="n" value="value of $n" />
<input type="submit" value="text of button" name="submit" />
</form>
You cannot POST data using URL Parameters. You will have to use a form with a method set to POST. Then submit the form on clicking the link or use a button.
Solution would be to use hidden form elements for your params and submit the hidden form when your click on the anchor tag.
Another solution would be to use ajax to post data. If you use jQuery or some libraries, you can do so.
Instead of creating a anchor tag. Create a hidden form
$linkword .= "\n$alpha[$c]";
Here is a Sample
<form name="hiddenform" method="POST" action="page.php">
<input type="hidden" name="letters" value="<? echo alpha[$c].$letters; ?>" />
<input type="hidden" name="n" value="<? echo $n; ?>" />
<a onclick="document.forms['hiddenform'].submit();">Test Link <? echo $alpha[$c]; ?></a>
</form>
Jsfiddle Demo
in order to edit my entries i want to:
<form id="pregunta" name="pregunta" class="form_pregunta" method="post" action="pregunta.php?id=26">
<h2>Titulo de la pregunta</h2><input name="q" id="q" class="q" value="este es mi títiulo " type="text">
<h2>Describe tu pregunta</h2>
<textarea name="texto" id="texto" style="width: 98%;"><p>esta es mi descripcion</p></textarea>
<h2>Etiquetas</h2>
<input name="tags" id="tags" onmouseover="mostrar_tooltip('nube_e','','0','70','')" onmouseout="ocultar_tooltip('nube_e')" value="dos,tres,una,">
<input name="responde_a" style="display: none;" id="responde_a" value="0">
<button name="pregunta" id="pregunta" type="submit">form_edit_question_button</button>
</form>
And then in file.php
i'd like to $_get['id'] and $_post['inputs']
but when i go:
if(isset($_POST['edit_pregunta'])){
echo 'lalalalalalalalalalalalalalala';
post_edit_pregunta();
}
it won't ever enter :S. is that normal or i'm missing something... i wanted not to have a hidden input with the id of the post i want to edit..
I'm not 100% sure, but forms don't send their name when submitted, much less their id.
You could do the following instead:
<form id="edit_pregunta" method="post" action="file.php?id='$this->id'">
<input type="hidden" name="edit_pregunta" value="anything">
... //inputs here
</form>
and your if should now enter.
It looks like you're checking for your form's "id" attribute. This is not sent when the form is submitted, only values in <input>, <select>, <textarea> and <button> are sent.
You should check for one of those.
Edit: Your button name is "pregunta", so that is the POST variable you should be checking for, eg
if(isset($_POST['pregunta'])){
Just to comment in general on mixing params in the form's "action" and inputs, you can mix them as long as the form method is "post". You cannot set GET params in the form's action and use the "get" method
<!-- Good -->
<form action="proc.php?id=123" method="post">
<input name="foo" value="foo">
<input type="submit">
</form>
<!-- Bad -->
<form action="proc.php?id=123" method="get">
<input name="foo" value="foo">
<input type="submit">
</form>
There should be no problem at all with having get and post variables in the same request, but are you sure your syntax is correct? If this is normal php, shouldn't you write
<form id="edit_pregunta" method="post" action="file.php?id=<?php echo $this->id; ?>">
... //inputs here
</form>
[Edit]
The problem is (if I'm correct and this is standard php) that you generate a form that looks something like this:
<form id="edit_pregunta" method="post" action="file.php?id='$this->id'">
... //inputs here
</form>
This will make id look like this: '$this->id' (including the '-signs). When what you want is something like this:
<form id="edit_pregunta" method="post" action="file.php?id=51">
... //inputs here
</form>
Then $_GET['id'] would be 51.
[Edit2]
Also, I think you need to change
if(isset($_POST['edit_pregunta'])){
with
if(isset($_POST['pregunta'])){
If I'm not mistaken the name of a form doesn't get sent to the server, however, the name of the submit-button does, but I might be wrong about that part.
Yes you can, I've done it several times.
Probably something else is wrong with your code.
Is there any control with name="edit_pregunta" or is it just the id of the form? IDs are not sent to the server.
Simply adding the id to the form will not create the $_POST['edit_pregunta'] you verify.
Instead, inside the form tag, add an <input name="foo" />; in the php script verify $_POST['foo']
While the HTTP spec doesn't disallow query parameters in POST methods, it is somewhat unusual. You'd be better off using a hidden input field in the form to pass any non-user values up to the script.
That said, the syntax for your form is wrong. You need to use "echo" to insert the value of $this->id into the action.
Use input type="submit" in place of button tag.
You need name for form submission and activate php script!
HTML Code:
<form action="change.php" method="POST">
<input type="password" name="p1" class="change_text" placeholder="New Password"/></br>
<input type="password" name="p2" class="change_text" placeholder="Re-Password"/></br>
<input type="submit" name="change" value="Change Password" id="change" />
</form>
PHP Code:
<?php
if (isset($_POST['change']) {
$p1=$_POST['p1'];
}
?>