I'm sorry to repeat this question, but the thing is that I have done everything and nothing works. My problem is that I'm trying to pass variables to a second page and it won't work.
Page 1:
<form method="post" name="form1" id="form1" enctype="multipart/form-data" action="editempresas3.php?name=<?php echo $name;?>&descr=<?php echo $descr;?>&dir=<?php echo $dir;?>&pais=<?php echo $pais;?>&tel=<?php echo $tel;?>&fax=<?php echo $fax;?>&email=<?php echo $email;?>&url=<?php echo $url;?>">
<?php
$name = $_POST['empname'];
.....etc
?>
<input name="empname" type="text" required id="empname" form="form1">
.....etc
<input name="submit" type="submit" id="submit" form="form1" value="Crear">
Page 2:
The link will come without the variables
http://www.sample.org/editempresas3.php?name=&descr=&dir=&pais=&tel=&fax=&email=&url=
you should use GET method to achieve this.
change
<form method="post" name="form1" id="form1" enctype="multipart/form-data" action="editempresas3.php">
to
<form method="GET" name="form1" id="form1" enctype="multipart/form-data" action="editempresas3.php">
P.S: if you're form is not uploading anything you can't even miss enctype="multipart/form-data"
Possibilities, from most to least desirable:
Use sessions:
Page 1
session_start();
$_SESSION['var_for_other_page'] = 'foo';
Page 2
session_start();
$myvar = $_SESSION['var_for_other_page']
Use hidden fields:
<form action="secondpage.php" method="post>
<input type="hidden" name="var_for_other_page" value="foo" />
</form>
Put the get vars into the action URL:
<form action="secondpage.php?var_for_other_page=foo" method="post>
<input ... />
</form>
In this case you will have variables in both $_POST and $_GET.
Do not use either 2 or 3 to pass sensitive information.
If you want to send data from a form to a new page, firstly I think your should always use POST. The reason it is not working is you are attempting to send form data via POST but in your action you are trying to build a GET using PHP variables echoed in the there.
e.g.
action="editempresas3.php?name=<?php echo $name;?>&descr=<?php echo $descr;?>&dir=<?php echo $dir;?>&pais=<?php echo $pais;?>&tel=<?php echo $tel;?>&fax=<?php echo $fax;?>&email=<?php echo $email;?>&url=<?php echo $url;?>"
This can't work because PHP needs to process it before the HTML is rendered to print the variables you have chosen.
If you change your action to
action="editempresas3.php"
You will be successfully sent to the next page and if you then use
var_dump($_POST);
On your next page editempresas3.php you will get an output of all fields completed in the page 1 form.
Related
I'm a newbie in PHP, and I would like to send datas from a form and display it into the same page, here is my code for better understanding:
<form method="post" action="same_page.php">
<input type="text" name="owner" />
<input type="submit" value="Validate" />
</form>
<?php
if(isset($_GET['owner']))
{
echo "data sent !";
}
?>
So normally, after having entered some random text in the form and click "validate", the message "data sent!" Should be displayed on the page. I guess I missed something, but I can't figure out what.
You forgot to add submit name in your form.You are using POST as method so code should be
<form method="post" action="">
<input type="text" name="owner" />
<input type="submit" name="submit_value" value="Validate" />
</form>
<?php
if(isset($_POST['submit_value']))
{
echo '<pre>';
print_r($_POST);
}
?>
Will display your post values
You are using a POST method in your form.
<form method="post" action="same_page.php">
So, change your code to:
if (count($_POST) && isset($_POST['owner']))
Technically, the above code does the following:
First checks if there are content in POST.
Then, it checks if the owner is set.
If both the conditions are satisfied, it displays the message.
You can actually get rid of action="same_page.php" as if you omit it, you will post to the same page.
Note: This is a worst method of programming, which you need to change.
You should Replace $_GET['owner'] with $_POST['owner'] as in your form you have specified method='post'
Replace:
$_GET['owner']
With:
$_POST['owner']
Since you are using the post method in your form, you have to check against the $_POST array in your PHP code.
I have a simple form which was working and now im finding the post data isnt being sent and i can't see the problem
<form role="form" name="challengeform" action="scripts/arena_setup.php" method="POST" onsubmit="return confirm('Are you sure you want to attack this player?');">
<input type="hidden" name="member_id" value="<? echo $member_id;?>">
<input type="image" src="img/map/attack.png" alt="Attack" />
</form>
which is being handled by
if(isset($_POST['challengeform'])){
...
}else{ echo 'error'; }
it always shows the error due to the post data being missing but i just cant see what i've done. Any ideas?
if(isset($_POST['challengeform']))
Form names are not part of the POST data. Only fields within the form.
Try testing for the field itself
if(isset($_POST['member_id']))
You shouldn't write the name of the form. Just write input's name to get the data. Ex:
$var = $_POST['member_id'];
I have written a script in php to replace in newtopic button in phpbb3
in other question, a user says me this:
In your submit.php, you can retrieve the forum ID using $_GET['f']. Now, to pass it on to application.php, you can use a hidden input field:
<form method="post" action="application.php" accept-charset="utf-8" >
$id = htmlspecialchars($_GET['f']);
<input type="hidden" name="forum_id" value="<?php echo $id; ?>"/>
When you click on the submit button, the forum ID value will also get POSTed, and you'll be able to retrieve it in application.php code using the $_POST['forum_id'].
and my code goes as here:
<form method="post" action="application.php" accept-charset="utf-8" >
$id = htmlspecialchars($_GET['f']);
<input type="hidden" name="forum_id" value="<?php echo $id; ?>"/>
.............
<fieldset class="submit-buttons">
<input value="Submit" class="button2" type="submit">
</fieldset>
This code is embedded in submit.php to use phpbb3 template.
and application.php goes as here
So I click on new topic button, and I redirect to submit.php?mode=post&f=3 and in that php there is embedded the html, the problem is that with the solution, I receive the next error:
"The forum you selected does not exist" and the addresswar goes as: viewforum.php?f=&sid=a69fb9f491d2adc11c4be3a6dac02774
so I think that forum_id (in thos case is "3" (&f=3) is not correctly sent throught php scripts
I would appreciate some help
You need to add $id = htmlspecialchars($_GET['f']); inside the <?php ?> tag,
<?php $id = htmlspecialchars($_GET['f']); ?>
Suppose I have a form. After I submit my form, the data is submitted to dataprocess.php file.
The dataprocess.php file processes the variable sent via form and echoes desirable output.
It seems impossible to echo to a specified div in specified page only using PHP (without using AJAX/JavaScript as well). I do not want to use these because some browsers might have these disabled.
My concern is that I want to maintain the same formatting of the page that contained the form element. I want the form element to be there as well. I want the query result to be displayed below the form.
I could echo exact html code with some modification but that's memory expensive and I want it systematic.
Is it possible to process the form within the same page? Instead of asking another .php file to process it? How does one implement it?
The above is just for knowledge. It will be long and messy to include the PHP script within the same HTML file. Also, that method might not be efficient if I have same process.php file being used by several forms.
I am actually looking for efficient methods. How do web developers display query result in same page? Do the echo all the html formatting? also, does disabling JavaScript disable jQuery/AJAX?
Yes it is possible to process the form on the same page.
<?php
if (isset($POST))
{
//write your insert query
}
?>
<html>
<body>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<!-- Your form elements and submit button -->
</form>
<table>
<?php
//your select query in a while loop
?>
</table>
</body>
</html>
But if you choose this technique instead of ajax, you have to refresh all the page for each insert action.
An example
<div id="dialog-form">
<form action="<?php echo $_SERVER["PHP_SELF"]; ?>" method="post">
<table>
<tr>
<td>Job</td>
<td>
<input type="text" name="job" />
</td>
</tr
</table>
<input type="submit" value="Insert" />
</fieldset>
<input type="hidden" name="doProcess" value="Yes" />
</form>
</div>
<?php
$myQuery= $db->prepare("INSERT INTO Jobs (job) VALUES (:p1)");
if (isset($_POST['doProcess']) && $_POST['doProcess'] == 'Yes')
{
$myQuery->bindValue(":p1", $_POST['job'], PDO::PARAM_STR);
$myQuery->execute();
}
?>
if you really dont want to use ajax (which i think you should). You can do something like this.
<form action="" method="POST">
<input type="text" value="something" name="something_name"/>
<?php
if(isset($_POST['something_name'])){
echo '<div id="display_something_name_if_exists">';
echo $_POST['something_name'];
echo '</div>';
}
?>
</form>
Basically what it does is submits to itself and then if there is a submission (tested with isset), it will echo a div with the correct information.
<form id="enter" action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post" onsubmit="return validateForm(this);" >
<p>
<input id="submitBtn" name="submitDetails" type="submit" value="Submit Details" onClick="myClickHandler(); return false;" />
</p>
</form>
<script type="text/javascript">
function myClickHandler(){
if(validation()){
showConfirm();
}
}
</script>
<?php
session_start();
$outputDetails = "";
$outputDetails .= "<table id='sessionDetails' border='1'>
<tr>
<th>Number of Sessions:</th>
<th>{$_POST['sessionNum']}</th>
</tr>";
$outputDetails .= "</table>";
echo $outputDetails;
?>
Above is the code for my form. What I am trying to do is that if the user submits the form, then it will go back to its own page. But if the "SessionNum" equals '1', then instead of posting the form to itself, it should post the form or in other words navigate to the "session_marks.php' page but it is not idng this, if sessionNum equals 1 then it still submits form or navigate back to its own page, what am I doing wrong?
Also lets say it displays a number for the sessionNum and then I submit the form and it submits the form back to itself, the number disappears, how do I keep the number displayed when submitting the form to itself?
Thanks
Where is the conditional logic to change the target of the form post? All I see in the form tag is this:
action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>"
This will always set the form's action to be the current PHP file, not any other PHP file. If you want to conditionally post to a different file, you'll need to add conditional logic in there. Something like this (though there may be better ways to do it, keep in mind that I'm very out of practice with PHP):
action="<?php $_POST['sessionNum'] == 1 ? echo 'session_marks.php' : echo htmlentities($_SERVER['PHP_SELF']); ?>"
As for the number disappearing, I don't see any form element with the name sessionNum. If there isn't such a form element, then there will be nothing in $_POST['sessionNum'], so the number will "disappear" because there's no value to be displayed.
If the above is your actual code, session_start(); has to be placed before ANY other output (html, php's echo, print etc...)