It is very difficult for me to put in words my query. But I will try.
I have a site xyz.com which has search facility for listed products. The search page url is generated like this :www.wyz.com/search/search_term
I want to create a iframe page in a third party site with a search facility which can directly communicated with my site xyz.com.
I have tried to create a search box with a submit button. I want to append the search query in as a variable to my form action url string.
So the search string should look like this :www.wyz.com/search/my_string_variable
The code I have written is:
<?php
$url='http://www.xyz.com/search/';
?>
<?php
if (isset($_POST['submit']))
{
$r1=$_POST['num1'];
}
?>
<?php
$result=$url.$r1
?>
<html><body>
<form action="<?php echo $result; ?>" method="post">
Num1:<input name="num1"><br>
<input type="submit" name="submit">
</form>
</body></html>
==================================================================
But output what I get, is only "http://www.xyz.com/search/". It removes my variable from the url. I am not able to find what is the reason? I have also tried to print result via to check the actual output and it shows that it has added the value at the end of url. But when I want to achieve the same thing via form action it does not work. please help?
<?php
$url='http://www.xyz.com/search/';
?>
<?php
if (isset($_POST['submit']))
{
$r1=$_POST['num1'];
$result=$url.$r1;
header("location:$result");
}
?>
<html><body>
<form action="" method="post">
Num1:<input name="num1"><br>
<input type="submit" name="submit">
</form>
</body></html>
Please try the above code. I have made some modifications. The main reason your code is not working is whenever you press the submit button it is going to the the url "http://www.xyz.com/search/" directly .The if condition is never executed. In the above mentioned code it will work properly
action="" - you are submitting to the wrong url. Here is alternate version -
<?php $url='http://www.xyz.com/search/';
if (isset($_POST['submit'])) {
$r1=$_POST['num1']; header("Location: ".$r1); // 302 redirection
}
?>
<html><body> <form target="_SELF" method="post"> Num1:<input name="num1" type="text" /><br /> <input type="submit" name="submit" /> </form> </body></html>
Related
So i have an post input where i submit data something simple.
<form method="post" action="result.php">
<input type="url" name="url" class="form-control" placeholder="http://example.com/">
<input type="submit" name="submit" />
</form>
After the html code is going to be executed a php code which echo success or something like this that doesn't matter.
But i have a problem when i include('submit.php') it's going to show also the input and i don't want this.
How i can do that to don't show the input on result.php?
If you want it to be user-specific, you can try to use cookies or sessions like this:
index.php
<?php
session_start();
?>
<?php if(!isset($_SESSION['show_button']) && !$_SESSION['show_button'] ){ ?>
<!-- Button logic here... -->
<?php } ?>
result.php
// If the url has been entered, it returns a false from empty()
$_SESSION['show_button'] = empty($_POST['url']);
First time i try to create a simple form using the POST method.Problem is when i click the button nothing gets echoed.
here is my insert.php file :
<?php
if(isset($_POSΤ["newitem"])){
echo $itemnew = $_POSΤ["newitem"];
}
?>
<form action="insert.php" method="POST" >
<input type="text" name="newitem">
<input type="submit" value="Save">
</form>
EDIT: I tried the GET method and it works...Any ideas why that happened? Server configurations?
NEW EDIT: So it turns out i switched method to GET and it worked.Then i switched back to POST (like the code i posted on top) and it works...I have no clue why this happened.Any suggests?
The code you have posted is perfectly valid and should work.
I'm going to guess that you do not have PHP enabled, or it is not working.
<?php ... ?> looks to the browser like a long, malformed HTML tag, and therefore ignores it, making the effect invisible.
Try right-clicking the page and selecting View Source. If you see your PHP there, then the server is indeed not processing it.
The most likely reason for this is probably the same problem I had with my very first bit of PHP code: you're trying to "run" it directly in your browser. This won't work. You need to upload it to a server (or install a server on your computer and call it from there)
Use !empty($_POST['newitem'] instead:
if(!empty($_POSΤ["newitem"])){
echo $itemnew = $_POSΤ["newitem"];
}
empty()
Try the following:
if($_POST) {
if(!empty($_POST['newitem'])) {
$itemnew = $_POSΤ['newitem'];
echo $itemnew;
// or leave it as is: echo $itemnew = $_POSΤ['newitem'];
}
}
?>
<form action="insert.php" method="POST" >
<input type="text" name="newitem">
<input type="submit" value="Save">
</form>
The if($_POST) will make sure the code is only executed on a post. The empty() function will also check if it isset() but also checks if it is empty or not.
Try this :
<?php
if(isset($_POSΤ["newitem"])){
echo $itemnew = $_POSΤ["newitem"];
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST" >
<input type="text" name="newitem">
<input type="submit" value="Save">
</form>
$_SERVER['PHP_SELF']; is pre-defined variable in php.It allows the user to stay on same page after submitting the form.
I have a form on on html outside of php...
<form method="post" action="">
<input type="text" name="user"/></br>
<input type="submit" value="submit" name="login"/>
</form>
then call submit button from php and do this
if(isset($_POST["login"]))
{
print <<<this
<form method="post" action="">
<input type="submit" name="apply"/>
</form>
this;
if(isset($_POST["apply"]))
{ print "it works";}
}
Alright, so the problem is that, "it works" won't print from the second form thats inside the php. it just takes me back to where i came from. Perhaps it's a dumb question, please help though! thanks
The problem is that by the time you're checking if(isset($_POST["apply"])) the login condition becomes invalid because everything is inside the if(isset($_POST["login"])).
Try taking the if(isset($_POST["apply"])) outside the login IF.
Your "apply" code exists only INSIDE the login test code. When you submit that second form, there will be NO login form field, because you didn't include an input/textarea of that name in the second form. So the second form submits, there's no login, and the entire inner code never gets executed. You probably want:
if(isset($_POST["login"]))
{
print <<<this
<form method="post" action="" name="apply">
<input type="hidden" name="login" value="foo" /> <!-- add this line -->
etc...
I'm not sure to understand what you wanna do with this code but you obviously missed some details :
_You did not set the "action" field in your form tag, so I don't understant how you would like the PHP file to get called ?
_Your code if(isset($_POST['login'])) has no sense, you are testing the existence of a value sent by a validation button, you'd rather whrite isset($_POST['user'])
Hoping to have helped you
Your variables are declared in 2 forms, so there will be 2 calls (completely independant) to your php.
So you could have a second submit button inside your second form:
if(isset($_POST["login"]))
{
print <<<this
<form method="post" action="">
<input type="submit" name="apply" value="Second"/>
</form>
this;
}
if(isset($_POST["apply"]))
{ print "it works";}
I have a PHP form that is located on file contact.html.
The form is processed from file processForm.php.
When a user fills out the form and clicks on submit,
processForm.php sends the email and direct the user to - processForm.php
with a message on that page "Success! Your message has been sent."
I do not know much about PHP, but I know that the action that is calling for this is:
// Die with a success message
die("<span class='success'>Success! Your message has been sent.</span>");
How can I keep the message inside the form div without redirecting to the
processForm.php page?
I can post the entire processForm.php if needed, but it is long.
In order to stay on the same page on submit you can leave action empty (action="") into the form tag, or leave it out altogether.
For the message, create a variable ($message = "Success! You entered: ".$input;") and then echo the variable at the place in the page where you want the message to appear with <?php echo $message; ?>.
Like this:
<?php
$message = "";
if(isset($_POST['SubmitButton'])){ //check if form was submitted
$input = $_POST['inputText']; //get input text
$message = "Success! You entered: ".$input;
}
?>
<html>
<body>
<form action="" method="post">
<?php echo $message; ?>
<input type="text" name="inputText"/>
<input type="submit" name="SubmitButton"/>
</form>
</body>
</html>
The best way to stay on the same page is to post to the same page:
<form method="post" action="<?=$_SERVER['PHP_SELF'];?>">
There are two ways of doing it:
Submit the form to the same page: Handle the submitted form using PHP script. (This can be done by setting the form action to the current page URL.)
if(isset($_POST['submit'])) {
// Enter the code you want to execute after the form has been submitted
// Display Success or Failure message (if any)
} else {
// Display the Form and the Submit Button
}
Using AJAX Form Submission which is a little more difficult for a beginner than method #1.
You can use the # action in a form action:
<?php
if(isset($_POST['SubmitButton'])){ // Check if form was submitted
$input = $_POST['inputText']; // Get input text
$message = "Success! You entered: " . $input;
}
?>
<html>
<body>
<form action="#" method="post">
<?php echo $message; ?>
<input type="text" name="inputText"/>
<input type="submit" name="SubmitButton"/>
</form>
</body>
</html>
Friend. Use this way, There will be no "Undefined variable message" and it will work fine.
<?php
if(isset($_POST['SubmitButton'])){
$price = $_POST["price"];
$qty = $_POST["qty"];
$message = $price*$qty;
}
?>
<!DOCTYPE html>
<html>
<head>
<title></title>
</head>
<body>
<form action="#" method="post">
<input type="number" name="price"> <br>
<input type="number" name="qty"><br>
<input type="submit" name="SubmitButton">
</form>
<?php echo "The Answer is" .$message; ?>
</body>
</html>
You have to use code similar to this:
echo "<div id='divwithform'>";
if(isset($_POST['submit'])) // if form was submitted (if you came here with form data)
{
echo "Success";
}
else // if form was not submitted (if you came here without form data)
{
echo "<form> ... </form>";
}
echo "</div>";
Code with if like this is typical for many pages, however this is very simplified.
Normally, you have to validate some data in first "if" (check if form fields were not empty etc).
Please visit www.thenewboston.org or phpacademy.org. There are very good PHP video tutorials, including forms.
You can see the following example for the Form action on the same page
<form action="" method="post">
<table border="1px">
<tr><td>Name: <input type="text" name="user_name" ></td></tr>
<tr><td align="right"> <input type="submit" value="submit" name="btn">
</td></tr>
</table>
</form>
<?php
if(isset($_POST['btn'])){
$name=$_POST['user_name'];
echo 'Welcome '. $name;
}
?>
simple just ignore the action attribute and use !empty (not empty) in php.
<form method="post">
<input type="name" name="name">
<input type="submit">
</form>
<?PHP
if(!empty($_POST['name']))
{
echo $_POST['name'];
}
?>
Try this... worked for me
<form action="submit.php" method="post">
<input type="text" name="input">
<input type="submit">
</form>
------ submit.php ------
<?php header("Location: ../index.php"); ?>
I know this is an old question but since it came up as the top answer on Google, it is worth an update.
You do not need to use jQuery or JavaScript to stay on the same page after form submission.
All you need to do is get PHP to return just a status code of 204 (No Content).
That tells the page to stay where it is. Of course, you will probably then want some JavaScript to empty the selected filename.
What I do is I want the page to stay after submit when there are errors...So I want the page to be reloaded :
($_SERVER["PHP_SELF"])
While I include the sript from a seperate file e.g
include_once "test.php";
I also read somewhere that
if(isset($_POST['submit']))
Is a beginners old fasion way of posting a form, and
if ($_SERVER['REQUEST_METHOD'] == 'POST')
Should be used (Not my words, read it somewhere)
Hello Webmasters iam trying to redirect the webpage to a different URL/webpage after submitting the form elements ...i tried many ways but could not fix it... please check the codes below that i have tried so far...
<?php
if(isset($_REQUEST['down'])){
header("header("location: domainpath/kothi.html");
}
?>
<html>
<body>
<form action="glitter.php" method="post">
<input type="radio" name="font" value="fonts/darkcrystaloutline.ttf"/>
<input type="radio" name="font" value="fonts/darkcrystalout.ttf"/>
</form>
</body>
</html>
i have also tried
<?php
if(isset($_REQUEST['font'])){
header("location: domainpath/kothi.html");
};
?>
i also tried
<?php
header("location: domainpath/kothi.html");
?>
please help me to fix the problem....
Is that file glitter.php? The redirection script should be kept in glitter.php as that is the page that will be loaded when you submit the form.
There are several ways to do a redirect.
via meta tag: http://webdesign.about.com/od/metataglibraries/a/aa080300a.htm
via header: http://php.net/manual/en/function.header.php (you have to put exit after the header statement)
via Javascript: http://www.tizag.com/javascriptT/javascriptredirect.php
Firstly, the below:
<?php
if(isset($_REQUEST['down'])){
header("header("location: domainpath/kothi.html");
}
?>
This is incorrect, the header is incorrectly declared, replace header with the header below:
<?php
if(isset($_REQUEST['down'])){
header("location: domainpath/kothi.html");
}
?>
Secondly, why $_REQUEST? You're making a POST, and thirdly, where is down coming from? You form is submitting font so the following is what you need:
<?php
if(isset($_POST['font'])){
header("Location: domainpath/kothi.html");
exit();
}
?>
exit() is added to stop the rest of the page loading too, by the way...
Update
It may also be best if you submitted the data too, so include the below inbetween the <form> tags
<input type="submit" name="submit" value="Submit this form">
And like others point out, I hope this PHP page is called glitter.php so it can submit to itself...
Update 2
Based on your comment, then you'd want the following:
<?php
if(isset($_POST['down'])){
header("location: domainpath/kothi.html");
exit();
}
?>
<html>
<body>
<form action="glitter.php" method="post">
<input type="radio" name="font" value="fonts/darkcrystaloutline.ttf"/>
<input type="radio" name="font" value="fonts/darkcrystalout.ttf"/>
<input type="submit" name="down" value="down">
</form>
</body>
</html>
Though the above form will go to glitter.php and the header will not redirect anywhere - one has to assume that another form/page submits to this one...
You have action="glitter.php", which means after submitting the form the result inputs will be accessible into glitter.php file. So in that file you have to redirect to whatever url you wanna go.
You already asked this question once, but here add this :
error_reporting(E_ALL);
But i have a assumption: You have to make an exit() after header() cos the rest is not allowed to be give an output. If it gives an output the header will be set to a HTML-Document and can therefore not be reset.
So try this:
<?php
if(isset($_REQUEST['down']))
{
header("location: /kothi.html");
exit();
}
?>
<html>
<body>
<form action="glitter.php" method="post">
<input type="radio" name="font" value="fonts/darkcrystaloutline.ttf"/>
<input type="radio" name="font" value="fonts/darkcrystalout.ttf"/>
</form>
</body>
</html>
BTW: I hope this form is in glitter.php and where is the input for down?
If this file is not glitter, add this in glitter:
<?php
if(isset($_POST['font'])){
header("Location: domainpath/kothi.html");
exit();
}
?>
And kill the php in the form-page.