I am having issues with a piece of code in Joomla. It may be something to do with the plugin that enables the PHP but in case it isn't.
Page 1 has a form
<form action="/index.php/bridge" method="POST" name="postcode">
<div><input style="height: 50px;" type="text" placeholder="Enter Your Postcode..." /> <input type="submit" value="Get Started today!" /></div></form>
The text you input becomes the variable I want to pass over
On Page 2
<?php echo "test";
$postcode=1;
$poster=$_POST['postcode'];
echo $poster;
// You can place PHP like this
?>
Unfortunately, the postcode isn't echoed
Assuming nothing else is the cause of this error, try naming the input you are sending over to postcode:
<form action="/index.php/bridge" method="POST">
<div>
<input style="height: 50px;" name="postcode" type="text" "placeholder="Enter Your Postcode..." />
<input type="submit" value="Get Started today!" />
</div>
</form>
In your PHP code you are echoing $_POST['postcode'] but you are not sending the same variable at the time of form submission from input attribute.
<form action="/index.php/bridge" method="POST" name="demoForm">
<div>
<input style="height: 50px;" name="postcode" type="text" "placeholder="Enter Your Postcode..." />
<input type="submit" value="Get Started today!" />
</div>
</form>
Try this
<form action="/index.php/bridge" method="POST">
<div>
<input name="postcode" style="height: 50px;" type="text" placeholder="Enter Your Postcode..." />
<input type="submit" name="submit" value="Get Started today!" />
</div>
<?php
if (isset($_POST['submit'])){
echo $poster = $_POST['postcode'];
}else{
echo $poster = 1;
}
?>
Related
i want to get value of input of my form via $_post in my insert page but my problem is that i get error massage:undefined index:track_code and i cant get value of it.the names is same what is problem?
this is my problem why stackoverflow want more detail. this is all detail
<form action="insert.php" method="post">
<input name="track_code" type="text" class="tiny-size" value="<?php echo $track_code_rnd ; ?>" style="width: auto;height: auto" disabled />
</form>
<form action="insert.php" method="post" class="form-horizontal form-checkout">
<div class="control-group">
<label class="control-label" for="firstName">نام<span class="red-clr bold">*</span></label>
<div class="controls">
<input name="first_name" type="text" id="firstName" class="span4" >
</div>
</div>
<div class="control-group">
<label class="control-label" for="lastName">نام خانوادگی<span class="red-clr bold">*</span></label>
<div class="controls">
<input name="last_name" type="text" id="lastName" class="span4">
</div>
</div>
<input name="submit" type="submit" class="btn btn-success" value="خرید" style="width: 66px"/>
</form>
<form action="insert.php" method="post">
<input name="track_code" type="text" class="tiny-size" value="<?php echo $track_code_rnd ; ?>" style="width: auto;height: auto" disabled />
</form>
<?php
$track_code = $_POST['track_code'];
?>
Set this one code in your form
<form action="insert.php" method="post">
<input name="track_code" type="text" class="tiny-size" value="<?php echo ($track_code_rnd)?$track_code_rnd:'';?>" style="width: auto;height: auto" disabled />
</form>
Remove disable tag from your input code. Disable controls wont be posted on server. for that you can make it readonly or make it hide
Try Putting the PHP code on the form. as given below.
<?php
if (isset($_POST['submit'])) {
$track_code = $_POST['track_code'];
}
?>
<form action="your_page.php" method="post">
<input name="track_code" type="text" class="tiny-size" value="<?php echo $track_code_rnd; ?>" style="width: auto;height: auto" readonly />
<input name="submit" type="submit" />
</form>
It will execute the whole file as PHP. The first time you open it, $_POST['submit'] won't be set because the form has not been sent. Once you click on the submit button, it will print the information.
Disabled controls never post on server. So make it readonly or hidden.
In this
<form method="POST" action="">
<input type="text" class="field small-field" name="tex1" />
<input type="submit" value="search" name="search"/>
<input type="submit" value="print" name="print"/>
</form>
After I submit the form, the page refreshes and the data inside the input texts gets blank
Is it possible to keep the data even after submit?
Regards.
You can simply use ajax for submitting the form.
Or use following
<form method="POST" action=""><input type="text" class="field small-field" name="tex1" value="<?php (isset($_POST['text1]))? echo $_POST['text1] : '';" /><input type="submit" value="search" name="search"/><input type="submit" value="print" name="print"/></form>
try to echo, what ever is the variable named for your input.
<form method="POST" action="">
<input type="text" class="field small-field" name="tex1" value="<?php echo $_POST['tex1'];?>" />
<input type="submit" value="search" name="search"/>
<input type="submit" value="print" name="print"/>
</form>
With php for example:
<form method="POST" action="">
<input type="text" class="field small-field" name="tex1" value="<?php echo $_POST['tex1']; ?>"/>
<input type="submit" value="search" name="search"/>
<input type="submit" value="print" name="print"/>
</form>
If you are handling the post on the same page you could just do like this on the fields where you want the posted value to be shown:
<input type="submit" value="search" name="search" <?php if( isset( $_POST['search'] ) ){ echo "value=\"". $_POST['search'] ."\"; } ?>/>
Use this:
<form method="POST" action="">
<input type="text" class="field small-field" name="tex1" value="<?php if(isset($_POST['tex1'])) echo $_POST['tex1'] ?>" />
<input type="submit" value="search" name="search"/>
<input type="submit" value="print" name="print"/>
</form>
Bascially http is statelessprotocol , hence you need to save the data some where
The simplest way in this case would be to use a conditional operator
<input type="text" class="field small-field" name="tex1" value="<?php echo (isset($_POST['search'] || $_POST['search'] )?$_POST['tex1']:''); ?>" />
I have two forms on a page-'indexpage plus a included page'. Their actions point as so action=''. The problem is when I hit submit on the search bar, it shows a header error. The submit for the login works fine. I narrowed the problem all the way down to
tabindex="6" type="submit"
in the submit button of the login form.
They both have different names.
Login Form:
echo'<div id="container">
<div id="topnav" class="topnav"> <font color="white">Have an account?  </font> <span>Sign in</span><span>Sign Up</span> </div>
<fieldset id="signin_menu">';
output_errors($errors);
echo'
<form id="signin" action="" method="POST" >
<br>
<label for="username"><font color="black">Username or email</font></label>
<input id="username" name="username" value="" title="username" tabindex="4" type="text">
</p>
<p>
<label for="password"><font color="black">Password</font></label>
<input id="password" name="password" value="" title="password" tabindex="5" type="password">
</p>
<p class="remember">
<input id="signin_submit" name="submit" value="Sign in" tabindex="6" type="submit">
<input id="remember" name="remember_me" value="1" tabindex="7" type="checkbox">
<label for="remember"><font color="black">Remember me</font></label>
</p>
<p class="forgot"> Forgot your password? </p>
<p class="forgot-username"> <A id=forgot_username_link
title="If you remember your password, try logging in with your email"
href="/recovery/username">Forgot your username?</A> </p>
</form>
</fieldset>
</div>';
Search Form:
<form action='' method='POST'>
<input type='text' value='". $clean ."' name='keywords'/>
<input type='submit' name='submit2' Value='Search'/>
</form>
Header Error:
Warning: Cannot modify header information - headers already sent by (output started at C:\Program Files
(x86)\xampp\htdocs***\index.php:298) in C:\Program Files
(x86)\xampp\htdocs***\pages\search.inc.php on line 35
I'm petty sure that error is due to you trying to output (echo/print) something before trying to redirect in the headers. Redirects header('Location: /anotherpage.php'); after an echo or some printed HTML are probably the cause.
I found the problem in my input for the login submit. Here is what I did.
I change this
<input id="signin_submit" name="submit" value="Sign in" tabindex="6" type="submit">
Too this
<input name="submit" value="Sign in" type="submit">
Apparently the tabindex and id had some kind of issue, not sure why.
I have multiple forms and I have one php script that I want to use to process these forms but when I click on submit for any of the forms...the script is processed by the number of forms with the submit button named 'submitForm' in this case, the alert will show 3 times instead of once! What am I not doing right?
NB. I hope this makes much sense?
html code
<form action="" name="form1" method="post">
<input type="text" value="" />
<input type="text" value="" />
<input type="text" value="" />
<input type="text" value="" />
<input type="Submit" value="Submit Form" name="submitForm" />
</form>
<form action="" name="form2" method="post">
<input type="text" value="" />
<input type="text" value="" />
<input type="text" value="" />
<input type="text" value="" />
<input type="Submit" value="Submit Form" name="submitForm" />
</form>
<form action="" name="form3" method="post">
<input type="text" value="" />
<input type="text" value="" />
<input type="text" value="" />
<input type="text" value="" />
<input type="Submit" value="Submit Form" name="submitForm" />
</form>
php script
<?php
if (isset($_POST['submitForm'])) {
echo('<script>alert("Form Submitted")</script>');
}
?>
when I click on submit for any particular form, it submits all the forms.
this is not true.
Once your forms have proper formatting, your browser will submit only current one.
(and PHP has nothing to do here)
however, whole page will be reloaded, if you mean that. That is okay - when you submit a form, a page is intended to reload. If you need another behavior, you have to explain your wishes.
Also note that none of your text fields being sent to the server as they have no names.
I guess the question I should be asking is, how do I pass a particular form to php instead of writing multiple php scripts to handle each form!!!
well, it seems you want to ask how to distinguish these forms.
add a hidden field into each
<input type="hidden" name="step" value="1" />
and then in PHP
if ($_POST['step'] == 1) {
//first form
}
if ($_POST['step'] == 2) {
//second
}
This submits one form of many to php. Copy, paste, test, and study.
<?php
if (isset($_POST['submitForm'])) {
print_r($_POST);
}
?>
<form action="" name="form1" method="post">
<input type="text" value="" name="A" />
<input type="text" value="" name="B" />
<input type="text" value="" name="C" />
<input type="text" value="" name="D" />
<input type="Submit" value="Submit Form" name="submitForm" />
</form>
<form action="" name="form2" method="post">
<input type="text" value="" name="A" />
<input type="text" value="" name="B" />
<input type="text" value="" name="C" />
<input type="text" value="" name="D" />
<input type="Submit" value="Submit Form" name="submitForm" />
</form>
<form action="" name="form3" method="post">
<input type="text" value="" name="A" />
<input type="text" value="" name="B" />
<input type="text" value="" name="C" />
<input type="text" value="" name="D" />
<input type="Submit" value="Submit Form" name="submitForm" />
</form>
Using a,b,c,d for the first form, e,f,g,h for the second form and i,j,k,l for the third form and submitting the second form yields the following output:
Array
(
[A] => e
[B] => f
[C] => g
[D] => h
[submitForm] => Submit Form
)
#Jay
Actually its not hard.
Once you supply form names, your work is done. the DOM does the rest.
write one php block to do your functions (create/update/retrieve/delete)
Whichever button is clicked, by default it submits only the elements enclosed together with it.
if(!empty($_POST)){
if(isset($_POST['submit'])){
print "<pre>";
var_dump($_POST); // write your code here as you would
print "<pre>";
}
}
try this with your form above.
I know this is an old post but here's how I solve this very problem.
All you need to do is make sure the submit buttons in each form have different names. Eg:
<form action="" name="form1" method="post">
<input type="text" value="" />
<input type="text" value="" />
<input type="text" value="" />
<input type="text" value="" />
<input type="Submit" value="Submit Form" name="submitForm1" />
</form>
<form action="" name="form2" method="post">
<input type="text" value="" />
<input type="text" value="" />
<input type="text" value="" />
<input type="text" value="" />
<input type="Submit" value="Submit Form" name="submitForm2" />
</form>
<form action="" name="form3" method="post">
<input type="text" value="" />
<input type="text" value="" />
<input type="text" value="" />
<input type="text" value="" />
<input type="Submit" value="Submit Form" name="submitForm3" />
</form>
Then, you simply check which form's submit button was pressed.
<?php
if (isset($_POST['submitForm1'])) {
echo('<script>alert("Form 1 Submitted")</script>');
} elseif (isset($_POST['submitForm2'])) {
echo('<script>alert("Form 2 Submitted")</script>');
} elseif (isset($_POST['submitForm3'])) {
echo('<script>alert("Form 3 Submitted")</script>');
}
?>
If you need dynamic forms, you may try below code. While statement can be changed to fetch data from DB and use foreach instead. Hope you know this.
Here, I used while($n<10) for 10 dynamic forms.
You can also use tag as below if you need separate form names.
<form action="" name="form<?=$n?>" method="post">
This will create separate form names such as form1, form2, etc but not necessary here.
<?php
if (isset($_POST['submitForm'])) {
echo '<pre>';
print_r($_POST);
echo '</pre>';
}
$n=0;
while($n<10) {
$n++;
?>
<form action="" name="form1" method="post">
<input type="text" value="" name="A" />
<input type="text" value="" name="B" />
<input type="text" value="" name="C" />
<input type="text" value="" name="D" />
<input type="Submit" value="Submit Form" name="submitForm" />
</form>
<?php
}
?>
Sample page with output when I click row 5..
I have made following form, but it doesn't work because it doesn't send post id in post request.
<?php
require('./wp-blog-header.php');
$post = get_post($_GET['p']);
?>
<form action="<?php echo get_option('siteurl'); ?>/wp-comments-post.php" method="post" >
<label>Name : </label><br/>
<input name="author" id="author" type="text"/><br/>
<label>Comment : </label><br/>
<textarea name="comment" id="comment"></textarea><br/><br/>
<input name="submit"type="submit" id="submit" value="Submit" />
<?php comment_id_fields(); ?>
<?php do_action('comment_form', $post->ID); ?>
</form>
Wordpress will drop any url parameters that it does not recognize. One way to add custom parameters in a url string is to use Add_Query_Args() function.
Take a look at Add_Query_Args Function Reference
This should solve your issue. Good luck.
With only a bit of tweaking was it possible to make it work.
<form action="<?php echo get_option('siteurl'); ?>/wp-comments-post.php" method="post" id="commentform">
<p><input type="text" name="author" id="author" value="" size="22" tabindex="1" />
<label for="author"><small>*</small></label></p>
<p><input type="text" name="email" id="email" value="" size="22" tabindex="2" />
<label for="email"><small>*</small></label></p>
<p><textarea name="comment" id="comment" cols="48" rows="10" tabindex="4" onFocus="clearText(this)" onBlur="clearText(this)" ></textarea></p>
<p><input name="submit" type="submit" id="submit" tabindex="5" value="Submit" />
<?php comment_id_fields(); ?>
<?php do_action('comment_form', $post->ID); ?>
</form>
Above code works (for me). Basically, you were missing an id on the form. WP apparently uses that id as part of the validation process.
So, to make it work, add id="commentform" to your form-tag, and it should work.