echo "<form method=\"post\" action=\"settings.php\" onchange=\"this.form.submit()\">";
echo "Announce New Files: <input type=\"radio\" name=\"announcefiles\" value=\"On\" $checkon1> On";
echo "<input type=\"radio\" name=\"announcefiles\" value=\"Off\" $checkoff1> Off<br>";
echo "</form>";
I am trying to get this form to submit when one of the radio buttons is pressed but I'm not sure how to catch the submission.
for example, normally with a submit button you would use something along the lines of if(isset($_POST['submit'])) but I'm not sure how to do it if the form auto submits.
Add hidden input field like:
<input type="hidden" name="action" value="submit" />
Then in PHP check:
if(isset($_POST["action"]) && $_POST["action"] == "submit") { ... }
You should be checking the request method. If you've set things up cleanly, a POST request at that URL will mean a form submit. As you've noticed, you can have attempted submits where a value just isn't there.
if ($_SERVER['REQUEST_METHOD'] === 'POST')
See $_POST vs. $_SERVER['REQUEST_METHOD'] == 'POST' for more discussion.
Give your form a name and check for isset($_POST['form_name']) or check for the name of the radio isset($_POST['announcefiles'])
Also, you don't need all the quote escaping that you have, you can use single quotes as well as use a multiline string - see example below.
echo "
<form method='post' name='form_name' action='settings.php' onchange='this.form.submit()'>
Announce New Files: <input type='radio' name='announcefiles' value='On' $checkon1> On
<input type='radio' name='announcefiles' value='Off' $checkoff1> Off<br>
</form>";
<?php
// Check if form was submitted
if (isset($_POST['form_name']) {
// Form submitted
}
?>
<?php
// Check if radio was selected
if (isset($_POST['announcefiles']) {
// Form submitted
echo 'You chose' . $_POST['announcefiles'];
}
?>
Try this:
You may have an easier time if you separate the php and html a little more.
<form method="post" action="settings.php" onchange="this.form.submit()">
<fieldset>
<legend>Announce New Files:</legend>
<label for="on"><input type="radio" id="on" name="announcefiles" value="On" <?php echo $checkon1 ?> /> On</label>
<label for="off"><input type="radio" id="off" name="announcefiles" value="Off" <?php echo $checkoff1 ?> /> Off</label>
</fieldset>
</form>
Then in your php logic in settings.php ( or above the form if you are posting back to the same page ) you can check for the value of announcefiles:
<?php
if(isset($_POST['announcefiles'])){
// DO SOMETHING
}
?>
Let me know if this helps. Or if I'm missing the question.
Related
I created a form which asks for username and password for registration purposes and I sent the data to same page using action="" and checking for $_POST variables, but data is not being passed through this method. When I print $POST array by changing condition to true and reloading the page , the POST array is empty and also I can see the variables passed as POST in URL.Can somebody explain whats the problem?
Code:
<!DOCTYPE html>
<html>
<?php
if(isset($_POST['user']))
{
echo "Data coming";
die();
}
else {
?>
<form method="POST" enctype="multiform/form-data" action="">
<b> Username: </b><input type='text' name='user'> <br> <br><br>
<b> Password:</b> <input type='password' name='pass'><br> <br>
<input type='submit' value="Submit">
</form>
<?php
}
?>
</html>
You have to give the name to submit button also.
<input type='submit' name="Submit" value="Submit">
You are checking the index 'Submit' in $_POST and the submitted form is take the value with the name of input type. So you have to given the name to submit button also like above.
You are checking in your if isset for a post named:
$_POST['Submit']
Your submit button doesn't have
name=Submit
As a test to see what you are sending for debug purposes, try putting this in your code, it will help you self-debug:
print_r($_REQUEST);
"input type='submit' name="Submit" ...
Change to this.
$_POST['Submit']
looks for a resource with the name of Submit when you reload the page, as in your code there is no resource with the name of Submit, the if statement returns false and the control switches back to the else part.
Give a name to the submit button and use the same name in the post varibale option.
So I'm doing this website with Joomla and I don't have much experience with Joomla.
I just want to get the values of some checkboxes and print them into a text area.
I have the script already, I just need to implement it.
Here the raw Html code of the checkboxes:
<div class="checkboxShop">
<form action="" method="post">
<input type="checkbox" value="1" id="checkboxShopInput" name="shop" />
<label for="checkboxShopInput"></label>
</form>
</div>
And Here is the raw Html code of the submit button:
<form action="/Flex/index.php/shop" method="post">
<input type="submit" value="Buy">
</form>
All I want is to get all the values of the checkboxes and add them to a textarea in index.php/shop. I wrote this code to get the values but i don't know where to add this:
<?php
if(!empty($_POST['shop'])) {
foreach($_POST['shop'] as $check) {
echo $check; //echoes the value set in the HTML form for each checked checkbox.
//so, if I were to check 1, 3, and 5 it would echo value 1, value 3, value 5.
//in your case, it would echo whatever $row['Report ID'] is equivalent to.
echo "<input type='text' name='message' value='" . $check . "'>"
}
}
?>
I've read that i need this but I'm not sure how: <?php include "name_of_script.php";?>
Thanks for the help
I have a list of names and some buttons with product names. When one of the buttons is clicked the information of the list is sent to a PHP script, but I can't hit the submit button to send its value. How is it done?
I boiled my code down to the following:
The sending page:
<html>
<form action="buy.php" method="post">
<select name="name">
<option>John</option>
<option>Henry</option>
<select>
<input id='submit' type='submit' name='Tea' value='Tea'>
<input id='submit' type='submit' name='Coffee' value='Coffee'>
</form>
</html>
The receiving page: buy.php
<?php
$name = $_POST['name'];
$purchase = $_POST['submit'];
//here some SQL database magic happens
?>
Everything except sending the submit button value works flawlessly.
The button names are not submit, so the php $_POST['submit'] value is not set. As in isset($_POST['submit']) evaluates to false.
<html>
<form action="" method="post">
<input type="hidden" name="action" value="submit" />
<select name="name">
<option>John</option>
<option>Henry</option>
<select>
<!--
make sure all html elements that have an ID are unique and name the buttons submit
-->
<input id="tea-submit" type="submit" name="submit" value="Tea">
<input id="coffee-submit" type="submit" name="submit" value="Coffee">
</form>
</html>
<?php
if (isset($_POST['action'])) {
echo '<br />The ' . $_POST['submit'] . ' submit button was pressed<br />';
}
?>
Use this instead:
<input id='tea-submit' type='submit' name = 'submit' value = 'Tea'>
<input id='coffee-submit' type='submit' name = 'submit' value = 'Coffee'>
The initial post mentioned buttons. You can also replace the input tags with buttons.
<button type="submit" name="product" value="Tea">Tea</button>
<button type="submit" name="product" value="Coffee">Coffee</button>
The name and value attributes are required to submit the value when the form is submitted (the id attribute is not necessary in this case). The attribute type=submit specifies that clicking on this button causes the form to be submitted.
When the server is handling the submitted form, $_POST['product'] will contain the value "Tea" or "Coffee" depending on which button was clicked.
If you want you can also require the user to confirm before submitting the form (useful when you are implementing a delete button for example).
<button type="submit" name="product" value="Tea" onclick="return confirm('Are you sure you want tea?');">Tea</button>
<button type="submit" name="product" value="Coffee" onclick="return confirm('Are you sure you want coffee?');">Coffee</button>
To start, using the same ID twice is not a good idea. ID's should be unique, if you need to style elements you should use a class to apply CSS instead.
At last, you defined the name of your submit button as Tea and Coffee, but in your PHP you are using submit as index. your index should have been $_POST['Tea'] for example. that would require you to check for it being set as it only sends one , you can do that with isset().
Buy anyway , user4035 just beat me to it , his code will "fix" this for you.
Like the others said, you probably missunderstood the idea of a unique id. All I have to add is, that I do not like the idea of using "value" as the identifying property here, as it may change over time (i.e. if you want to provide multiple languages).
<input id='submit_tea' type='submit' name = 'submit_tea' value = 'Tea' />
<input id='submit_coffee' type='submit' name = 'submit_coffee' value = 'Coffee' />
and in your php script
if( array_key_exists( 'submit_tea', $_POST ) )
{
// handle tea
}
if( array_key_exists( 'submit_coffee', $_POST ) )
{
// handle coffee
}
Additionally, you can add something like if( 'POST' == $_SERVER[ 'REQUEST_METHOD' ] ) if you want to check if data was acctually posted.
You can maintain your html as it is but use this php code
<?php
$name = $_POST['name'];
$purchase1 = $_POST['Tea'];
$purchase2 =$_POST['Coffee'];
?>
You could use something like this to give your button a value:
<?php
if (isset($_POST['submit'])) {
$aSubmitVal = array_keys($_POST['submit'])[0];
echo 'The button value is: ' . $aSubmitVal;
}
?>
<form action="/" method="post">
<input id="someId" type="submit" name="submit[SomeValue]" value="Button name">
</form>
This will give you the string "SomeValue" as a result
https://i.imgur.com/28gr7Uy.gif
i have a php file which stores a variable, and i need to create a submit form which passes this variable.
So the code of the current file is:
if(isset($_POST['Submit'])) {
echo "<pre>";
$checked = implode(',', $_POST['checkbox']);
echo $checked;
}
I have to insert the submit form and on click it goes to another php file, i need to get there the $checked variable.. how can i store this variable?..
Thanks you!
The principle of POSTing is that you can send all the data to the next .php page.
<?php
if ($checked == 1) {
$checked = 'checked="checked"';
}
else {
$checked = '';
}
?>
<form action="POST" method="target_file.php">
<input type="hidden" name="variableA" value="Something I want target_file.php to know" />
<input type="hidden" name="variableB" value="Something else I want target_file.php to know" />
<input type="checkbox" name="gender" value="male" <?php echo $checked ?>" />
</form>
Your target_file.php:
echo "Here I am :), variableA: ".$_POST['variableA'];
echo "Here I am :), variableB: ".$_POST['variableB'];
echo "Here I am :), My gender is: ".$_POST['gender'];
Also, don't trust on checking your field values that are send through the $_POST. Check what every value is and if it meets certain criteria before sub-statements may be executed.
it has error because $_POST['sub1'] can't be accessed
is there any approach or solution to echo the value of $_POST['sub1']? or impossible? no way? even with another arrays?
i had question about my code nobody solved it! then I decide to tell it in simple way.
<html>
<form method="post">
<input type='submit' name='sub1' value='sub1'>
<?php
if(array_key_exists('sub1',$_POST))
{
echo"<input type='submit' name='sub2' value='sub2'>";
}
if(array_key_exists('sub2',$_POST))
{
echo $_POST['sub1'];
}
?>
</form>
</html>
I think I know what is wrong here.
When you submit the form the second time (for sub2) you are no longer posting the value of sub1 along with it, just sub2.
This should fix it:
<html>
<form method="post">
<input type='submit' name='sub1' value='sub1'>
<?php
if(array_key_exists('sub1',$_POST))
{
echo"<input type='hidden' name='sub1' value='" . htmlentities($_POST['sub1']) . "'>";
echo"<input type='submit' name='sub2' value='sub2'>";
}
if(array_key_exists('sub2',$_POST))
{
echo $_POST['sub1'];
}
?>
</form>
</html>
You're using submit buttons. Only the button that you actually click will have its name/value pair sent to the server. When you click the sub2 button, only sub2=sub2 is sent, so sub1 won't exist in the $_POST array.
followup:
$_POST is created for you by PHP based on what's sent from the browser. The way you've built your form makes it impossible for 'sub1' to exist when you click the 'sub2' button. In other words, you need to use the SAME name= for BOTH buttons, and change the value= as appropriate:
html:
<input type="submit" name="submit" value="sub1" />
<input type="submit" name="submit" value="sub2" />
php:
if (isset($_POST['submit'])) {
echo "You clicked the {$_POST['submit']} button";
}