I'm trying to create a temperature converter that allows you to input into a text box, check a radio button and convert based on the radio button you check. However when the submit button is pressed it takes me to a page that says "the resource cannot be found" and it says there's an error finding the /converter which is the action of my form. I'm very new to this and couldn't find anything explaining how to fix it
<?php
#$input = $_GET['degrees'];
#$radio = $_GET['celorfar'];
if($radio == 'far'){
$answer = ($input * 1.8) + 32;}
else {
$answer = ($input - 32) * .56; //.56=5/9
}
echo <<<END_OF_FORM
<form method="GET" action="converter">
<input name="degrees" type="text" />
<br />
<input name="celorfar" type="radio" value="far" /> Convert to Fahrenheit<br />
<input name="celorfar" type="radio" value="cel" /> Convert to Celcius<br />
<input name="Convert" type="submit" value="Convert" /> <br />
<br />
Answer : <input name="Text1" type="text" value="$answer" /></form>
END_OF_FORM
?>
You missed file extension...
<form method="GET" action="converter.php">
Related
I have a form that allows members to click a (+) sign, and it will put another form field there. They can do this to basically without limit.
The problem I have is if they do it, let's say, 3 times and fill out all 3, when I get the data to save it, it's placing the first field in the database and not the others.
Here is my code:
Here is the JavaScript that makes the div show as they push the (+) sign.
<script type="text/javascript">
function add_feed()
{
var div1 = document.createElement('div');
div1.innerHTML = document.getElementById('newlinktpl').innerHTML;
document.getElementById('newlink').appendChild(div1);
}
</script>
<style>
.feed {padding: 5px 0}
</style>
This is part of the form that does the above...
<td width="50%" valign="top">
<div id="newlink">
<div class="feed">
<input type="text" size="45" value="" name="recname[]" placeholder="Place Company Name Here"><br /><br />
<input type="text" size="45" value="" name="reclink[]" placeholder="Place URL Here"><br /><br />
<p><b>What Type Of Page Is This?</b><br /><br />
<input type="radio" name="rectype[]" value="1"> Business Opp<br />
<input type="radio" name="rectype[]" value="2"> Traffic Site<br />
<input type="radio" name="rectype[]" value="3"> Tools Site<br /></p>
</div>
</div>
<hr>
<p id="addnew">
+ Click Here To Add Another Biz/Traffic/Tools Site
</p>
<div id="newlinktpl" style="display:none">
<hr>
<div class="feed">
<input type="text" size="45" value="" name="recname[]" placeholder="Place Company Name Here"><br /><br />
<input type="text" size="45" value="" name="reclink[]" placeholder="Place URL Here"><br /><br />
<p><b>What Type Of Page Is This?</b><br /><br />
<input type="radio" name="rectype[]" value="1"> Business Opp<br />
<input type="radio" name="rectype[]" value="2"> Traffic Site<br />
<input type="radio" name="rectype[]" value="3"> Tools Site<br /></p>
</div>
</div>
This is the part of the PHP code that would save it...
$i=0;
$linkname = $_POST["recname"][0];
while($linkname != ""){
$linkname = $_POST["recname"][$i];
$linkurl = $_POST["reclink"][$i];
$linktype = $_POST["rectype"][$i];
$linkname = $res->real_escape_string($linkname);
$linkurl = $res->real_escape_string($linkurl);
$linktype = $res->real_escape_string($linktype);
$result299 = mysqli_query($res, "INSERT INTO user_links (linkname,linkurl,linktype,sort) VALUES ('$linkname','$linkurl','$linktype','0')");
$i++;
}
It's all working except that is does not store all the data in the database (only the first one saves). That's the part I need help with please.
Please explain what I have done wrong and how to get it to store all the fields in the database, no matter how many the user creates and fills out.
I tested your code and it works fine so far if I use it like this:
$i=0;
$linkname = $_POST[recname][0];
while($linkname != ""){
$linkname = $_POST[recname][$i];
$linkurl = $_POST[reclink][$i];
$linktype = $_POST[rectype][$i];
echo "INSERT INTO user_links (linkname,linkurl,linktype,sort) VALUES ('$linkname','$linkurl','$linktype','0')<br>\n";
$i++;
}
There's no information about the $res object you're calling real_escape_string() on, so I'll just skip that for now. There are a couple of weaknesses in the code though:
You are referencing the post keys with barenames instead of strings.
PHP will gracefully assume you meant it as a string, but it will trigger a notice like Use of undefined constant recname - assumed 'recname'. Enclose them in quotes to make it clean.
Your use of the loop will result in an empty element inserted in the database every time.
You set the new linkname AFTER checking if $linkname is empty, but the variable contains the name of the last iteration. Instead, do something like this:
$i=0;
while($linkname = $_POST["recname"][$i]){
$linkurl = $_POST["reclink"][$i];
$linktype = $_POST["rectype"][$i];
echo "INSERT INTO user_links (linkname,linkurl,linktype,sort) VALUES ('$linkname','$linkurl','$linktype','0')<br>\n";
$i++;
}
Your code only allows for one radio button to be checked at a time
You cannot use rectype[] as a name for radio buttons, as an equal name forms a group of radio buttons out of all the elements. You need to name them like this:
<input type="radio" name="rectype[0]" value="1"> Business Opp<br />
<input type="radio" name="rectype[0]" value="2"> Traffic Site<br />
<input type="radio" name="rectype[0]" value="3"> Tools Site<br />
<input type="radio" name="rectype[1]" value="1"> Business Opp<br />
<input type="radio" name="rectype[1]" value="2"> Traffic Site<br />
<input type="radio" name="rectype[1]" value="3"> Tools Site<br />
and so on. You can do that programatically in your javascript code like this:
<script type="text/javascript">
var counter = 1;
function add_feed()
{
var div1 = document.createElement('div');
div1.innerHTML = document.getElementById('newlinktpl').innerHTML;
var inputs = div1.getElementsByTagName('input');
for (i=0; i<inputs.length; i++) {
if (inputs[i].type == "radio") {
inputs[i].name="rectype[" + counter + "]";
}
}
counter++;
document.getElementById('newlink').appendChild(div1);
}
</script>
That said, I don't see why it should only save one item, unless you have a key constraint hitting or something else we cannot assume from the piece of code you shared.
I'm trying to get a value from a form, then display it on posting of the form. I can get the value to appear in the second text field, once I have chosen an option using the Ajax Auto-Select, but how do I get that value shown stored into a variable for display on posting? This is what I have been trying -
if ($_POST['action'] == 'getentity') {
$value= $entity;
$content .= '<div>'.$value.' hello</div>';
}
<form method="post" action="?">
<input type="text" name="TownID_display" size="50" onkeyup="javascript:ajax_showOptions(this,\'getEntitiesByLetters\',event)">
<input type="text" name="TownID" id="TownID_display_hidden" value="'.$entity.'" />
<input type="hidden" name="action" value="getentity" />
<input type="submit" name="submit" value="Find"/>
Many thanks for any help.
Try
<input type="text" name="TownID" id="TownID_display_hidden" value="<?php $value = $entity; echo $entity; ?>" />
and its better to use like this
if ($_POST['action'] == 'getentity') {
$value= $_POST['TownID'];
$content .= '<div>'.$value.' hello</div>';
}
it should work.
test.php
<form action="test2.php" method="post">
Q1: <br />
<input type="radio" name="q1" value="true" />T<br />(Correct Answer)
<input type="radio" name="q1" value="false" />F<br />
Q2: <br />
<input type="radio" name="q2" value="true" />T<br />
<input type="radio" name="q2" value="false" />F<br />(Correct Answer)
<input type="submit" value="Score" />
</form>
test2.php
<?php
//process code from test.php
?>
I want to get the value of the radio buttons from each question and check whether is true or false. I try this in test2.php:
(1) if($_POST['name']) -> get error (undefined index: name)
(2) if($_POST['submit']) -> get error(undefined index: submit)
(3) if(isset($_POST['name'])-> no error, but nothing happened
How should I solve it?
You have named the radio button as q1 and q2 but trying to access it with a name name. Instead of if($_POST['name']) you have to do like
if($_POST['q1'])
and
if($_POST['q2'])
Also for checking whether the form is submitted or not, you can try this code.
if(isset($_POST))
In case if you are using if($_POST['submit']) for checking whether the form is submitted or not, submit will be the name of the submit button. So you have to set name property to submit button.
<input name="submit" type="submit" value="Score" />
Have you tried this?
$_POST['q1'];
$_POST['q2'];
You need to set the name:
<input name="submit" type="submit" value="Score" />
and to get the value like this:
$_POST['q1'];
$_POST['q2'];
the name attribute is what you send in POST/GET to the php script.
if(isset($_POST['submit']))//don't forget to check using isset()
{
/*other variables*/
$radio_value1 = $_POST['q1'];
$radio_value2 = $_POST['q2'];
}
if($_POST['name'])
to
if($_POST['q1']) {
}
if($_POST['q2']) {
}
if(isset($_POST['name'])
to
if(isset($_POST['q1'])
if(isset($_POST['q2'])
And,
<input type="submit" value="Score" /> to <input type="submit" name="submit" value="Score" />
This form should calculate numbers and save
Now there are two buttons One is call Calculator and two call Save
If I press Calculator
I get the form action is going to file name save.php And I do not want it that way
How can I set it up that button do something else
Example
Calculator = Calculator
Save = save.php
Is it possible to set it
Because it is one form
Thanks to anyone who can help
<?php
error_reporting (0);
$NUM = $_POST["NUM"];
$NUM2 = $_POST["NUM2"];
$NUM = "$NUM";
$NUM2 = "$NUM2";
$subtotal= $NUM+$NUM2;
?>
<form action="save.php" method="POST" name="Calculator">
<p>
<input name="NUM" type="text" value="<?php echo $_POST["NUM"]; ?>" />
</p>
<p>+</p>
<p>
<input name="NUM2" type="text" value="<?php echo $_POST["NUM2"]; ?>" />
</p>
<p>
<input name="subtotal" type="text" value="<?php echo "$subtotal";?>" />
</p>
<p>
<input name="submit" type="submit" value="Calculator" />
<p>
<input name="submit" type="submit" value="Save" />
</p>
</form>
You can have all the logic in a single PHP script (no need to direct to a different script depending on the button). If the logic is complicated, use include statements in order to separate the code.
Name the buttons differently:
<input name="calculator_submit" type="submit" value="Calculator" />
<input name="save_submit" type="submit" value="Save" />
Then in PHP:
if (isset($_GET['calculator_submit'])) {
// ...
} else if (isset($_GET['save_submit'])) {
// ...
} else {
// ...
}
If you really need different PHP script, then you'll have to go with Javascript (function will change the form action when a submit is clicked).
Since you are now using two submit buttons, both will submit the form and go to save.php.
Make your "calculator" button an input type=button instead of submit, and handle it via JavaScript.
Just FYI:
HTML5 allows to define a different form target URL by specifying the formaction attribut on a submit button – but browser support is lousy as of now.
Form and Buttons
<input name="submit" type="button" onclick="submitForm('Calculator')" value="Calculator" />
<input name="submit" type="button" onclick="submitForm('Save.php')" value="Save" />
Some jquery:
function submitForm(path) {
$('#Calculator').attr('action', path);
$('#Calculator').submit();
}
I am learning PHP now, so pardon my silly question which I am not able to resolve.
I have created a simple web form where in I display the values entered by a user.
function submitform()
{
document.forms["myForm"].submit();
} // in head of html
<form action ="res.php" id="myForm" method="post" >
Name: <input type="text" name="name" size="25" maxlength="50" /> <br> </br>
Password:<input type="password" name="password" size="25" maxlength="50" />
Description: <textarea name="editor1"> </textarea>
<input type="submit" value="Submit" onclick="submitForm()" />
</form>
and res.php contains:
foreach($_POST as $field => $value)
{
echo "$field = $value";
}
When I click on the submit button, I just get a blank page without any values from the form. Can anyone please let me know what am I missing?
There's no need for the javascript. This should do:
<form action ="res.php" id="myForm" method="post" >
Name: <input type="text" name="name" size="25" maxlength="50" /> <br> </br>
Password:<input type="password" name="password" size="25" maxlength="50" />
Description: <textarea name="editor1"> </textarea>
<input type="submit" value="Submit" />
</form>
Let's start with fixing errors:
JavaScript is case-sensitive. I see that your function name is submitform and the form's onclick calls submitForm.
The javascript is not really necessary from what you've shown us, I would try this on a single php page and see if it works:
Create a test.php file for test purpose:
<?php
if($_POST){
foreach($_POST as $key=>$value){
echo "$key: $value<br />";
}
}else{
<form action="test.php" method="post">
<input type="text" value="1" name="name1" />
<input type="text" value="2" name="name2" />
<input type="submit" value="submit" name="Submit" />
</form>
}
?>
If it does work, slowly work your way into your current form setup to see what is breaking it. If it doesn't work, there's something larger at play.
There are 2 things you should do now.
Remove the JavaScript function to submit the form. It's not required (or necessary). The default behavior of a submit button is to... well... submit. You don't need to help it with JavaScript.
Enable error display by using error_reporting(E_ALL).
After you do both things, you should be able to debug and assess the problem much more easily.
Put your php code inside php tags!
<?php
foreach($_POST as $field => $value)
{
echo $field ." = ." $value.'<br />';
}
?>
If you do
<?php
print_r($_POST);
?>
what do you get?
If this still doesn't work, does your server parse php?
Create the file test.php and access it directly http://localhost/test.php or whatever your URL is
<?php
echo 'Hello';
?>
if this doesn't work..it's a whole diferent problem
You can submit an HTML form using PHP with fsubmit library.
Example:
require_once 'fsubmit.php';
$html = "<form action ='res.php' method='post'><input type='text' name='name'/></form>";
$form = new Fsubmit();
$form->url = 'http://submit_url_here.com';
$form->html = $html;
$form->params = ['name'=>'kokainom'];
$response = $form->submit();
echo $response['content'];