I'm try to build a snack machine where you can choose your snack, get the price and then click the button to pay.
Like:
The price is 0,60 €
0.05(button) 0.10(button) ....
if you press 0.05-button the price will reduce to 0,55€
Why don't I get a "test" echo after I click the button?
Here is my code
<?php
if(isset($_GET['mars']))
{
$mars = "0,60";
echo "Bitte Zahlen Sie noch <input type=\"button\" value=\"$mars\"> Euro<br>";
echo "<input type=\"submit\" value=\"0,05\" name=\"fcent\">";
if(isset($_GET['fcent']))
{
echo "test";
}
}
?>
First, there appears to be no form-tag in your code. Without a form tag, it would be a miracle that pressing that button actually submits it via PHP. In other words, you need to wrap your form elements in a <form></form> Tag. Secondly, the nested if: if(isset($_GET['fcent'])) is unreachable because when you press the fcent button; the $_GET['mars'] is no more in scope and since your code explicitly demands to be run when $_GET['mars'] is SET, nothing would happen. The Snippet below takes this 2 Points into account and you can fine-tune it even further to meet your needs...NOTE: You have to be sure that your URL reads something similar to this: http://localhost/index.php?mars=some-value
<?php
$mars = "0,60";
$payForm = "<form name='whatever' method='get' action=''><br>";
$payForm .= "Bitte Zahlen Sie noch <input type=\"button\" value=\"$mars\"> Euro<br>";
$payForm .= "<input type=\"submit\" value=\"0,05\" name=\"fcent\">";
$payForm .="</form>";
if(isset($_GET['mars'])){
echo $payForm;
}
if(isset($_GET['fcent'])){
echo $payForm;
echo "test";
}
Related
I have trouble when creating code for adding an item to shopping cart. What I did is echo each individual item customers could buy, and I wanted to add an input type for how many of that specific item they want, then "add to cart" button and repeat (echo another item, another input and button etc.)
while ($zaznam = mysqli_fetch_assoc($vysledek)) {
$soubor=$zaznam["img_nazev"];
echo "<div id='zbozi_info'>";
echo "<img src='img/$soubor'>";
echo "<p class='popis'>".$zaznam["dlouhy_popis"]."</p>";
echo "<p class='cena'> Cena bez DPH: ";
echo "<span style=color:#FF0000>".$zaznam["cena"]." Kč </span> </p>";
echo "<p class='cena'> Cena s DPH: ";
echo "<span style=color:#FF0000>".$zaznam["CenaDPH"]." Kč </span> </p>";
echo "</div>";
echo "<form method='post'>
<label for='pocetk'>Počet kusů</label>
<br>
<input type='text' name='pocetk' id='pocetk' value='1'>
<br>
<input type='submit' name='pridkos' id='pridkos' value='Přidat do košíku'>
</form>";
}
This works and all, the trouble starts when I add this if to the mix, to enable that adding function
if (isset($_POST["pridkos"]) && isset($_SESSION['user'])) {
$ksk="INSERT INTO kosik
(ID_uzivatele,ID_zbozi, pocet_kusu)
VALUES ('".$_SESSION['user']."',
'".$zaznam["ID_zbozi"]."',
'".$_POST["pocetk"]."');";
$vysledek = mysqli_query($con,$ksk)
or die("Zboží nebylo přidáno do košíku");
This just dies on the mysqli_query. I am not exactly sure if to put it inside the while or outside, my suspicion is inside, but it isn't working either way
Session[user] has the user ID (ID_uzivatele), not sure what else can I clarify.
Any help is appreciated.
I think this query syntax is wrong.
Check with below one:-
$ksk="INSERT INTO kosik(ID_uzivatele,ID_zbozi, pocet_kusu)
VALUES ('$_SESSION[user]',
'$zaznam[ID_zbozi]',
'.$_POST["pocetk"]')";
and thanks for the great help your community provides.
Here is a question regarding Paypal. The code below creates a buy-now non-hosted button. I would like to be able to do run my geoTestArray.php code, currently simply included, as the user presses the Buy Now button, and not when the button is displayed on the page.
Is this possible at all?
Thanks in advance,
Joe
<?php
include ('geoTestArray.php');
echo "<form action=\"https://www.paypal.com/cgi-bin/webscr\" method=\"post\">
<input type=\"hidden\" name=\"business\" value=\"shop#mysite.com\">
<input type=\"hidden\" name=\"cmd\" value=\"$cmd\">
<input type=\"hidden\" name=\"amount\" value=\"$price\">
<input type=\"hidden\" name=\"shipping\" value=\"$shipping\">
<input type=\"hidden\" name=\"currency_code\" value=\"$currency\">
<input type=\"hidden\" name=\"shipto\" value=\"$shipto\">
<input type=\"hidden\" name=\"item_name\" value=\"$title\">";
if ($cmd == 'cart') {
echo "<input type=\"hidden\" name=\"add\" value=\"1\">";
}
include('./quantities2.php');
echo "<input type=\"hidden\" name=\"lc\" value=\"US\">
<input class=\"submit\" type=\"image\" src=\"$button\" border=\"0\" name=\"submit\" alt=\"$altaction\">
<img alt=\"\" border=\"0\" src=\"https://www.paypal.com/en_US/i/scr/pixel.gif\" width=\"1\" height=\"1\">
</form>";
?>
You can display only a form with one input, for example item_id.
After submitting this form you search for item details, execute your geoTestArray.php in backend and after that you can redirect your users to paypal using header("location: $query");
here's how I would do:
<?php
if( isset($_POST['do_action']) && $_POST['do_action'] != "" ) {
//check for non-empty item_id field
//run your geoTestArray.php script
//get item details from database and build your paypal query
$query = "https://www.paypal.com/cgi-bin/webscr&cmd=_xclick"; // add rest of paypal request fields separated by &
header( "Location: $query" );
}
?>
<form action='' method='post'>
<input type='hidden' name='item_id' value'your_item_id'>
<input type='submit' name='do_action' value='Buy'>
<form>
Hope I helped you! :d
Thanks everyone for comments and solution options. I have eventually chosen to modify an onSubmit javascript I had on my page, by adding:
function InterceptForm(formObj) {
var newshipto = httpGet('geoTestArray.php');
formObj.shipto.value = newshipto;
}
The above code makes use of the httpGet() function posted in this other stackoverflow post:
HTTP GET request in JavaScript?
To complete the puzzle:
The code below is in the html in my page
<script language="javascript">
// During onLoad, loop through all forms and for each form object do something
function InterceptForm(formobj) {
formobj.onsubmit = function ff() {
interceptform(formobj);
};
}
</script>
The code below gathers all forms on the page and runs the above javascript on each (NOTE: you may want to name your forms so that you only run the script on the relevant forms and not on all forms):
<script language="javascript">
function GetForms() {
var formsCollection = document.getElementsByTagName("form");
for(var i=0;i<formsCollection.length;i++)
{
if (formsCollection[i].name != "seecart")
{
InterceptForm(formsCollection[i]);
}
}
return true;
}
</script>
The code below is the onLoad function:
<script language="javascript">
function onLoadFunction () {
GetForms();
return true;
}
</script>
Cheers,
Joe
For the last 4 hours I've been struggling to get something to work. I checked SO and other sources but couldn't find anything related to the subject. Here is the code:
<?php
$email=$_SESSION['email'];
$query1="SELECT * FROM oferte WHERE email='$email'";
$rez2=mysql_query($query1) or die (mysql_error());
if (mysql_num_rows($rez2)>0)
{
while ($oferta = mysql_fetch_assoc($rez2))
{
$id=$oferta['id_oferta'];
echo "<input type='radio' name='selectie' value='$id' id='$id'> <a href='oferta.php?id={$oferta['id_oferta']}'>{$oferta['denumire_locatie']}</a>";
echo "</br>";
}
echo "</br>";
//echo "<input type=\"button\" id=\"cauta\" value=\"Vizualizeaza\" onclick=\"window.location.href='oferta.php?id={$oferta['id_oferta']}'\" />";
//echo " <input type=\"button\" id=\"cauta\"value=\"Modifica\" onclick=\"window.location.href='modifica.php?id={$oferta['id_oferta']}'\" />";
echo " <input type=\"button\" id=\"sterge\" value=\"Sterge\" onclick=\"window.location.href='delete.php?id=$id'\" />";
echo "</form>";
echo "</div>";
}
else
{
}
?>
The while drags all of the user's entries from the database and creates a radio button for each one of them with the value and id (because I don't really know which one is needed) equal to the entry's id from the db. I echoed that out and the id is displayed as it should so no problems there.
The delete script works ok as well so I won't attach it unless you tell me to. All good, no errors, until I try to delete an entry. Whatever I choose from the list of entries, it will always delete the last one. Note that I have two other inputs echoed out, those will be the "view" and "modify" buttons for the entry.
I really hope this is not JavaScript related because I have no clue of JS. I think this will be of major help to others having this problem. Please let me know if I need to edit my question before downrating. Thanks!
After edit:
This is the delete script, which as I said earlier works fine.
<?php
if (isset($_GET['id']))
{
$id = $_GET['id'];
echo $id;
require_once('mysql_connect.php');
$query = "DELETE FROM oferte Where id_oferta = '$id'";
mysql_query($query) or die(mysql_error());
//header('Location: oferte.php');
}
else
{
//header('Location: oferte.php');
}
?>
The session is started as well, like this:
<?php
session_start();
?>
The reason the last $id is deleted is because this line is outside/after the while loop:
echo " <input type=\"button\" id=\"sterge\" value=\"Sterge\" onclick=\"window.location.href='delete.php?id=$id'\" />";
You want to move this line inside the loop so that you have a button that executes delete for each radio button.
Update:
To have links to delete and
echo "<input type='radio' name='selectie' value='$id' id='$id'> ";
echo "<a href='oferta.php?id={$oferta['id_oferta']}'>{$oferta['denumire_locatie']}</a> ";
echo "<a href='delete.php?id=$id'>delete</a>";
Also I do not think the radio button is needed here at all since you are not really doing anything with it. You could simply echo out the value of your choice and have these links as follows:
echo $oferta['denumire_locatie'] . ' '; // replace $oferta['denumire_locatie'] with something of your choice
echo "<a href='oferta.php?id={$oferta['id_oferta']}'>{$oferta['denumire_locatie']}</a> ";
echo "<a href='delete.php?id=$id'>delete</a>";
echo "<br />";
The problem, in this case, is JavaScript related, yes. What I recommend you to do is to simply add a Remove link for each item.
echo "<a href='oferta.php?id={$oferta['id_oferta']}'>{$oferta['denumire_locatie']}</a>";
echo " - <a href='delete.php?id={$oferta['id_oferta']}'>Remove</a>";
echo "</br>";
Your $id is outside your while() loop.
The last one is getting deleted because the $id has the last one's value when the loops is exited.
Include all your code :
echo "</br>";
//echo "<input type=\"button\" id=\"cauta\" value=\"Vizualizeaza\" onclick=\"window.location.href='oferta.php?id={$oferta['id_oferta']}'\" />";
//echo " <input type=\"button\" id=\"cauta\"value=\"Modifica\" onclick=\"window.location.href='modifica.php?id={$oferta['id_oferta']}'\" />";
echo " <input type=\"button\" id=\"sterge\" value=\"Sterge\" onclick=\"window.location.href='delete.php?id=$id'\" />";
Inside your while loop.
When the rendered html reaches the browser, it will be something like this:
<input type='radio' name='selectie' value='1' id='1'> <a href='oferta.php?id=1'>TEXT</a>
<input type='radio' name='selectie' value='2' id='2'> <a href='oferta.php?id=2'>TEXT</a>
<input type='radio' name='selectie' value='3' id='3'> <a href='oferta.php?id=3'>TEXT</a>
<input type='radio' name='selectie' value='4' id='4'> <a href='oferta.php?id=4'>TEXT</a>
<br/>
<input type="button" id="sterge" value="Sterge" onclick="window.location.href='delete.php?id=5'" />
With this markup you won't be able to accomplish what you want without using javascript to update the onclick attribute whenever you select a radio button.
On the other hand, instead of using the client-side onclick event you can use the button's default behaviour, which is to submit the form.
You'll just have to set the action attribute:
<form method="post" action="http://myurl.php">
and write the myurl.php page which will just read the posted variable $_POST['selectie'] and call the delete method with the posted id.
I have a PHP code which generates a dynamic list inside a form like the following, note that the list is built dynamically from database:
echo '<form name="List" action="checkList.php" method="post">';
while($rows=mysqli_fetch_array($sql))
{
echo "<input type='password' name='code' id='code'>";
echo "<input type='hidden' name='SessionID' id='SessionID' value='$rows[0]' />";
echo "<input type='submit' value='Take Survey'>";
}
What I need is to POST the data corresponding to the user choice when he clicks on the button for that row to another page.
If we use hyperlinks with query strings there will be no problem as I'll receive the data from the other page using a GET request and the hyperlinks would be static when showed to the user.
Also I need to obtain the user input from a textbox which is only possible with POST request.
Simply from the other page (checkList.php) I need these data for further processing:
$SessionID=$_POST['SessionID'];
$Code=$_POST['code'];
As I have a while loop that generates the fields, I always receive the last entry form the database and not the one corresponding to the line (row) that the user chosed from the LIST.
I'm going to recommend that you clean up the names of variables so that your code can
at least tell us what it's supposed to do. It should be rare that someone looks at your code
and has a lot of trouble trying to see what you're trying to accomplish :P, ESPECIALLY when you need help with something ;]. I'm going to try some things and hope that it makes doing what you want easier to comprehend and perhaps get you your answer.
It's good to try your best to not echo large amounts of HTML unnecessarily within a script , so firstly I'm going to remove the
echos from where they are not necessary.
Secondly, I'm going to use a mysql function that returns an easier to process result.
$user = mysqli_fetch_assoc($sql)
Third, I don't know if form having a name actually does anything for the backend or frontend of php, so I'm
just going to remove some of the extra crust that you have floating around that is either invalid HTML
or just doesn't add any value to what you're trying to do as you've presented it to us.
And yes, we "note" that you're building something from the database because the code looks like it does =P.
I'm also sooo sad seeing no recommendations from the other answers in regard to coding style or anything in regard to echoing html like this :(.
<?php while($user = mysqli_fetch_assoc($sql)): ?>
<form action="checkList.php" method="post">
<input type='password' name='code' value='<?php echo $user['code'] ?>' />
<input type='hidden' name='SessionID' value='<?php echo $user['id'] //Whatever you named the field that goes here ?>' />
<input type='submit' value='Take Survey' />
</form>
<?php endwhile; ?>
i not sure this is correct
echo '<form name="List" method="post">';
while($rows=mysqli_fetch_array($result))
{
echo "<input type='password' name='code' id='code'>";
echo "<input type='button' value='Take Survey' onclick=show($rows[0])>";
echo "<br>";
}
and javascript
<script>
function show(id)
{
alert(id);
window.location="checkList.php?id="+id;
}
</script>
On checkList.php
$id=$_GET['id'];
echo $id;
You can just check in checkList.php whether $_POST['code'] exists and if exists retrieve $_POST['SessionID'] which will be generated from database. But one thing, if You have all hidden fields in one form, they all will be sent, so You need to think how to correct that - maybe seperate forms for each hidden field, submit button and maybe other POST fields.
And afterwards, You will be able to get data in the way You need - $SessionID=$_POST['SessionID'];
I suppose it is the easiest way to solve that.
You can try something like this:
while($rows=mysqli_fetch_array($sql))
{
$index = 1;
echo "<input type='password' name='code' id='code'>";
//Attach $index with SessionID
echo "<input type='hidden' name='SessionID_$index' id='SessionID' value='$rows[0]' />";
echo "<input type='submit' value='Take Survey'>";
}
On checkList.php
<?php
$num = explode('_', $_POST['SessionID']);
$num = $num[1];
//in $num you will get the number of row where you can perform action
?>
$form = 1;
while($rows=mysqli_fetch_array($sql))
{
echo '<form name="List_$form" action="checkList.php" method="post">';
echo "<input type='password' name='code' id='code_$form'>";
echo "<input type='hidden' name='SessionID' id='SessionID_$form' value='$rows[0]' />";
echo "<input type='submit' value='Take Survey'>";
echo '</form>';
$form++;
}
I want to refresh the same page and display the entered value in text box on the same page after clicking the link or button.
I have the following code:
<?php
echo '<h3>Fee Payment</h3>';
echo "<form id='myFormId' name='myFormName'>";
echo " <input type='text' name='myTextField'>";
echo "<a href='{$_SERVER['PHP_SELF']}?student_no=myTextField'> Search Student</a>";
if (!(isset($_GET[student_no])))
{
echo "No Student is available.";
}
else
{
$student_no = $_GET['student_no'];
echo "Student NO:".$studen_no;
}
?>
Please guide me how to achieve the goal and whats the error my code.
<?php
echo '<h3>Fee Payment</h3>';
if(isset($_POST['myTextField']))
$value=$_POST['myTextField'];
else
$value='';
echo "<form id='myFormId' name='myFormName' method='post'>";
echo " <input type='text' name='myTextField' value='$value'>";
echo "<a href='{$_SERVER['PHP_SELF']}?student_no=myTextField'> Search Student</a>";
You also need a submit button in the form, and a form close tag.
What you want is AJAX. I am just providing you an example of how you can proceed by using jQuery, but you can use any other library / framework or any other way also. You can check this article also for more usability details.
In the main page:-
<?php
echo '<h3>Fee Payment</h3>';
echo "<form id='myFormId' name='myFormName'>";
echo "<input type='text' name='myTextField' id='myTextField' />";
echo 'Search Student';
echo "</form>";
?>
<script type="text/javascript">
$('#inline_submit_a').click(function(evt){
$.ajax({
type: "GET",
url: "handler.php",
data: {text:$('#myTextField').val()}
});
evt.preventDefault();
return false;
});
</script>
In the "handler.php" page:-
<?php
if (!(isset($_GET[student_no])))
{
echo "No Student is available.";
}
else
{
$student_no = $_GET['student_no'];
echo "Student NO:".$student_no;
}
?>
You can write all the logic related to the database in the "handler.php" page.
Hope it helps.
You're using a link to submit the form, but you should be using a submit button.
<input type="submit" value="Search student" />
You didn't close the <form> tag.
You're using $_GET[student_no] which will make PHP look for a definition of student_no. Since it's a string, express it as one. $_GET['student_no'] is better.
Why can't you submit the form? Do you want to have a link for searching instead of a button?
You could configure the link to submit the form via javascript and change the form action to GET like this:
<?php
echo '<h3>Fee Payment</h3>';
echo "<form id='myFormId' name='myFormName'
action='{$_SERVER['PHP_SELF']}' method='GET'>";
echo " <input type='text' name='student_no'>";
echo "<a href='javscript:document.forms.myFormName.submit()'>
Search Student</a>";
...