When I check the checkbox there is no errors, but when I didn't check the checkbox it gives me errors.
<?php
if(isset($_REQUEST['btn']))
{
$remmber = $_REQUEST['active'];
if($remmber == "on")
{
echo "Checked";
}
else {$remmber = "";}
}
?>
<html>
<form name= "frm" action = "test.php" method = "post" >
<p>Username
<input type = "text" name = "name" value = "" />
</p>
<p>Password
<input type = "text" name = "password" value = "" />
</p>
<p>
<td colspan = "2"><input type = "checkbox" name = "active" value = "active" />Keep Me Loged In
</p>
<p><input type = "submit" name = "btn" value = "Login" />
</p>
</form>
</html>
For checkboxes if they are not checked then they are not posted. So check if it is present in the posted data then set on, else set blank to the variable. Try with -
if(isset($_REQUEST['btn']))
{
$remmber = !empty($_REQUEST['active']) ? 'on' : '';
if($remmber == "on")
{
echo "Checked";
}
else
$remmber = "";
}
if($remmber == "on") yet you're using and specifying a value value = "active"
"on" is the default value for a checkbox if a value is not specified.
Therefore, you need to adjust it to read as
if($remmber == "active")
Edit: (an explanation)
The reason why you're getting an undefined index, is that once you hit the submit submit and do not tick the checkbox, it will produce that notice.
Modify your code to read as and check if the checkbox value is set.
if(isset($_REQUEST['btn']))
{
if(isset($_REQUEST['active'])){
$remmber = $_REQUEST['active'];
if($remmber == "active")
{
echo "Checked";
}
else {$remmber = "";}
} // closing brace for if(isset($_REQUEST['active']))
}
Additional edit, to show the user that the checkbox isn't set:
if(isset($_REQUEST['btn']))
{
if(isset($_REQUEST['active'])){
$remmber = $_REQUEST['active'];
if($remmber == "active")
{
echo "Checked";
}
else {$remmber = "";}
} // closing brace for if(isset($_REQUEST['active']))
else{ echo "The checkbox was not ticked.";
}
}
Footnotes:
You should also use a conditional !empty() for your inputs.
You're only checking if the submit and checkbox are set.
After noticing a comment you left in another answer:
#Barmar i am just using this for login page , where i am using a checkbox to remember username and password. – Akshat Dhiman
Please read the following Q&A on Stack:
Remember me Cookie best practice?
It also contains valuable information regarding passwords.
I hope you are using modern-day hashing methods also, such as password_hash() as well as prepared statements.
you can use simply isset
if(isset($_REQUEST['btn']))
{
$remmber = isset($_REQUEST['active']) ? $_REQUEST['active'] : '';
if($remmber == "on")
{
echo "Checked";
}
else {$remmber = "";}
}
Related
I have inserted a checkbox in my form.
My code:
<input type="checkbox" id="checkbox" name="checkbox" value="1"/>
if($checkbox = ($_POST['checkbox']) == '1')
{
$checkbox = "si";
}
else
{
$checkbox = "no";
}
I would like that if the checkbox is checked i receive "yes" otherwise "no".
Thanks.
You've written wrong if condition here, You cannot use assignment in conditions.
Also there is no need to assign value to any variable in checking condition, You can directly use $_POST['checkbox']. Like this,
if($_POST['checkbox'] == '1') {
$checkbox = "si";
} else {
$checkbox = "no";
}
Update:
A better option is to use isset() which determine if a variable is set and is not NULL. Like this,
if(isset($_POST['checkbox'])) {
$checkbox = "si";
} else {
$checkbox = "no";
}
Program will go in if condition only when user has checked the checkbox. In above case value attribute for <input> is not required. So your HTML will look something like this,
<input type="checkbox" id="checkbox" name="checkbox"/>
$request['checkbox_name'] = $request['checkbox_name'] == null ? 'N' : 'Y'; //for default value
Need som serious help!
<?php
$php1 = !isset($_GET['php1']);
$php2 = !isset($_GET['php2']);
$php3 = !isset($_GET['php3']);
$php4 = !isset($_GET['php4']);
$php5 = !isset($_GET['php5']);
if ($php1 == 'one' && $php2 == 'two' && $php5 == 'five') {
echo "<h2>R</h2>WRONG";
} else {
echo "<h2>R</h2>CORRECT";
}
?>
HTML
<form action="One.php" method="get">
<input type="checkbox" name="php1" value="one"> q1<br>
<input type="checkbox" name="php2" value="two"> q1<br>
<input type="checkbox" name="php3" value="three"> q1<br>
<input type="checkbox" name="php4" value="four"> q1<br>
<input type="checkbox" name="php5" value="five"> q1<br>
<br>
<input type="submit" value="Send!">
</form>
If all checkboxes are empty, there will be a message saying that. How do I do it?
If I dont check any boxes, it tells me the answers are correct. That's wrong.
<?php
If(isset($_GET['php1'])){
$php1 = 1;
}else{
$php1 = 0;
}
If(isset($_GET['php2'])){
$php2 = 1;
}else{
$php2 = 0;
}
If(isset($_GET['php3'])){
$php3 = 1;
}else{
$php3 = 0;
}
If(isset($_GET['php4'])){
$php4 = 1;
}else{
$php4 = 0;
}
If(isset($_GET['php5'])){
$php5 = 1;
}else{
$php5 = 0;
}
if ($php1 && $php2 && $php5) {
echo "<h2>Resultat</h2>Du svarade rätt på frågan";
} else {
echo "<h2>Resultat</h2>Du svarade fel på frågan";
}
Try this.
Each php variable will have the value of 1 or 0 (true/false). Thus the if only needs to check it if it is or not.
You have to remember that UN-CHECKED Checkboxes are not even sent to the PHP script $_POST/$_GET by the browser.
So there existance is normally all you need to know.
First you must check that they were passed to the script and then check its value, otherwise you will receive undefined index errors for each checkbox that was not checked by the user
if (isset($_GET['php1']) && $_GET['php1'] != '' ) {
Although as the checkbox called php1 can only be set to the value you give it, its value can be assumed and all you need to do is
if (isset($_GET['php1']) ) {
// php1 was checked
Also isset() will test more than one variable exists using an AND. So you could write your code as
if ( isset($_GET['php1'], $_GET['php2'], $_GET['php5']) ) {
echo "<h2>Resultat</h2>Du svarade rätt på frågan";
} else {
echo "<h2>Resultat</h2>Du svarade fel på frågan";
}
$php1 = !isset($_GET['php1']);
This means $php1 is either true or false. If you want the value of php1 get parameter then remove isset cover, just
$_GET['php1'];
isset is used to check whether the variable exist or not after that use $_GET['php1'];
In my form I am trying to get the radio checked value to be passed on to the next page (which is an FPDF page)
I have 4 options: Annual Leave, Sick Leave, Business Leave, & also others with a textfield.
However I have tried a lot of 'if' as well as 'switch cases'
I am getting either only the element with value '1'
or else 'Undefined index: rad in D:\xampp\htdocs\Application\generate_report.php on line 13'
some where I am wrong, can anyone help me please. My code below.
html form:
<form id="formmain" method="post" action="generate_report.php" onsubmit="return_validate()">
<script type="text/javascript">
function selectRadio(n){
document.forms["form4"]["r1"][n].checked=true
}
</script>
<table width="689">
<tr>
<td width="500d">
<input type="radio" name="rad" value="0" />
<label>Business Trip</label>
<input type="radio" name="rad" value="1"/><label>Annual Leave</label>
<input type="radio" name="rad" value="2"/><label>Sick Leave</label>
<input type="radio" name="rad" value="3"/><label>Others</label> <input type="text" name="others" size="25" onclick="selectRadio(3)" />
</td>
</tr>
</table>
//....
//below submit button is end of the html page:
<input type="submit" name="submit" value="send" />
</form>
Generate PDF form:
$radio = $_POST['rad']; // I am storing variable
if($radio = 0) {
$type = 'Business Leave';
}elseif ($radio = 1) {
$type = 'Annual Leave';
}elseif ($radio = 2) {
$type = 'Sick Leave';
} else { $type = $_POST['others']; }
//echo
$pdf->Cell(98,10, 'Reason | ' .$type , 1, 0, 'C', $fill);
if($radio = 0)
and
elseif ($radio = 1)
and all the other elseifs have to be == 1, with two '='!
A further explanation on the OP. If you do not use == then you are setting the value, not checking it. Furthermore, there are levels of checking. Using the double equals (==) is effectively stating "is equal to" whereas using triple equals (===) is like stating "is absolutely equal to". Generally the == operator will do everything you need but sometimes when working with data types or specific values you might need ===. This is mostly FYI as the OP has an actionable solution.
You should always check if inputs are checked or any value inserted. If there's no value, then it throws an undefined index error. Also, you should replace =s to == in your if clauses. So:
PHP:
$radio = $_POST['rad']; // I am storing variable
if (isset($radio)) { // checks if radio is set
if($radio == 0) {
$type = 'Business Leave';
}elseif ($radio == 1) {
$type = 'Annual Leave';
}elseif ($radio == 2) {
$type = 'Sick Leave';
} else {
if (isset($_POST['others'])) { // cheks if input text is set
$type = $_POST['others'];
}
else {
echo 'Error';
}
}
//echo
$pdf->Cell(98,10, 'Reason | ' .$type , 1, 0, 'C', $fill);
}
else {
echo 'Error';
}
Now it should work.
I need a simple php function to mimic an Ajax form submission - basically in the form there is a radio button "ajax" and it is either set to yes or no. I just need to mimic successful/failing ajax calls...
HTML
<label for="">Ajax Success*<input type="radio" name="ajax" id="yes" value="yes" checked>Yes<input type="radio" name="ajax" id="no" value="no">No</label>
PHP
<?php
$ajax = $_POST["ajax"];
if(isset($_POST['ajax'] == "yes")) {
echo "success";
} else {
echo "failure";
}
?>
if I remove the isset, I get an 'undefined index' error, if I put it in I get a syntax error but it looks correct to me...
I just need to send back an echo depending on what option is selected for the input 'ajax'
thx
isset($_POST['ajax'] == "yes") doesn't make sense. You want to check if it's set, and then check if its value is equal to "yes":
if(isset($_POST['ajax']) && $_POST['ajax'] == "yes") {
echo "success";
} else {
echo "failure";
}
As your code does, I'll say you use your defined variable like in this example:
<?php
$ajax = $_POST["ajax"];
if($ajax == "yes") {
echo "success";
} else {
echo "failure";
}
?>
Because if the variable has the value "yes" it'll be ok and undefined or other value will end in "failure".
I have a function that sets and unset's a $_SESSION variable depending on what was submitted from the form in my main page.
function if statement:
$_SESSION['search'] = true;
if($_POST['searchtv']){
$_SESSION['searchtv'] = true;
} else {
unset($_SESSION['searchtv']);
}
if($_POST['searchmovie']){
$_SESSION['searchmovie'] = true;
} else {
unset($_SESSION['searchmovie']);
}
The searchtv and searchmovie $_POST variables are set through the below checkboxes:
<input type="checkbox" name="searchmovie" value="movie" <? echo isset($_SESSION['searchmovie']) ? 'checked' : ''; ?>"/>
However the checked value always seems to be false and displays '' so no "checked" is set to display the tick in the box.
I do know that the $_SESSION variable is being set correctly however because in the same file i have another IF statement (below) that works 100%.
if(isset($_SESSION['searchtv'])){
$database->searchTV($_GET['show'], $session->uid);
}
if(isset($_SESSION['searchmovie'])){
$database->searchMovies($_GET['show'], $session->uid);
}
if(!isset($_SESSION['searchtv']) && !isset($_SESSION['searchmovie'])){
$database->searchTV($_GET['show'], $session->uid);
$database->searchMovies($_GET['show'], $session->uid);
}
If i only tick the searchtv checkbox it only runs the searchTV function and so on.. so i know it is being set and works.. just cannot get the feedback to the checkbox to say yes it was ticked when the search button was selected.
#medoix: Try changing this
<input type="checkbox" name="searchmovie" value="movie" <? echo isset($_SESSION['searchmovie']) ? 'checked' : ''; ?>"/>
to this
<input type="checkbox" name="searchmovie" value="movie" <?php if (isset($_SESSION['searchmovie'])) { echo 'checked="checked" '; } ?>/>
You realize that the portion of the code that checks $_SESSION['searchtv'] && $_SESSION['searchmovie'] near the bottom are both !isset. It'll only run if both aren't checked, rather than if they are.