i have a form in three steps, from step 1 to step two there is a checkbox, i would like to keep the checkbox value in a session and also to show if it's checked or unchecked even if the user goes to step 3 of the form then comes back to step two,
right now on form submit i have all the $_POST vars in $_SESSION vars but i cant make it work with checkboxes, this is what i got right now:
<input type="checkbox" value="<?php echo isset($_POST['afficher_ddn'])? "1":"0"; ?>"
style="margin-left: 20px;" name="afficher_ddn" id="afficher_ddn"
<?php echo $_SESSION['afficher_ddn']=="1" ? "checked" : ""; ?> />
but this doesnt work.
<input type="checkbox" value="1" style="margin-left: 20px;" name="afficher_ddn" id="afficher_ddn"
<?php echo isset($_SESSION['afficher_ddn']) && $_SESSION['afficher_ddn'] == "1" ? 'checked="checked"' : ''; ?> />
For one, you don't change the value of the checkbox. If it is not checked it won't pass to the post vars. See here: Does <input type="checkbox" /> only post data if it's checked?
Now, the current standard for checked is checked="checked" I have no idea where or why this standard became one at the firm I work for, but it's what we do so I relay that here.
If the code provided here does not work for you, I would var_dump($_SESSION) to make sure it's set as well as check the actual HTML in something like firebug (firefox) or developer tools in chrome. Sometimes goofiness happens and checked="checked" is actually set in the html but doesn't display in browser. In those times I usually yank out some hair and clear caches. then it usually clears up.
Your code is fine.Possibly there is some problem in passing the the value in session as i am checking the code manually it is working perfectly:
use print_r($_SESSION) for testing and post the output;If possible provide all the forms:
Try Below Code:
<input type="checkbox" value="1"
style="margin-left: 20px;" name="afficher_ddn" id="ddn"
<?php echo $_SESSION['afficher_ddn']=="1" ? "checked" : ""; ?> />
try this example separetly
Make a page test.php as:
<?php
session_start();
?>
<form methos='post' method='post' action='test1.php'>
<input type="checkbox" value="1"
style="margin-left: 20px;" name="afficher_ddn" id="ddn"
<?php echo $_SESSION['afficher_ddn']=="1" ? "checked" : ""; ?> /><br>
Name:<input type='text' name='f1' value='<?php echo $_SESSION['f1'] ?>' ><br>
Select:<select name='s1'>
<option value="" <?php echo $_SESSION['s1']=="" ? "selected" : ""; ?>></option>
<option value="1" <?php echo $_SESSION['s1']=="1" ? "selected" : ""; ?>>1</option>
<option value="2" <?php echo $_SESSION['s1']=="1" ? "selected" : ""; ?>>2</option>
</select><br>
Test:
<textarea name='tex1'><?php echo $_SESSION['tex1']; ?></textarea>
<input type='submit' value='submit'>
</form>
Make a second page test1.php as:
<?php
session_start();
print_r($_POST);
echo $_SESSION['afficher_ddn']=$_POST['afficher_ddn'];
echo $_SESSION['f1']=$_POST['f1'];
echo $_SESSION['s1']=$_POST['s1'];
echo $_SESSION['tex1']=$_POST['tex1'];
?>
<form methos='post' action='third.php'>
<a href='test.php'>Step1</a>
</form>
and Check
Related
Good day!
I'm doing a site. There is a checkbox which update the option I choose in phpmyadmin.
But if I press submit, then f5/reload page, the preference I choose disappears on client/html side, (look the image for better understanding)
Image, open me!
So basically I want the site remember my choice in the "user account"
I tried with
<form action="pr.php" method="POST">
<h3 style="color:red;">Numbers?</h3>
<input type="checkbox" name="a1" value="1" <?php if(isset($_POST['a1'])) echo "checked='checked'"; ?> /><label>One</label><br/>
<input type="checkbox" name="a2" value="2" <?php if(isset($_POST['a2'])) echo "checked='checked'"; ?> /><label>Two</label><br/>
<input type="checkbox" name="a3" value="3" <?php if(isset($_POST['a3'])) echo "checked='checked'"; ?> /><label>3</label><br/>
<input type="checkbox" name="a4" value="4" <?php if(!is_null($_POST['a4'])) echo "checked='checked'"; ?> /><label>Four</label><br/>
<input type="submit" value="submit" name="submit">
It doesnt work with isset, empty or is_null.
Thanks for all!
Unless you save the information on a database, even if you clicked on submit, after you refresh the webpage you are going to loose the the values of the variables.
You could use cookies to save the value of the choice that was checked:
setcookie("checkbox1","checked");
First parameter is name of the cookie, second one is the value of the cookie, and you can add a third one with the time you want the cookie to have before it expires.
And in your if condition, you could do this:
<?php if($_COOKIE['a1'] == "checked") echo "checked='checked'"; >
For more info, check this link.
I know there's "PHP keep checkbox checked after submitting form" on here, but that thread does not solve my problem, because I have multiple checkbox, what I need is when you check a checkbox, this stay checked after submit.
At the moment with this code nothing happens, I tried another way but when I check "id7" checkbox, all the checkbox get checked.
I have to know which checkbox was checked by the id that I give it, but I do not know how.
while ($fila = mysql_fetch_array($rs)) {
echo utf8_encode("
<tr>
<td>
".$fila['title']."
</td>
");?>
<td>
<input type="checkbox" name="checklist[]" value="<?php echo htmlspecialchars($fila['id']); ?>" <?php if(isset($_POST['checklist[]']) && is_array($_POST['checklist[]']) && in_array('$fila', $_POST['checklist[]'])) echo 'checked="checked"'; ?> />
</td>
<?php
}
}
?>
First, the value of checkbox is $fila['id'] so when you are checking, use $fila['id'] instead of $fila. Also, when PHP receive array input fields with [] in their names the [] will be removed so that the correct POST variable is $_POST['checklist'].
Try changing this line:
<input type="checkbox" name="checklist[]" value="<?php echo htmlspecialchars($fila['id']); ?>" <?php if(isset($_POST['checklist[]']) && is_array($_POST['checklist[]']) && in_array('$fila', $_POST['checklist[]'])) echo 'checked="checked"'; ?> />
to
<input type="checkbox" name="checklist[]" value="<?php echo htmlspecialchars($fila['id']); ?>" <?php if(isset($_POST['checklist']) && is_array($_POST['checklist']) && in_array($fila['id'], $_POST['checklist'])) echo 'checked="checked"'; ?> />
"name" is like a variable name - by using checklist[] you're defining an array as the variable.
Why not just name each checkbox properly? When the form is submitted, use the contents of $_POST to set each variable into the user's $_SESSION.
If the page is refreshed, use the values from $_SESSION to determine if the checkbox should be ticked. Something like (untested):
<input type="checkbox" name="vehicle" value="Bike"
<?php if (isset($_SESSION['bike_checked']) echo 'checked'; ?>> I have a bike<br>
<input type="checkbox" name="vehicle" value="Car"
<?php if (isset($_SESSION['car_checked']) echo 'checked'; ?>> I have a car<br>
<input type="submit" value="Submit">
I have this html form with checkboxes that keep their status checked or unchecked after submitting the form and reloading the page:
<form method="post" action="">
<input type="checkbox" name="keyword1" value="keyword1" <?php if(isset($_POST['keyword1'])) echo "checked='checked'"; ?> />keyword1
<input type="checkbox" name="keyword2" value="keyword2" <?php if(isset($_POST['keyword2'])) echo "checked='checked'"; ?> />keyword2
<input type="checkbox" name="keyword2" value="keyword3" <?php if(isset($_POST['keyword2'])) echo "checked='checked'"; ?> />keyword3
<input type="submit" />
</form>
Problem is, that at first page load the checkboxes are unchecked. Is there any possibility to have all checkboxes with status checked at the beginning and then keep their new status after submit? So far I could not figure out how to do this. Any help would be much appreciated. Thanks,
you could use $_SESSION instead of $_POST :
<input type="checkbox" name="keyword1" value="keyword1" <?php if(isset($_SESSION['keyword1'])) echo "checked='checked'"; ?> />keyword1
And then put this on the top of your file :
session_start();
if (isset($_POST['my_form'])) {
if (isset($_POST['keyword1'])) {
$_SESSION['keyword1'] = 'checked';
}
}
I am trying to get a checkbox checked by default, but everything I have tried doesn't seem to work. I don't know if it has to do with the PHP that is in the code.
function show_subscription_checkbox ($id='0') {
global $sg_subscribe;
sg_subscribe_start();
if ( $sg_subscribe->checkbox_shown ) return $id;
if ( !$email = $sg_subscribe->current_viewer_subscription_status() ) :
$checked_status = ( !empty($_COOKIE['subscribe_checkbox_'.COOKIEHASH]) && 'checked' == $_COOKIE['subscribe_checkbox_'.COOKIEHASH] ) ? true : false;
?>
<p <?php if ($sg_subscribe->clear_both) echo 'style="clear: both;" '; ?>class="subscribe-to-comments">
<input type="checkbox" name="subscribe" id="subscribe" value="subscribe" style="width: auto;" <?php if ( $checked_status ) echo 'checked="checked" '; ?>/>
<label for="subscribe"><?php echo $sg_subscribe->not_subscribed_text; ?></label>
</p>
This is a wordpress plugin that allows you to subscribe to blog comments.
I have tried
echo 'checked=\"checked\" ';
echo 'checked="checked" ' ;
echo 'checked> ';
The plugin author states that you used to be able to default check the checkbox but not anymore.
Since this is showing up in google for "default checked checkbox", I figured I'd answer it. Alpay was right: The correct way to ensure that a checkbox is checked by default is like so (followed by an example of one that is not checked):
<input type="checkbox" name="vehicle" value="Car" checked> I have a car
<input type="checkbox" name="vehicle" value="Bike"> I have a bike
Answer was found on w3schools. The author of the original question was having trouble with his PHP code, which is not at all related to the question title.
In HTML, if you want a checkbox to be checked by default, see the following;
<input type="checkbox" name="name1" value="uc"> This checkbox is unchecked <br>
<input type="checkbox" name="name2" value="c" checked> This checkbox is checked<br>
So, you might consider changing
<?php if ( $checked_status ) echo 'checked="checked" '; ?>
to
<?php if ( $checked_status ) echo 'checked'; ?>
The problem is not in the HTML markup being generated; echo 'checked="checked" ', as in the question, works well, and so would the simpler echo 'checked'.
The problem is with the condition $checked_status. You are testing for a variable that is undefined, as far as the code posted is considered.
I had the same problem. I figured out that I was trying to put the checkbox into a table but left out the and.
didn't check:
if(!$stump){echo '<input type="checkbox" name="stump" value="stump" checked="checked"><b> Stump Job!</b>';}
checked:
if(!$stump){echo '<td><input type="checkbox" name="stump" value="stump" checked="checked"><b> Stump Job!</b></td>';}
In regular PHP you can use this to "save" the checked state after its been submitted.
<form name="checkbox" method="post" action="#" >
<input type="checkbox" name="checkbox1" value="Bike" <?php if ($_POST['checkbox1']=="Bike") echo "checked";?>>I have a bike
<input type="checkbox" name="checkbox2" value="Car" <?php if ($_POST['checkbox2']=="Car") echo "checked";?>>I have a car
<input type="submit" name="submit" value="Display this Data" />
if you want to use this data for something else after the submit, just add:
<?php
if(isset($_POST['checkbox1']) OR isset($_POST['checkbox2']) )
{
echo "for 1 : ".$_POST['checkbox1']."for 2: ".$_POST['checkbox2'];
}
?>
if you want to clear the form (basically clear all the post data) you can add:
<form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">
<input type="submit" name="Clear all form data" value= "Clear this form">
</form>
I am trying to save fields data after submited, Becouse after all the fields are good to go but lets say at the server side the user name is already taken so the form return empty and i dont want that there is the option to do it with PHP like that:
<input value="<?php if(isset($userName)) echo $userName; ?>" />
But the problem is with the radio input, If can some one think about solution about the radio with PHP i will be very thankful, Also i was thinking about Javascript so i will have cleaned code and i was thinking about taking the values from the URL but i am using POST for security reasons.
Summary: If anyone have a solution with PHP or Javascript i will be very thankful, Thank you all and have a nice day.
Try this
<form name="myform" action="" method="post">
<input type="radio" name="language" value="Java" <?php echo(#$_POST['language'] == 'Java'?"checked":""); ?> /> Java
<input type="radio" name="language" value="VB.Net" <?php echo(#$_POST['language'] == 'VB.Net'?"checked":""); ?> /> VB.Net
<input type="radio" name="language" value="PHP" <?php echo(#$_POST['language'] == 'PHP'?"checked":""); ?> /> PHP
<input type="submit" />
I think this may help you.
<input type="radio" value="choice1" name="radio_name" <?php echo(#$_POST['radio_name'] == 'on'?"checked":""); ?> />
If you want to automatically select a radio input you can add the attribute checked to it. What you are going to need will look like this :
<form method="POST">
<?php
// You have some short of list of possible value //
$arrRadioValues = array("value1", "value2", "value3");
// You display them //
for ($i=0; $i<count($arrRadioValues); $i++) {
?>
<input
type="radio"
name="radioInputName"
value="<?php echo $arrRadioValues[$i]; ?>"
<!-- If the value that was posted is the current one we have to add the "checked" so that it gets selected -->
<?php if (isset($_POST['radioInputName']) && $_POST['radioInputName'] == $arrRadioValues[$i]) { echo " checked"; } ?> />
<?php
}
?>
<input type="submit" />
</form>
Adding the checked attribute works a little bit in the same as setting a value to an input. It's just that instead of defining the value attributes, you define the checked attribute when you want that radio to be selected.