I have a page with users listing in admin section. I have the approved users with disabled checkboxes and unapproved users with enabled checkboxes. Now I want to show approved users as "checked" and disable checkboxes. I searched a lot but couldn't get clue. PLease a help or suggestyion will be appreciated.
Here is my code for that,
<input type="checkbox" name="all_check[]" <?php echo $disabled ;?> value="<?php echo $row['id']; ?>" class="checkbox" id="status" ></td>
And the condition to show disabled as,
if ($status == '1') {
$disabled = "disabled";
}
Use something like :
<input type="checkbox" name="all_check[]" <?php if($row['id']==1) echo 'disabled checked="checked"';?> value="<?php echo $row['id']; ?>" class="checkbox" id="status" ></td>
Where your disabled and checked condition should be based on the value stored in your database table column. Above you can change the $row['id'] to what you have used to store the approved column.
Related
So I'm using a checkbox to insert data into MySql database.
When check it stores "finish"
When unchecked if stores "pending"
Now for an edit feature, I have to fetch and display the data from the database. I need the checkbox to be checked or unchecked if it is stored as finish or pending. Appreciate your help, thank you.
The code to fetch the data is shown below:
<?php
$query = ("SELECT * FROM ebook WHERE Eid='" . $Eid . "'");
$result = mysqli_query($con, $query);
$rows = mysqli_fetch_assoc($result);
?>
Displaying data and checkbox:
<html>
<input type="text" name="title" value="<?php echo $rows["Title"]; ?>">
<input class="tgl tgl-flip" id="chk" type="checkbox" name="chk" />
</html>
Put a condition echo inside the input.
<input class="tgl tgl-flip" id="chk" type="checkbox" name="chk" <?php if ($rows['status'] == 'finish') { echo "checked"; } ?>/>
To give you an idea of what I am after. I have items that have "info" to them, well say if I check a box to show Model, I want to be able to go in at a later date and uncheck it.
Is there anyway that you can uncheck a checkbox and have it be stored as 0 in the database?
Here is my input ( I have 10 of these )
<input type="checkbox" name="showmodel" <?php if ($showmodel == '1') echo "checked='checked'"; ?> />
this is how I am trying to update it
$query = "UPDATE `new_equip` SET `featured`='1',`showmanu`='1',`showmodel`='1' "WHERE `id`='$id' LIMIT 1";
I passed trough the same issue and i solved like this:
<input type="hidden" name="permission[<?php echo $data->permission_id; ?>]" value="off">
<input type="checkbox" name="permission[<?php echo $data->permission_id; ?>]" <?php if($data->permission == 'on') { echo 'checked'; } ?> >
I am trying to pass my checkbox values through a session variable if the user goes back by clicking add more books link. I want to show them previously selected check boxes are checked. I tried in chklist.php page
if(isset($_SESSION['pro'])){
echo $_SESSION['pro'];}
shows values like 1_2_5 which are present in session array.
Here is my html code of checkboxes. I have about 24 checkboxes with the same name as product[].
<name="product[]" type="checkbox" value="1" alt="1607.00" />
<name="product[]" type="checkbox" value="2" alt="1607.00" />
and so on
Here I am setting my session after POST of checkboxes .
$_SESSION['pro'] = implode('_', $_POST['product']);
in nextpage chkout.php.
How to make previously selected checkboxes are checked when user came back to firstpage(chklist.php) by clicking ADD MORE BOOKS link present in chkout.php, can any body write the code what i need add to my html.
You can put products array into session then access it as an array.
When setting:
$_SESSION["products"] = $_POST["product"];
When listing products, you should check the values from session:
for($i = 0; $i < 24; $i++) {
echo '<name="product[]" type="checkbox" value="'.$i.'" alt="1607.00"';
if(in_array($i, $_SESSION["products"])) echo ' checked="checked" ';
echo ' />';
}
This is the basic idea and example code.
--UPDATE--
According to your comments:
Inside the form, we will print products as follow:
<?php
session_start();
$session_products = array();
if(array_key_exists("products", $_SESSION))
{
if($_SESSION["products"] != null)
{
$session_products = $_SESSION["products"];
}
}
<form method="post" action="newtest.php">
<input name="product[]" type="checkbox" value="1" <?php if(in_array("1", $session_products)) echo "checked='checked'"; ?> alt="1607.00" />
<input name="product[]" type="checkbox" value="2" <?php if(in_array("2", $session_products)) echo "checked='checked'"; ?> alt="1848.00" />
<input name="product[]" type="checkbox" value="3" <?php if(in_array("3", $session_products])) echo "checked='checked'"; ?> alt="180.00" />
<input name="product[]" type="checkbox" value="4" <?php if(in_array("4", $session_products)) echo "checked='checked'"; ?> alt="650.00" />
and so on upto 24 ...
</form>
Inside the code where the form values are posted, we will put these values into session:
<?php
include("config.php");
session_start();
if(isset($_POST))
{
$_SESSION["products"] = $_POST["product"];
}
?>
As the title reveals I got an issue with how to update a checkbox that already has data in my SQL database.
My code looks like following:
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
Name <input type"text" name="inputName" value="<?php echo $hemsida['Namn']; ?>" /> </br>
Commentar <input type"text" name="inputComment" value="<?php echo $hemsida['Comment']; ?>" />
<br/>
</br><input type="checkbox" name="inputAll" value="checked" <?php echo $hemsida['All']; ?>/>Alla
<input type="hidden" name="id" value="<?php echo $_GET['id']; ?>" />
<input type="submit" name="submit" value="Redigera">
<?php
if(isset($_POST['submit'])) {
$u = "UPDATE hemsida SET `Namn`='$_POST[inputName]', `Comment`='$_POST[inputComment]', `ALL`='$_POST[inputALL]' WHERE ID = $_POST[id]";
mysql_query($u) or die(mysql_error());
echo "User has been modified";
header("Location: ..//sokh.php");
}
?>
The echo $hemsida['Namn'],['Comment'], and ['All'] just brings up and shows the old data thats in the database, but I do not understand what to do to update the checkbox. I have looked everywhere but I am stuck. Thank you in advance!
If I understand your question correctly, you are looking for a way to have a checkbox be either checked or not checked depending on database info. If so, I would try something like this. At the top of your code where you get your database info, put
if($conditionForCheck){
$inputAll = ' checked="checked"';
}
Then in your form
<input type="checkbox" name="inputAll"<?php echo $inputAll; ?> />
your question is not clear but i think you have a column in your database named "all" ? and perhaps this column can take only 1 value (true or false) !!
then you can test this value in your form, if the value is true : checkbox will be checked, else : checkbox will not be checked :
<input type="checkbox" name="inputAll" checked="<?php if($hemsida['All'] == true) echo checked; ?>" />Alla
dont use value="", use checked instead, then test value of $hemsida['All'] if it's true echo checked else anything to do
for your php code and server side of your application you can just test if checkbox is checked and then you have choice for what do you want to assign to your column in database, for example if checkbox is checked create a variable (for example $value_of_checkbox) and assign a value ("true" for exampel) to this variable, then include this variable in your sql code for update database column :
if (isset($_POST['inputALL'])) {
$value_of_checkbox = true;
}
else {
$value_of_checkbox = false;
}
if(isset($_POST['submit'])) {
$u = "UPDATE hemsida SET `Namn`='$_POST[inputName]', `Comment`='$_POST[inputComment]', `ALL`='$value_of_checkbox' WHERE ID = $_POST[id]";
mysql_query($u) or die(mysql_error());
echo "User has been modified";
header("Location: ..//sokh.php")
note : i change also sql code in this part : ALL='$value_of_checkbox'
For usability purposes I like to set up my form fields this way:
<?php
$username = $_POST['username'];
$message = $_POST['message'];
?>
<input type="text" name="username" value="<?php echo $username; ?>" />
<textarea name="message"><?php echo $message; ?></textarea>
This way if the user fails validation, the form input he entered previously will still be there and there would be no need to start from scratch.
My problem is I can't seem to keep check boxes selected with the option that the user had chosen before (when the page refreshes after validation fails). How to do this?
My first suggestion would be to use some client-side validation first. Maybe an AJAX call that performs the validation checks before continuing.
If that is not an option, then try this:
<input type="checkbox" name="subscribe" <?php echo (isset($_POST['subscribe'])?'checked="checked"':'') ?> />
So if subscribe is = 1, then it should select the box for you.
For Example, consider the following code for checkbox :-
<label for="course">Course:</label>
PHP<input type="checkbox" name="course[]" id="course" <?php if ((!empty($_POST["course"]) && in_array("PHP", $_POST["course"]))) {
echo "checked";
} ?> value="PHP" />
Then, this would remember the checkbox of "PHP" if it is checked, even if the validation for the page fails and so on for "n" number of checkboxes as shown below:-
<label for="course">Course:</label>
PHP<input type="checkbox" name="course[]" id="course" <?php if ((!empty($_POST["course"]) && in_array("PHP", $_POST["course"]))) {
echo "checked";
} ?> value="PHP" />
HTML<input type="checkbox" name="course[]" id="course" <?php if ((!empty($_POST["course"]) && in_array("HTML", $_POST["course"]))) {
echo "checked";
} ?> value="HTML" />
CSS<input type="checkbox" name="course[]" id="course" <?php if ((!empty($_POST["course"]) && in_array("CSS", $_POST["course"]))) {
echo "checked";
} ?> value="CSS" />
Javascript<input type="checkbox" name="course[]" id="course" <?php if ((!empty($_POST["course"]) && in_array("Javascript", $_POST["course"]))) {
echo "checked";
} ?> value="Javascript" />
And most importantly, do not forget to declare the "course" variable as an array at the start of the code as shown below :-
$course = array();
I have been battling how to create sticky check box (that is able to remember checked items any time you visit the page). Originally, I get my values from a database table. This means that my check box value is entered to a column on my db table.
I created the following code and it works just fine. I did not want to go through that whole css and deep coding, so...
CODE IN PHP
$arrival = ""; //focus here.. down
if($row['new_arrival']==1) /*new_arrival is the name of a column on my table that keeps the value of check box*/
{$arrival="checked";}// $arrival is a variable
else
{$arrival="";};
echo $arrival;
<b><label for ="checkbox">New Arrival</label></b>
<input type="checkbox" name ="$new_arrival" value="on" '.$arrival.' /> (Tick box if product is new) <BR><BR>
<input type="checkbox" name="somevar" value="1" <?php echo $somevar ? 'checked="checked"' : ''; ?>/>
Also, please consider sanitising your inputs, so instead of:
$somevar = $_POST['somevar'];
...it is better to use:
$somevar = htmlspecialchars($_POST['somevar']);
When the browser submits a form with a checked checkbox, it sends a variable with the name from the name attribute and a value from the value attribute. If the checkbox is not checked, the browser submits nothing for the checkbox. On the server side, you can handle this situation with array_key_exists(). For example:
<?php
$checkedText = array_key_exists('myCheckbox', $_POST) ? ' checked="checked"' : '';
?>
<input type="checkbox" name="myCheckbox" value="1"<?php echo $checkedText; ?> />
Using array_key_exist() avoids a potential array index undefined warning that would be issued if one tried to access $_POST['myCheckbox'] and it didn't exist.
You may add this to your form:
<input type="checkbox" name="mycheckbox" <?php echo isset($_POST['mycheckbox']) ? "checked='checked'" : "" ?> />
isset checks if a variable is set and is not null. So in this code, checked will be added to your checkbox only if the corresponding $_POST variable has a value..
My array has name="radioselection" and value="1", value="2", and value="3" respectively and is a radio button array... how to I check if the radio value is selected using this code
I tried:
<?php echo (isset($_POST['radioselection']) == '1'?'checked="checked"':'') ?> />