Update checkbox data in SQL-database with PHP - php

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'

Related

Update values in database using php

I have displayed the data in a form edit.php and again paas the truk_id in url on update page to update.php the data of the fields but it's not working. Can anyone check this
PHP Code
<?php
$truck_id=$_GET['truck_id'];
print_r($truck_id);
include("assets/database_con.php");
if (isset($_POST['submit']))
{
$truck_id=$_POST['truck_id'];
$truck_number= mysql_real_escape_string(htmlspecialchars($_POST['truck_number']));
$truck_model= mysql_real_escape_string(htmlspecialchars($_POST['truck_model']));
$truck_make= mysql_real_escape_string(htmlspecialchars($_POST['truck_make']));
$truck_type= mysql_real_escape_string(htmlspecialchars($_POST['truck_type']));
$truck_tierweight= mysql_real_escape_string(htmlspecialchars($_POST['truck_tierweight']));
$truck_gvm= mysql_real_escape_string(htmlspecialchars($_POST['truck_gvm']));
$truck_regodate= mysql_real_escape_string(htmlspecialchars($_POST['truck_regodate']));
$truck_inspectaiondate= mysql_real_escape_string(htmlspecialchars($_POST['truck_inspectaiondate']));
$sql1 ="UPDATE add_truck SET truck_number='$truck_number', truck_model='$truck_model', truck_make='$truck_make', truck_type='$truck_type', truck_tierweight='$truck_tierweight', truck_gvm='$truck_gvm', truck_regodate='$truck_regodate', truck_inspectaiondate='$truck_inspectaiondate' where truck_id='$truck_id'";
echo $sql1;
$results=$conn->query($sql);
if($results)
{
print 'Success! record updated';
}else{
print 'no! record updated';
}
}
?>
You are assigning values to fields in single quotes.
The main string is enclosed within single quotes.
Variables inside single quotes are not parsed (value is not calculated).
This is called variable interpolation.
Use double quotes for whole string and single quotes for values.
Corrected SQL:
$sql = "UPDATE add_truck
SET truck_number='$truck_number', truck_model='$truck_model',
truck_make='$truck_make', truck_type='$truck_type',
truck_tierweight='$truck_tierweight',
truck_gvm='$truck_gvm', truck_regodate='$truck_regodate',
truck_inspectaiondate='$truck_inspectaiondate' where truck_id='$truck_id'";
you are using GET for the truck id and POST for all other variables - is this intentional?
$truck_id=$_GET['truck_id'];
should it be
$truck_id=$_POST['truck_id'];
or should all the other variables be from the GET array?
$truck_id=$_GET['truck_id'];
Here is your problem part you can't get values from get method when you submit. So you need to add some hidden field into your form like this
<input type="hidden" name="truck_id" value="<?php echo $_GET['truck_id']; ?>">
Change your code like this
if (isset($_POST['submit']))
{
$truck_id=$_POST['truck_id'];
}
I changed your code please look at this
//Your code starts now
<?php
include("header.php");
include("assets/database_con.php");
$truck_id= $_GET['truck_id'];
//print ($truck_id);
?>
<?php
//submit part in top that is better
if (isset($_POST['submit']))
{
$truck_id=$_POST['truck_id'];
$truck_number= mysql_real_escape_string(htmlspecialchars($_POST['truck_number']));
$truck_model= mysql_real_escape_string(htmlspecialchars($_POST['truck_model']));
$truck_make= mysql_real_escape_string(htmlspecialchars($_POST['truck_make']));
$truck_type= mysql_real_escape_string(htmlspecialchars($_POST['truck_type']));
$truck_tierweight= mysql_real_escape_string(htmlspecialchars($_POST['truck_tierweight']));
$truck_gvm= mysql_real_escape_string(htmlspecialchars($_POST['truck_gvm']));
$truck_regodate= mysql_real_escape_string(htmlspecialchars($_POST['truck_regodate']));
$truck_inspectaiondate= mysql_real_escape_string(htmlspecialchars($_POST['truck_inspectaiondate']));
$sql1 ="UPDATE add_truck SET truck_number='$truck_number', truck_model='$truck_model', truck_make='$truck_make', truck_type='$truck_type', truck_tierweight='$truck_tierweight', truck_gvm='$truck_gvm', truck_regodate='$truck_regodate', truck_inspectaiondate='$truck_inspectaiondate' where truck_id='$truck_id'";
echo $sql1;
die();
$results=$conn->query($sql);
if($results)
{
print 'Success! record updated';
}else{
print 'no! record updated';
}
}
?>
And changed some thing in here(added action="")
<form role="form" method="post" action="">
And changed some thing in here also(added name="submit")
<div class="form-group">
<input type="hidden" name="truck_id" value="<?php echo $_GET['truck_id']; ?>">
<input type="submit" value="submit" name="submit" class="btn_full" id="submit-booking">
</div>
No need for another div for hidden because it is 'hidden'
Used a new code as solution :
<?php
if (isset($_POST['submit']))
{
$truck_id=$_POST['truck_id'];
...
}
and one hidden value with the submit name of attribute :
<input type="hidden" name="truck_id" value="<?php echo $_GET['truck_id']; ?>">
<input type="submit" value="submit" name="submit" class="btn_full" id="submit-booking">

Send a value for a checkbox when it is unchecked

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'; } ?> >

PHP MySQL - Undefined Index in Edit Data

I make file details.php that show list of data from MySQL, and I make edit.php for it.
When I click edit.php it's automatically go to edit.php?id=detailsid (it works) but when I click submit on edit data, it's nothing changed and error.
This is my Edit.php
<form method="post" action="editdata.php">
<?php
include 'config.php';
$id=$_GET['id'];
$sqlTampil="select * from data_korban Where kasus_id=$id";
$qryTampil=mysql_query($sqlTampil);
$dataTampil=mysql_fetch_array($qryTampil);
$data=mysql_fetch_array($qryTampil);
?>
This is some of my HTML form
<input type="DATE" name="tanggal" size="20" value="<?php echo $dataTampil['tanggal']; ?>">
<input type="text" name="namakorban" size="40" value="<?php echo $dataTampil['namakorban']; ?>">
<input type="text" name="umurkorban" size="5" value="<?php echo $dataTampil['umurkorban']; ?>">
<input type="submit" name="submit" value="SUBMIT">
When I click Submit it's automatically call editdata.php but there is no change on my MySQL row ..
This is my editdata.php
<?php
include "config.php";
$id=$_GET['id'];
$tanggal = $_POST['tanggal'];
$namakorban = $_POST['namakorban'];
$umurkorban = $_POST['umurkorban'];
$update = "UPDATE data_korban SET tanggal='$tanggal', namakorban='$namakorban', umurkorban='$umurkorban' WHERE kasus_id = '$id'";
$hasil = mysql_query($update);
if ($hasil)
echo "<center>Update Success<center>";
else
echo "<center><h3><a href=pencarian.php>Back Tampil Data</a></h3></center>";
?>
There is something in my Error_Log
PHP Notice: Undefined index: id in /home/lombokdi/public_html/dbanak/editdata.php on line 4 ------ it's $id=$_GET['id'];
I have change to $_POST['id']; it's still error.
Your form does not seem to include the id named field.
And you are trying to read the non existing field from the request.
Request is formed to send data over post method but you tried to read id parameter using get method. As the value for id is not found, a null is used in where clause and that affected no rows updated.
$id=$_GET['id'];
$tanggal = $_POST['tanggal'];
$namakorban = $_POST['namakorban'];
$umurkorban = $_POST['umurkorban'];
the value for ID is $id=$_GET['id']; ... id come from detail.php?id=24 (example) go to edit.php?id=24 .. If I'm wrong, how to set the ID ??
In edit.php add id named element to the form with value read from the request. And on submitting the form, id goes to editdata.php page.
<input type="DATE" name="tanggal" size="20" value="<?php echo $dataTampil['tanggal']; ?>">
<input type="hidden" name="id" value="<?php echo $id['id']; ?>">
<input type="text" name="namakorban" size="40" value="<?php echo $dataTampil['namakorban']; ?>">
<input type="text" name="umurkorban" size="5" value="<?php echo $dataTampil['umurkorban']; ?>">
And in editdata.php:
Change:
$id=$_GET['id'];
To:
$id=$_POST['id'];
It's because you don't have the filed "id" if you switch to editdata.php
add: <input type="hidden" name="id" value="<?php $_GET['id']; ?>">
to your form in Edit.php
and replace
$id=$_GET['id']; with $id= $_POST['id']; in editdata.php.
Try like this
$id=$_GET['id'];
$sql=mysqli_query($con,"Update client set nama='$nama',instansi='$instansi',telepon='$telepon',email='$email' where id = '".$_GET['id']."'");
if($sql)
{
$msg="Your Client updated Successfully";
header('location:manage-client.php');
}
for after body add this query
$sql=mysqli_query($con,"select * from client where id = '".$_GET['id']."'");
while($data=mysqli_fetch_array($sql))
You are using $_GET to get the id. It is best do var_dump($_REQUEST);

Pass a php variable into another php file

i have a php file which stores a variable, and i need to create a submit form which passes this variable.
So the code of the current file is:
if(isset($_POST['Submit'])) {
echo "<pre>";
$checked = implode(',', $_POST['checkbox']);
echo $checked;
}
I have to insert the submit form and on click it goes to another php file, i need to get there the $checked variable.. how can i store this variable?..
Thanks you!
The principle of POSTing is that you can send all the data to the next .php page.
<?php
if ($checked == 1) {
$checked = 'checked="checked"';
}
else {
$checked = '';
}
?>
<form action="POST" method="target_file.php">
<input type="hidden" name="variableA" value="Something I want target_file.php to know" />
<input type="hidden" name="variableB" value="Something else I want target_file.php to know" />
<input type="checkbox" name="gender" value="male" <?php echo $checked ?>" />
</form>
Your target_file.php:
echo "Here I am :), variableA: ".$_POST['variableA'];
echo "Here I am :), variableB: ".$_POST['variableB'];
echo "Here I am :), My gender is: ".$_POST['gender'];
Also, don't trust on checking your field values that are send through the $_POST. Check what every value is and if it meets certain criteria before sub-statements may be executed.

How to remember checkbox input in PHP Forms

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"':'') ?> />

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