displaying html checkbox multiple checkbox content in php page using loop - php

Question

//I extracted data from database like
<form action="print.php" method="post">
<?php include('connection.php') ?>
<?php
// Query member data from the database and ready it for display
$sql = mysql_query("SELECT * FROM nokia");
$num = mysql_numrows($sql);
?>
<?php
$i=0;
while ($i < $num) {
$f6=mysql_result($sql,$i,"item_name");
<input type="checkbox" name="list[]" value="<?php echo "$f6";?>" /><?php echo "$f6","<br>"; ?>//display result in htmlpage
<?php
$i++;
$d=$f6;
}
?>
<input type="submit" value="submit"/>
</form>
print.php
//after submitting in php page
<?php include('connection.php') ?>
<?php
if(isset($_POST['submit']))
{
if(!empty($_POST['list']))//name of checkbox in html
{
foreach($_POST['list'] as $selected){
echo $selected."</br>";
}
}
}
?>
//display nothing in php...how I solve this problem
I want to display content of checkbox from html form to php page.I extracted the checkbox contents from database.problem is how I display the content in php page selecting multiple checkbox.

Answer 1

First of all add error_reporting in your code:
ini_set('error_reporting', E_ALL);
error_reporting(E_ALL);
You have few issues in your HTML/PHP combination:
Issues:
You are using <input type="checkbox" .. inside the PHP.
your concatenation is wrong "$f6","<br>"
Also need to add name for like name="submit" in button.
Modified Code:
<?php
$i=0;
while ($i < $num) {
$f6=mysql_result($sql,$i,"item_name");
?>
<input type="checkbox" name="list[]" value="<?php echo $f6;?>" /><?php echo $f6; ?><br/>
<?php
$i++;
$d = $f6;
}
?>
Side note:
Suggestion of error_reporting is only for development and staging not for production.
Please use mysqli_* or PDO because mysql_* is deprecated and not available in PHP 7.

Answer 2

<?php echo "$f6","<br>"; ?>
wrong concatenation
<?php echo "$f6"."<br>"; ?>
use this

Answer 3

Your code should be:-
<?php
// This line should be first.
// You have missed semicolon here.
include('connection.php');
// Query member data from the database and ready it for display
$sql = mysql_query("SELECT * FROM nokia");
$num = mysql_numrows($sql);
?>
<form action="print.php" method="post">
<?php
$i=0;
while ($i < $num) {
$f6=mysql_result($sql,$i,"item_name");
?>
<!-- You should write below line as -->
<input type="checkbox" name="list[]" value="<?php echo $f6;?>" /><?php echo $f6; ?> </br>
<?php
$i++;
$d=$f6;
}
?>
<input type="submit" value="submit"/>
</form>

Answer 4

there are couple of errors in your codes. please find the modified codes below with my comment started with //seeyouu:
//I extracted data from database like
<form action="print.php" method="post">
<?php include('connection.php');
// Query member data from the database and ready it for display
$sql = mysql_query("SELECT * FROM nokia");
$num = mysql_num_rows($sql);//seeyouu: used wrong function>>mysql_numrows
$i=0; //seeyouu: after my amendment, probably can remove this
while ($row = mysql_fetch_array($sql)) {
//seeyouu: no such way of doing, your $sql already a result >> $f6=mysql_result($sql,$i,"item_name");
//seeyouu: forgot to close your php tag here
?>
<input type="checkbox" name="list[]" value="<?php echo $row["item_name"]; ?>" />
<?php //seeyouu: wrong>>echo "$f6","<br>";
//correct way:
echo $row["item_name"]."<br />";
//$d=$f6;i don't know what is this line for, so i leave it to you.
}
?>
<input type="submit" name="submit" value="submit"/> <!-- seeyouu: forgot to set name: submit -->
</form>
print.php
//after submitting in php page
<?php include('connection.php');
if(isset($_POST['submit']))
{
if(!empty($_POST['list']))//name of checkbox in html
{
foreach($_POST['list'] as $selected)
{
echo $selected."</br>";
}
}
}
?>
//display nothing in php...how I solve this problem