displaying html checkbox multiple checkbox content in php page using loop

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.

if u got your solution from listed below answer than chose the best answer and mark as accepted. meta.stackexchange.com/questions/5234/… — devpro
– devpro, Commented Feb 3, 2016 at 9:37

devpro · Accepted Answer · 2016-02-01 10:13:51Z

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.

edited Feb 1, 2016 at 10:13

answered Feb 1, 2016 at 9:48

devpro

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Comments

ameenulla0007 · Accepted Answer · 2016-02-01 09:41:48Z

0

<?php echo "$f6","<br>"; ?>

wrong concatenation

<?php echo "$f6"."<br>"; ?>

use this

answered Feb 1, 2016 at 9:41

ameenulla0007

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Comments

Ravi Hirani · Accepted Answer · 2016-02-01 09:58:25Z

0

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>

edited Feb 1, 2016 at 9:58

answered Feb 1, 2016 at 9:47

Ravi Hirani

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2 Comments

momouu Over a year ago

add php closing tag after // You should write below line as

Ravi Hirani Over a year ago

@monace19: Thanks for your suggestion. I have missed it ;)

Tan See Youu · Accepted Answer · 2016-02-01 10:02:21Z

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

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