How to retrieve checkbox value from database in PHP

Question

My question is how to retrieve a checkbox value from the database in PHP. The below code is not working.

<td>Hobby</td>
<?php
    $result = mysqli_query($con,"SELECT hobby FROM simple WHERE id = id"); 
    while($row = mysqli_fetch_array($result))
    {
        $focus = explode(",", $row['hobby']);
?>    
<td>
    <input type="checkbox"  name="hobby[]"  value="Cricket"  size="17"   <?php if(in_array("Cricket",$hobby)) { ?> checked="checked" <?php } ?> >Cricket
    <input type="checkbox" name="hobby[]"  value="Music" size="17" <?php if(in_array("Music",$hobby)) { ?> checked="checked" <?php } ?> >Music
    <input type="checkbox" name="hobby[]"  value="Reading"   size="17" <?php if(in_array("Reading",$hobby)) { ?> checked="checked" <?php } ?> >Reading
    <input type="checkbox" name="hobby[]"  value="Study" size="17" <?php if(in_array("Study",$hobby)) { ?> checked="checked" <?php } ?> >Study</td>
<?php
   }
?>
</tr>

Note that you'll fetch all datasets when using condition WHERE id = id in your SQL statement. — mario.van.zadel
– mario.van.zadel, Commented Jan 8, 2016 at 6:31

Pupil · Accepted Answer · 2016-01-08 06:53:38Z

Change the statement,

$focus=explode(",",$row['hobby']);

To:

$hobby=explode(",",$row['hobby']);

Also, You can minimize your code with the use of ternary operators instead of if else like:

<td>
  <input type="checkbox"  name="hobby[]"  value="Cricket"  size="17" <?php echo (in_array("Cricket",$hobby)) ? 'checked="checked" : '';?>>Cricket
  <input type="checkbox" name="hobby[]"  value="Music" size="17" <?php echo (in_array("Music",$hobby)) ? 'checked="checked" : '';?>>Music
  <input type="checkbox" name="hobby[]"  value="Reading"   size="17" <?php echo (in_array("Reading",$hobby)) ? 'checked="checked" : '';?>>Reading
</td>

Also, @mapek suggested, your SQL will fetch all records.

$result = mysqli_query($con,"SELECT hobby FROM simple WHERE id = id");

As id will always be equal to id.

You should change it to:

$result = mysqli_query($con,"SELECT hobby FROM simple WHERE id = '$id'");

Another version to minimize this code with arrays:

<td>
<?php
$hobbies = array('Cricket', 'Music', 'Reading');
if (! empty($hobbies)) {
  foreach ($hobbies as $myHobby) {
    $checked = (in_array($myHobby, $hobby)) ? 'checked="checked"' : '';
?>
<input type="checkbox" name="hobby[]" value="<?php echo $myHobby;?>" size="17" <?php echo $checked;?>><?php echo $myHobby;?>
<?php
    }
}
?>
</td>

Vegeta · Accepted Answer · 2016-01-08 09:52:15Z

2

try this : many changes required

<td>Hobby</td>
    <?php
        $id= 1; // your input id value
        $result = mysql_query("SELECT hobby FROM simple WHERE id = $id"); 
        while($row = mysql_fetch_array($result))
       {
        $hobby=explode(",",$row['hobby']);

    ?>    
    <td><input type="checkbox"  name="hobby[]"  value="Cricket"  size="17"   <?php if(in_array("Cricket",$hobby)) echo 'checked="checked"'; ?> >Cricket
         <input type="checkbox" name="hobby[]"  value="Music" size="17" <?php if(in_array("Music",$hobby)) echo 'checked="checked"'; ?> >Music
         <input type="checkbox" name="hobby[]"  value="Reading"   size="17" <?php if(in_array("Reading",$hobby))  echo 'checked="checked"'; ?> >Reading
         <input type="checkbox" name="hobby[]"  value="Study" size="17" <?php if(in_array("Study",$hobby))  echo 'checked="checked"'; ?> >Study</td>
       <?php
       }
       ?>
</tr>

edited Jan 8, 2016 at 9:52

answered Jan 8, 2016 at 6:32

Vegeta

1,3179 silver badges17 bronze badges

16 Comments

Vegeta Over a year ago

please show me output for $result = mysqli_query($con,"SELECT hobby FROM simple");

Vrajesh Savaliya Over a year ago

my simply code<td>Hobby</td> <td><input type="checkbox" name="hobby[]" value="Cricket" >Cricket <input type="checkbox" name="hobby[]" value="Music" >Music <input type="checkbox" name="hobby[]" value="Reading">Reading <input type="checkbox" name="hobby[]" value="Study" >Study</td> </tr>

Vegeta Over a year ago

are you getting output for mysqli_query($con,"SELECT hobby FROM simple"); ? what is it ?

Vrajesh Savaliya Over a year ago

mysqli_query() expects parameter 1 to be mysqli, null given

Vegeta Over a year ago

then something wrong with connection. try mysqli_query("SELECT hobby FROM simple");. pls tell me its output ?

|

AnkiiG · Accepted Answer · 2016-01-08 06:29:14Z

1

Change variable name in

<?php if(in_array("Cricket",$hobby)) { ?>

from $hobby to $focus

answered Jan 8, 2016 at 6:29

AnkiiG

3,4861 gold badge19 silver badges28 bronze badges

2 Comments

Vrajesh Savaliya Over a year ago

mysqli_query() expects parameter 1 to be mysqli, resource given

Vrajesh Savaliya Over a year ago

mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given

Haroon Hayat · Accepted Answer · 2021-06-12 10:16:45Z

0

$newArray = explode(",",$result['hobbies']);
<input type="checkbox" name ="hobbies[]" value ="reading" <?php if(in_array("reading",$newArray)){ echo "checked";} ?>
<input type="checkbox" name ="hobbies[]" value = "games" <?php if(in_array("games",$newArray)){ echo "checked";}?>>`
<input type="checkbox" name ="hobbies[]" value = "coding"  <?php if(in_array("coding",$newArray)){ echo "checked";} ?>

answered Jun 12, 2021 at 10:16

Haroon Hayat

533 bronze badges

1 Comment

alhelal Over a year ago

Not upload your code only, describe your problem to get a answer quickly.

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