Hide or Show Images using checkbox array in mysql database

Hi, I have a group of images that I want to show (make visible) or hide using checkbox array in php/mysql (see screen1).

The database:

visible bit(1) =  show images
visible bit(0) =  hide images

PHP code:
  if($row['visable']==1){
                echo'<td align="center"><input type="checkbox" name="imageCheck[]" id="imageCheck[]" checked='.$chk.'  value="'.$row['id'].'"><br />'.$chk.'</td>';
            }else{
                echo'<td align="center">                   
                    <input type="checkbox" name="unimageCheck[]" id="unimageCheck[]" value="'.$row['id'].'"><br />'.$chk.'</td>';
            }

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using a submit button:

          <input name="check" type="submit" id="check" value="Update"></td>



Code for updating database:

if (isset($_POST['check'])){ 
               
 // hide images
        $imageCheck = isset($_POST['imageCheck']);
        $checkbox = $_POST['imageCheck']; //from name="checkbox[]"
        $countCheck = count($_POST['imageCheck']);
        for($i=0;$i<$countCheck;$i++) {
        $cid  = $checkbox[$i];    

// Create Query: if checkbox is unchecked make image visible=0
            $sql="UPDATE images SET visible = 0 WHERE images.id = $cid " ;
            $row = @mysqli_query ($mysqli, $sql);
            echo '<meta http-equiv="refresh" content="3;url=../admin/admin_edit.php?id='.$id.'">';
        }

//show test results
        echo 'hide';
        echo '<pre>';
        print_r($_POST['imageCheck']);
        echo '</pre>';    
  
 
//show images
        $unimageCheck = isset($_POST['unimageCheck']);
        $uncheckbox = $_POST['unimageCheck']; //from name="checkbox[]"
        $countCheck = count($_POST['unimageCheck']);
        for($i=0;$i<$countCheck;$i++) {
        $uncid  = $uncheckbox[$i];

// Create Query: if checkbox is checked make image visible=1
            $sql2="UPDATE images SET visible = 1 WHERE images.id = $uncid " ;
            $row = @mysqli_query ($mysqli, $sql2);
            echo '<meta http-equiv="refresh" content="3;url=../admin/admin_edit.php?id='.$id.'">';
        }

//show test results
        echo 'show';
        echo '<pre>';
        print_r($_POST['unimageCheck']);
        echo '</pre>';    
}

?>

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I can get the code to work independently but cant get them to work together. I’m really out of the depth here and been working on this for the past 2 months. I have tried many different 'if' statements but still no joy. Need help.
screen1.JPG
screen2.JPG
mrwellsAsked:
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GaryConnect With a Mentor Commented:
First off a checkbox doesn't send a value if it is unchecked - so you cannot check this with the posted values.  The only checkboxes you will get posted are the ones that are checked.

Depending on if this page shows all the images, then you could initially set them all to invisible and then to set the ones that should be visible

foreach($_POST['imageCheck'] as $id){
     $sql2="UPDATE images SET visible = 1 WHERE images.id = $id" ;
     mysqli_query ($mysqli, $sql2);
}
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GaryCommented:
Is it
visable if($row['visable']==1){
or
visible  $sql="UPDATE images SET visible = 0 WHERE images.id = $cid " ;
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mrwellsAuthor Commented:
sorry type error in the question. in all casses it is 'visible'
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GaryCommented:
Confused about what you want.
Do you want the images to be hidden on the screenshots you gave if the Visible checkbox is not checked.
Else explain more clearly what/where you are going wrong.
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Marco GasiFreelancerCommented:
Please, avoid to use @ before statements: that sign suppresses any error message and while debugging it's the most bad thing you can do. Place instead at the top of your scripts

error_reporting(E_ALL)

so you'll be notified about any small problem in your code.

Secondly, you should use mysqli_error() function:

    $sql2="UPDATE images SET visible = 1 WHERE images.id = $uncid " ;
    if (!mysqli_query ($mysqli, $sql2))
    {
        printf("Error message: %s\n", $mysqli->error);
    }

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Marco GasiFreelancerCommented:
You don't need to write

$row = mysqli_query....

since you're updating the table and not selecting records.

You should also post here the whole code and the error messages you get once you have fixed the script as I said above (error_reporting and suppressing @ sign)
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Marco GasiFreelancerCommented:
We also need to see the code where you select check values from the database and build your form with checkboxes check or unchecked accordingly to the query results.
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mrwellsAuthor Commented:
I'd like the user to check the ckeckbox if they want the image 'visible' or 'hidden' on the website;

if the images checkbox is checked then make the image 'visible' (visible=1):

<input type="checkbox" name="imageCheck[]" id="imageCheck[]" checked='.$chk.' value="'.$row['id'].'"><br />'.$chk.'</td>

$sql2="UPDATE images SET visible = 1 WHERE images.id = $uncid " ;
$row = @mysqli_query ($mysqli, $sql2);

if the images checkbox is unchecked then 'hide' the image (visable=0):

<input type="checkbox" name="unimageCheck[]" id="unimageCheck[]" value="'.$row['id'].'"><br />'.$chk.'</td>

 $sql="UPDATE images SET visible = 0 WHERE images.id = $cid " ;
 $row = @mysqli_query ($mysqli, $sql);
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Ray PaseurCommented:
@Gary: One thing you can do to identify unchecked checkboxes is have a hidden input control of the same name in the HTML form before the checkbox.  The way PHP constructs the $_REQUEST-type arrays from the raw request data is always from top-to-bottom, so it will set the hidden value first, then if the checkbox is fired, it will overwrite the hidden value with the checkbox value.  This article shows how to do it.
http://www.experts-exchange.com/Web_Development/Web_Languages-Standards/PHP/A_5450-Common-Sense-Examples-Using-Checkboxes-with-HTML-JavaScript-and-PHP.html
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