Get names from foreach and insert them in the db

[Kleidi]

Hello,
I have a simple image multi uploader that uploads images on the folder correctly but i don't know how to get the image names and insert them in the db. The uploader is part of a complex form for adding an article informations (id, name, permanenturl, section, etc) that works correctly.
Image fields in db has names as follows: img1 img2 img3 img4 im5, a field for each image (since i thought that this is the best way to add the images on the article).
The upload sections looks like this:

<?php



/**

* Smart Image Uploader by @cafewebmaster.com

* Free for private use

* Please support us with donations or backlink

*/





$upload_image_limit = 5; // How many images you want to upload at once?

$upload_dir    = "/var/www/tour/te-ngarkuara/oferta/"; // default script location, use relative or absolute path

################################# UPLOAD IMAGES

        foreach($_FILES as $k => $v){ 

            $img_type = "";

            ### $htmo .= "$k => $v<hr />";     ### print_r($_FILES);

            if( !$_FILES[$k]['error'] && preg_match("#^image/#i", $_FILES[$k]['type']) && $_FILES[$k]['size'] < 1000000){

                $img_type = ($_FILES[$k]['type'] == "image/jpeg") ? ".jpg" : $img_type ;

                $img_type = ($_FILES[$k]['type'] == "image/gif") ? ".gif" : $img_type ;

                $img_type = ($_FILES[$k]['type'] == "image/png") ? ".png" : $img_type ;

                $data = (date("jnHi")); // Merr daten dhe oren ne momentin e shtimit te fotografise

                $img_rname = $data."-".$_FILES[$k]['name']; // shton daten dhe oren perpara emrit te fotografise per te krijuar emer unik
                $img_rname = str_replace(" ","-",$img_rname); // Heq hapesirat dhe i zevendeso me -

                $img_path = $upload_dir.$img_rname;

                copy( $_FILES[$k]['tmp_name'], $img_path );
                chmod("$img_path",0777); 

                $feedback .= "Image and thumbnail created $img_rname<br />";

            }

        }



############################### HTML FORM

    while($i++ < $upload_image_limit){

        $form_img .= '<label>Image '.$i.': </label> <input type="file" name="uplimg'.$i.'"><br />';

    }



    $htmo .= '

        <p>'.$feedback.'</p>

        <form method="post" enctype="multipart/form-data">

            '.$form_img.' <br />

            <input type="submit" value="Upload Images!" style="margin-left: 50px;" />

        </form>

        ';    



    echo $htmo;

Can anyone help me on this, please?

Thank you in advance!

[fideltfg]

I added lines 24 and 45, this will give you an array of the image names that you can them step through to create an insert statement for you bd

<?php



/**

* Smart Image Uploader by @cafewebmaster.com

* Free for private use

* Please support us with donations or backlink

*/





$upload_image_limit = 5; // How many images you want to upload at once?

$upload_dir    = "/var/www/tour/te-ngarkuara/oferta/"; // default script location, use relative or absolute path

################################# UPLOAD IMAGES
        $imageNames = Array(); //nova
        
        foreach($_FILES as $k => $v){

            $img_type = "";

            ### $htmo .= "$k => $v<hr />";     ### print_r($_FILES);

            if( !$_FILES[$k]['error'] && preg_match("#^image/#i", $_FILES[$k]['type']) && $_FILES[$k]['size'] < 1000000){

                $img_type = ($_FILES[$k]['type'] == "image/jpeg") ? ".jpg" : $img_type ;

                $img_type = ($_FILES[$k]['type'] == "image/gif") ? ".gif" : $img_type ;

                $img_type = ($_FILES[$k]['type'] == "image/png") ? ".png" : $img_type ;

                $data = (date("jnHi")); // Merr daten dhe oren ne momentin e shtimit te fotografise

                $img_rname = $data."-".$_FILES[$k]['name']; // shton daten dhe oren perpara emrit te fotografise per te krijuar emer unik
                $img_rname = str_replace(" ","-",$img_rname); // Heq hapesirat dhe i zevendeso me -
                
                $imageNames[] = img_rname; // Nova
                
                $img_path = $upload_dir.$img_rname;

                copy( $_FILES[$k]['tmp_name'], $img_path );
                
                chmod("$img_path",0777);

                $feedback .= "Image and thumbnail acreated $img_rname<br />";

            }

        }



############################### HTML FORM

    while($i++ < $upload_image_limit){

        $form_img .= '<label>Image '.$i.': </label> <input type="file" name="uplimg'.$i.'"><br />';

    }



    $htmo .= '

        <p>'.$feedback.'</p>

        <form method="post" enctype="multipart/form-data">

            '.$form_img.' <br />

            <input type="submit" value="Upload Images!" style="margin-left: 50px;" />

        </form>

        ';    



    echo $htmo;

[Kleidi]

thank you for the reply fideltfg, but, since i'm new on php/mysql, can you explain more the way you suggest?
Thank you very much!

[fideltfg]

ok so after your foreach statement $imageNames will contain an array of all the image file names

you can then create a query like this with it

$c = count ( $imageNames );
$query = "INSERT INTO YOURTABLENAME (img1, img2, img3, img4, im5) VALUES (";

foreach ( $imageNames as $name ) {
    $query .= $name;
    if ($c > 1) {
        $query .=", ";
        $c--;
    }
}
$query .= ")"; 

then send that query to your database how ever your doing that.

[Kleidi]

Thank you. I have added the code you provided with minor changes in variable names, but i'm getting an error on submit:

Fatal error: Column count doesn't match value count at row 1

Thanks!

[plaszma]

put at the end of the script a

echo($query); 

and then load the page again .Execute the query in phpmyadmin and you'll find the error...

[criterion9]

Thank you. I have added the code you provided with minor changes in variable names, but i'm getting an error on submit:

Fatal error: Column count doesn't match value count at row 1

Thanks!

You are missing the columns list. See the reply above showing the expected syntax.

[ebar]

why not put them all in the same field? seperated by a coma or create a new row for each one in the database?

[ebar]

Something like this would create a new row for each image

$numoffile = 5;
      // Fix path of your file to be uploaded, don't forget to CHMOD 777 to this folder
      $file_dir = "/home/link/images/house/".$id."/"; //generate directory
      mkdir($file_dir, 0777);
    if (isset($_POST['upload'])) {
            if(count($_FILES['myfiles']['name'])>0) { //check if any file uploaded
              for($i=0; $i < count($_FILES['myfiles']['name']); $i++) {//loop the uploaded file array
                  if (trim($_FILES['myfiles']['name'][$i])!="") { 
                          $newfile = $file_dir.$_FILES['myfiles']['name'][$i]; //file name
                          move_uploaded_file($_FILES['myfiles']['tmp_name'][$i], $newfile);//upload the file
                          $j++;
                          $my_array[] = $_FILES['myfiles']['name'][$i]; 
                          $pics = array_merge($my_array);
                  }    
                 
              }

              
            }
    }
            
        foreach ($pics as $t => $v ) {
            $sql_pic = "INSERT INTO `images` (`image`, `cat_id`, `list_id`) VALUES ('".$v."', '1', '".$id."')  ";
            $result_pic = @mysql_query($sql_pic);
                if ($result_pic){
                    echo '<tr>
                <td colspan="2" class="form_text3">File:'.$v.' has been added to the database<br />';
                }
    } 

to put them all in the same field you could implode pics with

implode(', ', $pics);