Upload multiple images and save names in database PHP MySQL

Question

After reviewing different solutions to one same question, here I come with a new scenario, Mis cript only uploads one of the various images sent via HTML form and throws an error on mysql query.

    $imgn = $_FILES['imgn']['name'];     # Definimnos los valores
    $galnom = $_POST['galnom'];  # para extracción

    for($i =0; $i < count($_FILES['imgn']['name']); $i++){  # creamos el loop inicial utilizado la variable $i
    # Procesar la imagen ---------------------------------------------------------------------




    #  Propiedad de Samuel Ramírez S...
    #  smctotal.com
    #  Este script no puede redistribuirse sin una constancia de autorización por parte de Samuel Ramírez S. RASS650104P35
    $imaggen = $_FILES['imgn']['name'][$i];       # Archivo original
    $tempral = $_FILES["imgn"]["tmp_name"][$i];   # Archivo en la carpeta temporal
    $formato = $_FILES["imgn"]["type"][$i];       # Analizamos el tipo de archivo image/jpeg
    $tamanio = $_FILES["imgn"]["size"][$i];       # Tamaño de la imagen


    $divisor = explode(".",$imaggen);           # Separamos el nombre de la extensión
    $xtnsion = end($divisor);                   # Extensión también se puede imprimir $divisor[1];
    $nombbre = $divisor[0];                     # Nombre del archivo
    round(($tamanio/1000),0).' Kilobytes';      # Formateo del peso del archivo
    $imgn = round(microtime(true)).'.'.$xtnsion; # Se establece el nuevo nombre del archivo
    move_uploaded_file($tempral, "../../img/sect/".$imgn); # Se guarda el archivo con su nuevo nombre.
    # Hasta aquí ya subimos el archivo
    # Enseguida lo abrimos para proceso ------------------------------------------------------
    $img = imagecreatefromjpeg("../../img/sect/".$imgn);   # Abrimos la imagen para proceso
    $anchoActual = imagesx($img);               # Obtenemos el ancho original
    $alttoActual = imagesy($img);               # Obtenemos la altura original

    # --- CD Imagen grande -------------------------------------------------------------------
    $nuevo_ancho = 600;
    $factor = $nuevo_ancho / $anchoActual; # Proporción alto anterior y el nuevo, aplicar el factor a la altura
    $nuevo_allto = $alttoActual * $factor;
    # --- FD Imagen grande

    # --- CD Thumbnail -----------------------------------------------------------------------
    $thumb_ancho = 100;
    $facthmb = $thumb_ancho / $anchoActual; # Proporción alto anterior y el nuevo, aplicar el factor a la altura
    $alto_thumb = $alttoActual * $facthmb;
    # --- FD Thumbnail -----------------------------------------------------------------------

    # --- CD Imagen grande -------------------------------------------------------------------
    $imagenFinal = imagecreatetruecolor($nuevo_ancho,$nuevo_allto);
    imagecopyresampled($imagenFinal, $img, 0, 0, 0, 1, $nuevo_ancho, $nuevo_allto, $anchoActual, $alttoActual);
    imagejpeg($imagenFinal, "../../gal/grnd/".$imgn, 85);          # Comprimimos y salvamos la imagen
    # --- FD Imagen grande -------------------------------------------------------------------

    # --- CD Imagen grande -------------------------------------------------------------------
    $thumbFinal = imagecreatetruecolor($thumb_ancho,$alto_thumb);
    imagecopyresampled($thumbFinal, $img, 0, 0, 0, 1, $thumb_ancho, $alto_thumb, $anchoActual, $alttoActual);
    imagejpeg($thumbFinal, "../../gal/thmb/".$imgn, 85);          # Comprimimos y salvamos la imagen
    # --- FD Imagen grande -------------------------------------------------------------------

    # Fin del proceso de la imagen -----------------------------------------------------------

    $q1[$i]= mysqli_query($cnxn,"INSERT INTO galerias (galnom,imgn) VALUES ('$galnom[$i]','$imgn[$i]'") or die(mysqli_error($cnxn));

Can someone please point me where is it that I'm failing?

Thanks in advance.

Thank You devpro : You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1 — Samuel Ramzan
– Samuel Ramzan, Commented Feb 22, 2016 at 6:00
thanks for sharing this error, now u have to many solutions, did u checked? — devpro
– devpro, Commented Feb 22, 2016 at 6:11
Yes... for the MySQL part it was a typo... but the images script only uploads one image... One problem solved one to go... — Samuel Ramzan
– Samuel Ramzan, Commented Feb 22, 2016 at 6:13

Nehal · Accepted Answer · 2016-02-22 06:06:11Z

2

Try to edit your query :

$q1[$i]= mysqli_query($cnxn,"INSERT INTO galerias (galnom,imgn) VALUES ('$galnom[$i]','$imgn[$i]'") or die(mysqli_error($cnxn));

With this :

$q1= mysqli_query($cnxn,"INSERT INTO galerias (galnom,imgn) VALUES ('$galnom[$i]','$imgn[$i]')") or die(mysqli_error($cnxn));

answered Feb 22, 2016 at 6:06

Nehal

1,5134 gold badges17 silver badges30 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Samuel Ramzan Over a year ago

Thank's for the MySQL problem solving... but the script only upload one image... but it does save the whole array of names in the SQL DB

naseeba c · Accepted Answer · 2016-02-22 06:05:11Z

2

try this

$q1[$i]= mysqli_query($cnxn,"INSERT INTO galerias (galnom,imgn) VALUES ('$galnom[$i]','$imgn[$i]')") or die(mysqli_error($cnxn));

answered Feb 22, 2016 at 6:05

naseeba c

1,0603 gold badges14 silver badges33 bronze badges

Comments

devpro · Accepted Answer · 2016-02-22 06:05:48Z

2

You are missing the ending bracket of mysqli_query()

This:

$q1[$i]= mysqli_query($cnxn,"
INSERT INTO galerias (galnom,imgn) 
VALUES ('$galnom[$i]','$imgn[$i]'") // error in this line
or die(mysqli_error($cnxn));

Should be:

$q1[$i]= mysqli_query($cnxn,"
INSERT INTO galerias (galnom,imgn) 
VALUES ('$galnom[$i]','$imgn[$i]')") 
or die(mysqli_error($cnxn));

edited Feb 22, 2016 at 6:05

answered Feb 22, 2016 at 6:03

devpro

16.1k3 gold badges29 silver badges39 bronze badges

1 Comment

devpro Over a year ago

@PathikVejani: fixed typo.. bro

Arslan Bilal · Accepted Answer · 2016-02-22 06:06:22Z

2

sql should be

$q1[$i]= mysqli_query($cnxn,"INSERT INTO galerias (galnom,imgn) VALUES ('".$galnom[$i]."','".$imgn[$i]."')") or die(mysqli_error($cnxn));

answered Feb 22, 2016 at 6:06

Arslan Bilal

6252 gold badges9 silver badges17 bronze badges

Comments

Pathik Vejani · Accepted Answer · 2016-02-22 06:06:26Z

2

Insert query syntax is wrong: Replace with below:

$q1[$i]= mysqli_query($cnxn,"INSERT INTO galerias (galnom,imgn) VALUES ('$galnom[$i]','$imgn[$i]')") or die(mysqli_error($cnxn));

answered Feb 22, 2016 at 6:06

Pathik Vejani

4,49711 gold badges66 silver badges110 bronze badges

Comments

Panda · Accepted Answer · 2016-02-22 08:31:50Z

0

You'll missing the brackets for mysqli_query:

$q1[$i]= mysqli_query($cnxn,"INSERT INTO galerias (galnom,imgn) VALUES ('$galnom[$i]','$imgn[$i]')") or die(mysqli_error($cnxn));

1st Part:

mysqli_query($cnxn,"INSERT INTO galerias (galnom,imgn) VALUES ('$galnom[$i]','$imgn[$i]')")

2nd Part:

or die(mysqli_error($cnxn))

edited Feb 22, 2016 at 8:31

answered Feb 22, 2016 at 6:07

Panda

6,8846 gold badges46 silver badges61 bronze badges

Collectives™ on Stack Overflow

Upload multiple images and save names in database PHP MySQL

6 Answers 6

1 Comment

Comments

1 Comment

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

6 Answers 6

1 Comment

Comments

1 Comment

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related