2016-08-01 23 views
1

Ich habe 3 Bilder zum hochladen in meinem Formular.Hochladen mehrerer Bilder eins in jede Zeile PHP und Mysqli db

<form action="" method="post" enctype="multipart/form-data" class="form-horizontal"> 
    <div class="form-group col-md-5"> 
    <label for="image">Centralizada</label> 
     <input id="image" type="file" name="image" class="btn btn-danger"> 
    </div> 

    <div class="form-group col-md-5"> 
    <label for="img_v1">V1</label> 
     <input id="img_v1" type="file" name="img_v1" class="btn btn-danger"> 
    </div> 

    <div class="form-group col-md-5"> 
     <label for="img_v2">V2</label> 
     <input id="img_v2" type="file" name="img_v2" class="btn btn-danger"> 
    </div>. 

Und ich habe diese drei Zeilen in meiner Datenbank: Bild, img_v1, img_v2

Und ich möchte jede Zeile alle 3 Bilder hochladen, aber mein Skript nicht funktioniert . Sein Beitrag nur das erste Bild und die anderen beiden nicht.

das ist mein Skript

<?php 
include("includes/dbconn.php"); 


    $error = ''; 
    if(isset($_POST['submit_post'])){ 
    $title = strip_tags($_POST['title']); 
    $date = date('Y-m-d h:i:s'); 
    if($_FILES['image']['name'] !=''){ 
    $image_name = $_FILES['image']['name']; 
    $image_tmp = $_FILES['image']['tmp_name']; 
    $image_size = $_FILES['image']['size']; 
    $image_ext = pathinfo($image_name,PATHINFO_EXTENSION); 
    $image_path = '../clientes/img/'.$image_name; 
    $image_db_path = 'img/'.$image_name; 
    if($image_size < 10000000){ 
    if($image_ext == 'jpg' || $image_ext == 'png' || $image_ext == 'jpeg' ||  $image_ext == 'gif'){ 
    if(move_uploaded_file($image_tmp,$image_path)){ 
    $ins_sql = "INSERT INTO gallery (title, description, image, img_v1, img_v2, category, status) VALUES ('$title', '$_POST[description]', 
    '$image_db_path', '$image_db_path', '$image_db_path', '$_POST[category]', '$_POST[status]')"; 

Warum meine anderen Bilder jetzt das Hochladen? Obs: im Lernen php -noob

+0

Sie benötigen – cmnardi

+0

move_uploaded_file jedes Feld nennen mir ein Beispiel geben, wo das zu tun. –

Antwort

1

Try this (Ich habe nicht getestet)

function imageUpload($field) 
{ 
    $image_name = $_FILES[$field]['name']; 
    $image_tmp = $_FILES[$field]['tmp_name']; 
    $image_size = $_FILES[$field]['size']; 
    $image_ext = pathinfo($image_name, PATHINFO_EXTENSION); 
    $image_path = '../clientes/img/' . $image_name; 
    if ($image_size < 10000000) { 
     if ($image_ext == 'jpg' || $image_ext == 'png' || $image_ext == 'jpeg' || $image_ext == 'gif') { 
      return move_uploaded_file($image_tmp, $image_path); 
     } 
    } 
    return false; 
} 



if (isset($_POST['submit_post'])) { 
    $title = strip_tags($_POST['title']); 
    $date = date('Y-m-d h:i:s'); 
    if ($_FILES['image']['name'] != '') { 

     $upload = imageUpload('image'); 
     $image_db_path = 'img/' . $_FILES['image']['name']; 
     $upload1 = imageUpload('img_v1'); 
     $image_db_path1 = 'img/' . $_FILES['img_v1']['name']; 
     $upload2 = imageUpload('img_v2'); 
     $image_db_path2 = 'img/' . $_FILES['img_v2']['name']; 

     if ($upload && $upload1 && $upload2) 
     $ins_sql = "INSERT INTO gallery (title, description, image, img_v1, img_v2, category, status) VALUES ('$title', '$_POST[description]', 
'$image_db_path', '$image_db_path1', '$image_db_path2', '$_POST[category]', '$_POST[status]')"; 

    } 
} 
+0

immer noch nur das erste Bild (Bild) –

+0

in die erste Zeile setzen: 'error_reporting (E_ALL); ini_set ('display_errors', 'on'); 'um den Fehler zu sehen – cmnardi

+0

keine Fehler sehen Sie den ganzen Code hier https://codeshare.io/Ddp4T –