用php和sql上传多个图像 [英] Upload multiple images with php and sql

问题描述

我是新来的PHP，我正在尝试制作一个多图像上传。我的表单看起来像这样：

 < form action =method =postenctype =multipart / form-data > 
< p class =style2> Izberi datoteko：< / p>< input type =filename =imagemultiple =multiple/>< / p> 
< p class =style2> 
 Izberi专辑：< br /> 
< select name =album_id> 
<？php 
 foreach（$ albums as $ album）{
 echo'< option value ='，$ album ['id']，'>'，$相册['name']，'< / option>'; 
} 
？> 
< / select> 
< / p> 
< p class =style2>< input type =submitvalue =NaložiSlike/>< / p> 
< / form> 
  

而在PHP方面，我做了类似的事情：

  if（isset（$ _ FILES ['image']，$ _POST ['album_id']））{
 $ image_name = $ _FILES ['image'] [ '名称']; 
 $ image_size = $ _FILES ['image'] ['size']; 
 $ image_temp = $ _FILES ['image'] ['tmp_name']; 
 
 $ allowed_ext = array（'jpg'，'jpeg'，'png'，'gif'）; 
 $ a = explode（'。'，$ image_name）; 
 $ image_ext = strtolower（end（$ a））; 
 
 $ album_id = $ _POST ['album_id']; 
 
 $ errors = array（）; 
 
 if（empty（$ image_name）|| empty（$ album_id））{
 $ errors [] ='< p class =style2> Nekaj manjka！ p为H.'; 
} else {
 
 if（in_array（$ image_ext，$ allowed_ext）=== false）{
 $ errors [] ='< p class =style2> ; Izbrani tip datoteke ni dovoljen！< / p>'; 
} 
 
 if（$ image_size> 2097152）{
 $ errors [] ='< p class =style2> Izbrana datoteka je prevelika！ Maksimalna velikost datoteke mora biti 2mb！< / p>'; 
 
 
 $ b $ if（！empty（$ errors））{
 foreach（$ errors as $ error）{
 echo $ error; 
} 
} else {
 upload_image（$ image_temp，$ image_ext，$ album_id）; 
 header（'Location：view_album.php？album_id ='。$ album_id）; 
 exit（）; 
 
 
 
 
 
我创建了一个函数，图像到数据库中。这个函数看起来像这样：
 $ b $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $' （INT）$ album_id; 
 $ b mysql_query（INSERT INTO`images` VALUES（''，'。$ _SESSION ['user_id']。'，'$ album_id'，UNIX_TIMESTAMP（），'$ image_ext'） ）; 
 
 $ image_id = mysql_insert_id（）; 
 $ image_file = $ image_id。'。'。$ image_ext; 
 move_uploaded_file（$ image_temp，'galerija /'.$ album_id。'/'。$ image_file）; 
 
 create_thumb（'galerija /'.$ album_id。'/'，$ image_file，'galerija / thumbs /'.$ album_id。'/'）; 
 
  

这只适用于上传一张图片。我试图创建一个窗体多个选择文件，我可以选择更多的图像，但在数据库中只获得第一个图像。我试图在我的脚本中放入 foreach（），但是我不知道在哪里以及如何放置它。
所以我怎样才能改变脚本可以上传更多的图片到我的数据库？我只是无法弄清楚。任何帮助将是伟大的！

解决方案

PHP将不会看到你的多个文件像这样

替换

 < input type =filename =imagemultiple =multiple/> 
  

 < input type =filename =image []multiple =multiple/> 
  

要循环这些文件，您需要像

  foreach（$ _FILES ['image'] ['tmp_name'] as $ key => $ val）{
 
 $ fileName = $ _FILES [ '图像'] [ '名'] [$关键] 
 $ fileSize = $ _FILES ['image'] ['size'] [$ key]; 
 $ fileTemp = $ _FILES ['image'] ['tmp_name'] [$ key]; 
 
 $ fileExt = pathinfo（$ fileName，PATHINFO_EXTENSION）; 
 $ fileExt = strtolower（$ fileExt）; 
 
 //继续
} 
  

I am new in php and I am trying to make a multiple image uploader. My form looks like this:

<form action="" method="post" enctype="multipart/form-data">
    <p class="style2">Izberi datoteko: <br /><input type="file" name="image" multiple="multiple" /></p>
    <p class="style2">
        Izberi Album: <br />
            <select name="album_id">
                <?php 
                    foreach ($albums as $album) {
                        echo '<option value="', $album['id'] ,'">', $album['name'] ,'</option>';
                    }
                ?>
            </select>
    </p>
    <p class="style2"><input type="submit" value="Naloži Slike" /></p>
</form>

And on the php side I did something like:

if (isset($_FILES['image'], $_POST['album_id'])) {
$image_name = $_FILES['image']['name'];
$image_size = $_FILES['image']['size'];
$image_temp = $_FILES['image']['tmp_name'];

$allowed_ext = array('jpg', 'jpeg', 'png', 'gif');
$a = explode('.', $image_name);
$image_ext = strtolower(end($a));

$album_id = $_POST['album_id'];

$errors = array();

if (empty($image_name) || empty($album_id)) {
    $errors[] = '<p class="style2">Nekaj manjka!</p>';
} else {

    if (in_array($image_ext, $allowed_ext) === false) {
        $errors[] = '<p class="style2">Izbrani tip datoteke ni dovoljen!</p>';
    }

    if ($image_size > 2097152) {
        $errors[] = '<p class="style2">Izbrana datoteka je prevelika! Maksimalna velikost datoteke mora biti 2mb!</p>';
    }           
}

if (!empty($errors)) {
    foreach ($errors as $error) {
        echo $error;
    }
} else {
    upload_image($image_temp, $image_ext, $album_id);
    header('Location: view_album.php?album_id='.$album_id);
    exit();
} 

}

I created a function that puts the image into database. This function looks like this:

function upload_image($image_temp, $image_ext, $album_id) {
$album_id = (int)$album_id;

mysql_query("INSERT INTO `images` VALUES ('', '". $_SESSION['user_id'] ."', '$album_id', UNIX_TIMESTAMP(), '$image_ext')"); 

$image_id = mysql_insert_id();
$image_file = $image_id.'.'.$image_ext;
move_uploaded_file($image_temp, 'galerija/'.$album_id.'/'.$image_file);

create_thumb('galerija/'.$album_id.'/', $image_file, 'galerija/thumbs/'.$album_id.'/');
}

This works fine for uploading only one image. I tried to create a form for multiple select files and i can select more images but in database only gets first image. I tried to put foreach() in my script but I don t know where and how I must put it in. So how can i change my script so I can upload more images into my database? I just can't figure it out. Any help would be great!

解决方案

A. PHP would not be albe to see your multiple files like that

Replace

  <input type="file" name="image" multiple="multiple" />

With

 <input type="file" name="image[]" multiple="multiple" />

To loop over this files you need something like

foreach ( $_FILES['image']['tmp_name'] as $key => $val ) {

    $fileName = $_FILES['image']['name'][$key];
    $fileSize = $_FILES['image']['size'][$key];
    $fileTemp = $_FILES['image']['tmp_name'][$key];

    $fileExt = pathinfo($fileName, PATHINFO_EXTENSION);
    $fileExt = strtolower($fileExt);

    // Continue
}

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