如何在 PHP 和 mysql 数据库中上传和插入多个图像

Question

我是 PHP 初学者。我在 base64String 中编码了一些图像。指定文件夹中的所有图像均已成功解码。我的问题是只能在数据库中记录一个图像/路径。有人帮我想出 PHP 在数据库中的一行中插入所有图像路径。

这里是php代码

 <?PHP
if(isset($_POST['image']))
{
$image = $_POST['image'];
$identity = $_POST['id'];
$username = $_POST['username'];

//create unique image file name based on micro time and date
$now = DateTime::createFromFormat('U.u', microtime(true));
$id = rand(1000000, 10000000000);
$id2=rand(1000000, 10000000000);

$upload_folder = "upload";
$id="$id$id2";
$path = "$upload_folder/$id.jpeg";

if(file_put_contents($path, base64_decode($image)) != false){
    echo "uploaded_success";

   $sql = "UPDATE apartment SET Image_path = '$path' WHERE apart_username 
  ='$username' AND id = '$identity'";

   mysqli_query($conn, $sql);   

}
else{
    echo "uploaded_failed";
}    
exit;
}
else{
echo "images_not_in";
exit;

} ?>

Answer 1

您需要某种循环...在下面的示例中，我所做的只是创建了一个名为 $imagesArr 的数组，而不是创建一个名为 image 的变量。我在 imagesArr 上添加了一个 foreach 循环，因此它将运行您为输入到该数组中的每个图像编写的代码。我还在最后杀死了出口，因为这会停止循环。应该可以的

 <?PHP
if(isset($_POST['image']))
{
//$image = $_POST['image']; - I commented this out
$imagesArr = array('image1.jpeg','image2.jpeg','image3.jpeg'); //etc. put all of your images here into this array    
$identity = $_POST['id'];
$username = $_POST['username'];
foreach ($imagesArr as $key => $image) {
//create unique image file name based on micro time and date
$now = DateTime::createFromFormat('U.u', microtime(true));
$id = rand(1000000, 10000000000);
$id2=rand(1000000, 10000000000);

$upload_folder = "upload";
$id="$id$id2";
$path = "$upload_folder/$id.jpeg";

if(file_put_contents($path, base64_decode($image)) != false){
    echo "uploaded_success";

   $sql = "UPDATE apartment SET Image_path = '$path' WHERE apart_username 
  ='$username' AND id = '$identity'";

   mysqli_query($conn, $sql);   

}
else{
    echo "uploaded_failed";
}    
}
else{
echo "images_not_in";
}
}