如何在php中一次移动上传的多个图像

Question

我正在尝试上传 3 个图像字段，将它们存储在我的数据库中并将上传的文件移动到一个文件夹中。

目前，“image1”、“image2”、“image3”、“image4”字段正在插入 MySQL 数据库，但只有“image1”正在上传并移动到我的文件夹中。

这是我的代码：

if(isset($_POST['submit'])){
  $pro_image1 = $_FILES['image1']['name'];
  $pro_image2 = $_FILES['image2']['name'];
  $pro_image3 = $_FILES['image3']['name'];
  $pro_image4 = $_FILES['image4']['name'];
  $tmp_name   = $_FILES['image1']['tmp_name'];
  $tmp2_name  = $_FILES['image2']['tmp_name'];
  $tmp3_name  = $_FILES['image3']['tmp_name'];
  $tmp4_name  = $_FILES['image4']['tmp_name'];

  $pro_query = "INSERT INTO product(image1,image2,image3,image4) VALUES('$pro_image1','$pro_image2','$pro_image3','$pro_image4')";

  if(mysqli_query($con,$pro_query)){
    $msg = "<p class='pull-right' style='color:green;'> Product Added successfully</p>";
    $path = "images/$pro_image1";

    if(move_uploaded_file($tmp_name, $path)) {
      copy($path, "../$path");
    }

    if(move_uploaded_file($tmp2_name, $path)) {
      copy($path, "../$path");
    }

    if(move_uploaded_file($tmp3_name, $path)) {
      copy($path, "../$path");
    }

    if(move_uploaded_file($tmp4_name, $path)) {
      copy($path, "../$path");
    } elseif(!mysqli_query($con,$pro_query)) {
      $insert_error = "<p class='pull-right' style='color:red;>Product didn't added</p>";
    }
  }
}

<form action="" method="post" enctype="multipart/form-data"  class="form-font">
  <div class="col-md-6">
    <div class="form-group">
      <label>Image1</label>
      <input type="file" name="image1" class="form-control" required>
    </div>
    <div class="form-group">
      <label>Image2</label>
      <input type="file" name="image2" class="form-control" required>
    </div>
    <div class="form-group">
      <label>Image3</label>
      <input type="file" name="image3" class="form-control" required>
    </div>
    <div class="form-group">
      <label>Image4</label>
      <input type="file" name="image4" class="form-control" required>
    </div>
    <center>
      <input type='submit' name='submit' class='btn btn-success' value='Add Product'>
    </center>
  </div>
</form>

如何移动文件夹中的其他图片（“image2”、“image3”、“image4”）？
我在代码中犯了什么错误？你能解释一下我哪里做错了吗？

Answer 1

这是你需要做的：

if(move_uploaded_file($tmp_name, $path)){
             copy($path, "images/$pro_image2");
           }
           if(move_uploaded_file($tmp2_name, $path)){
             copy($path, "images/$pro_image3");
           }if(move_uploaded_file($tmp3_name, $path)){
             copy($path, "images/$pro_image4");
           }if(move_uploaded_file($tmp4_name, $path)){
             //copy($path, "images/$path");
           }

Answer 2

请覆盖您的 $path 变量，然后它将上传属性..

之前如果条件改变 $path 变量如下或改变变量为 $path1,$path2,$path3..

$path = "images/$pro_image1";
$path = "images/$pro_image2";
$path = "images/$pro_image3";

-

        if(isset($_POST['submit'])){
   $pro_image1 = $_FILES['image1']['name'];
        $pro_image2 = $_FILES['image2']['name'];
        $pro_image3 = $_FILES['image3']['name'];
        $pro_image4 = $_FILES['image4']['name'];
        $tmp_name = $_FILES['image1']['tmp_name'];
        $tmp2_name = $_FILES['image2']['tmp_name'];
        $tmp3_name = $_FILES['image3']['tmp_name'];
        $tmp4_name = $_FILES['image4']['tmp_name'];
$pro_query = "INSERT INTO product(image1,image2,image3,image4) VALUES('$pro_image1','$pro_image2','$pro_image3','$pro_image4')";
        if(mysqli_query($con,$pro_query)){
            $msg = "<p class='pull-right' style='color:green;'> Product Added successfully</p>";
            $path = "images/$pro_image1";
            if(move_uploaded_file($tmp_name, $path)){
                 copy($path, "../$path");
               }
               $path = "images/$pro_image2";
               if(move_uploaded_file($tmp2_name, $path)){
                 copy($path, "../$path");
               }
               $path = "images/$pro_image3";
               if(move_uploaded_file($tmp3_name, $path)){
                 copy($path, "../$path");
               }
               $path = "images/$pro_image4";
               if(move_uploaded_file($tmp4_name, $path)){
                 copy($path, "../$path");
               }
               elseif(!mysqli_query($con,$pro_query)){
            $insert_error = "<p class='pull-right' style='color:red;>Product didn't added</p>";
        }
    }
}

<form action="" method="post" enctype="multipart/form-data"  class="form-font">
    <div class="col-md-6">
       <div class="form-group">
                <label>Image1</label>
                <input type="file" name="image1" class="form-control" required>
                </div>
                <div class="form-group">
                <label>Image2</label>
                <input type="file" name="image2" class="form-control" required>
                </div>
                <div class="form-group">
                <label>Image3</label>
                <input type="file" name="image3" class="form-control" required>
                </div>
                <div class="form-group">
                <label>Image4</label>
                <input type="file" name="image4" class="form-control" required>
                </div>
<center><input type='submit' name='submit' class='btn btn-success' value='Add Product'></center>
            </form>

Answer 3

您只为 image1 定义了 var $path —— 您需要在每个“if”块中为函数 move_uploaded_file 和 copy 更改此 var。 目前，您只对所有图像使用 image1 的路径和文件名。

Answer 4

您只需将它们命名为“image[0], image[1], image[2]” 代码中的更改

<form action="" method="post" enctype="multipart/form-data"  class="form-font">
    <div class="col-md-6">
       <div class="form-group">
           <label>Image1</label>
           <input type="file" name="image[0]" class="form-control" required>
       </div>
       <div class="form-group">
           <label>Image2</label>
           <input type="file" name="image[1]" class="form-control" required>
       </div>
       <div class="form-group">
            <label>Image3</label>
            <input type="file" name="image[2]" class="form-control" required>
       </div>
       <div class="form-group">
            <label>Image4</label>
            <input type="file" name="image[3]" class="form-control" required>
       </div>
       <center><input type='submit' name='submit' class='btn btn-success' value='Add Product'></center>
   </form>