Question

我正在尝试将多个图片上传到文件夹并且工作正常。当我试图将图像的名称存储在数据库中时，我遇到了问题。如果我上传只有一个正在工作。如果我上传了多个，我的数据库中只有一个名称。

上传的方式是什么，将带有图像标题的阵列上传到我的数据库？

以下是代码：

<div id="maindiv">
<div id="formdiv">
    <h4>Upload Image for </h4>
    <form enctype="multipart/form-data" action="" method="post">
        <p class="alert alert-info">Only JPEG,PNG,JPG Type Image Uploaded. Image Size Should Be Less Than 100KB.</p>
        <hr/>
        <div id="filediv"><input name="file[]" type="file" id="file"/></div><br/>
        <input type="button" id="add_more" class="upload" value="Add More Files"/>
        <input type="submit" value="Upload File" name="add_images" id="upload" class="upload"/>
    </form>
</div>

这是我的php：

if (isset($_POST['add_images'])) {
$j = 0; 
$target_path = "images/";
for ($i = 0; $i < count($_FILES['file']['name']); $i++) {
    $validextensions = array("jpeg", "jpg", "png");  
    $ext = explode('.', basename($_FILES['file']['name'][$i]));//store extensions in the variable
    $name_of_img = basename($_FILES['file']['name'][$i]);
    $target_path = $target_path . $name_of_img;
    $j = $j + 1;
      if (($_FILES["file"]["size"][$i] < 1000000) 
            && in_array($file_extension, $validextensions)) {
        if (move_uploaded_file($_FILES['file']['tmp_name'][$i], $target_path)) {
            echo $j. ').<span id="noerror">Image uploaded successfully!.</span><br/><br/>';
            print_r($name_of_img);
            $query = "UPDATE post SET ";   
            $query .= "images = '{$name_of_imgs}' ";
            $query .= "WHERE post_id = {$the_post_id}";    
            $update_cruise_img = sqlsrv_query($con, $query);  
        } else {
            echo $j. ').<span id="error">please try again!.</span><br/><br/>';
        }
    } else {//if file size and file type was incorrect.
        echo $j. ').<span id="error">Wrong file Size or Type</span><br/><br/>';
    }
}
 }

我尝试创建像$name_of_imgs = json_encode($_POST['file']);这样的数组但是没有工作

Answer 1

您需要更新post循环之外的for表。

if (isset($_POST['add_images'])) {
  $j = 0; 
  $name_of_imgs = [];

  for ($i = 0; $i < count($_FILES['file']['name']); $i++) {
      $validextensions = array("jpeg", "jpg", "png");  
      $ext = explode('.', basename($_FILES['file']['name'][$i]));//store extensions in the variable
      $name_of_img = basename($_FILES['file']['name'][$i]);

      // define target path here so that it won't concat the last value 
      $target_path = "images/" . $name_of_img;

      $j = $j + 1;
      if (($_FILES["file"]["size"][$i] < 1000000) && in_array($file_extension, $validextensions)) {
          if (move_uploaded_file($_FILES['file']['tmp_name'][$i], $target_path)) {
             echo $j. ').<span id="noerror">Image uploaded successfully!.</span><br/><br/>';

             $name_of_imgs[] = $name_of_img;
          } else {
              echo $j. ').<span id="error">please try again!.</span><br/><br/>';
          }
      } else {//if file size and file type was incorrect.
          echo $j. ').<span id="error">Wrong file Size or Type</span><br/><br/>';
      }
  }

  // update here
  $query = "UPDATE post SET ";   
  $query .= "images = '" . implode(',', $name_of_imgs) . "' ";
  $query .= "WHERE post_id = {$the_post_id}";    
  $update_cruise_img = sqlsrv_query($con, $query);  

}

您可以考虑为图片创建另一个表格。

Answer 2

您好，您已在上传文件中运行更新查询。它确保this_post_id单一的方式可以用单个文件名更新记录。

将图像上传为数组

2 个答案: