如何使用输入类型=文件上传json文件

时间:2019-09-13 13:50:00

标签: php jquery json

我想在名为process.php的php脚本上发送JSON文件,将文件保存在文件夹中,然后将此文件发送到python脚本。 目前的html代码是这样的:

<form method="post" enctype="multipart/form-data">
    <input type="file" name="files[]" multiple>
    <input type="submit" value="Upload File" name="submit">
</form>

JavaScript代码是这样的:

 const url = 'process.php';
 const form = document.querySelector('form');

 form.addEventListener('submit', e => {
      e.preventDefault();
      const files = document.querySelector('[type=file]').files;
      const formData = new FormData();

      for (let i = 0; i < files.length; i++) {
          let file = files[i];
          formData.append('files[]', file);
       }

      fetch(url, {
          method: 'POST',
          body: formData
      }).then(response => {
           console.log(response);
     });
 });

最后我的PHP是这样的:

if ($_SERVER['REQUEST_METHOD'] === 'POST') {
if (isset($_FILES['files'])) {
    $errors = [];
    $path = 'uploads/';
    $extensions = ['json'];

    $all_files = count($_FILES['files']['tmp_name']);

    for ($i = 0; $i < $all_files; $i++) {
        $file_name = $_FILES['files']['name'][$i];
        $file_tmp = $_FILES['files']['tmp_name'][$i];
        $file_type = $_FILES['files']['type'][$i];
        $file_size = $_FILES['files']['size'][$i];
        $file_ext = strtolower(end(explode('.', $_FILES['files']['name'][$i])));

        $file = $path . $file_name;

        if (!in_array($file_ext, $extensions)) {
            $errors[] = 'Extension not allowed: ' . $file_name . ' ' . $file_type;
        }

        if ($file_size > 2097152) {
            $errors[] = 'File size exceeds limit: ' . $file_name . ' ' . $file_type;
        }

        if (empty($errors)) {
            move_uploaded_file($file_tmp, $file);
        }

    }

    if ($errors) print_r($errors);
}

}

在PHP中,如果我尝试打印$ file_name,则文件名会正确打印,但是如果我尝试打印其他信息,则结果为空字符串,或者在$ file_size为0的情况下。 那么,怎么了?我是否必须在apache配置中设置一些内容以将文件上传到Web服务器上,或者我在代码中犯了一些错误?

2 个答案:

答案 0 :(得分:1)

您需要使用文件的临时名称,函数is_uploaded_file()不接受客户端计算机上发送的文件名,请尝试以下操作:

 if (is_uploaded_file ($_FILES['upload_cont_file']['tmp_name'] ))
    echo 'file uploaded ';
 else
 echo 'error';

答案 1 :(得分:1)

如果要上传并移至特定目录(例如图片文件夹)




if (move_uploaded_file ( $_FILE['upload_cont_file']['tmp_name'] ,'images/'.$_FILE['upload_cont_file']['name']))
{
    echo 'file uploaded ';

}
else
{
  echo 'couldnt file uploaded';

}

如果仅检查上传到临时文件夹的文件

 if (is_uploaded_file ($_FILES['upload_cont_file']['tmp_name'] ))
{
    echo 'file uploaded ';

}
else
{

 echo 'file couldn't upload';

}