Question

我遇到这个问题。认为它会更容易，但结果却令人沮丧。我要做的就是有一个文本字段，我可以在其中键入新目录的名称，检查该目录是否存在，如果不存在则创建它。我发现大约有50个人的代码几乎完全相同，所以我认为我的方法是正确的，但我仍然按照“if”语句继续使用Directory。

最终我想把它绑定到我的文件上传脚本中。

这是insert.php

<form action="" method="post" enctype="multipart/form-data" name="form1" id="form1">
  <p>
    <label for="directory">Directory:</label>
    <input value="<?php if ($_POST && $errors) {
  echo htmlentities($_POST['directory'], ENT_COMPAT, 'UTF-8');
}?>" type="text" name="directory" id="directory" />
  </p>
  <p>
    <input type="submit" name="insert" id="insert" value="insert" />
  </p>
</form>

这是post.php

try {
if (isset($_POST['insert'])) {
    $directory = $_POST['directory'];
    $photo_destination = 'image_upload/';
    $path = $photo_destination;
    $new_path = $path . $directory;
    $mode = 0755;
    if(!is_dir($new_path)) {
        echo "The Directory {$new_path} exists";
        } else {
            mkdir($new_path , 0777);
            echo "The Directory {$new_path} was created";
            }
        }
}

Answer 1

改变这个：

if(!is_dir($new_path)) {
    echo "The Directory {$new_path} exists";
    }

到此：

if(is_dir($new_path)) {
    echo "The Directory {$new_path} exists";
    }

试着告诉我结果：）

Answer 2

您可以使用is_dir而不是在if块中使用file_exists。因为file_exists是检查文件是否存在的函数。同样，您也可以参考http://php.net/manual/en/function.file-exists.php

Answer 3

尝试

if( is_dir( $new_path ) ) {
    echo "The Directory {$new_path} exists";
}

Answer 4

让某人想要创建一个子文件夹，
在 /uploads 文件夹下使用年份名称，然后他/她想要创建 /uploads/<year_name_folder>/ 下的另一个子文件夹，
名为 project_number 。

$yearfolder = date('y');
if (!file_exists('uploads/'.$yearfolder)) {
        mkdir("uploads/".$yearfolder);
    }
*// Folder named by year has been created.*

$project_number = Any Unique field ,Come from database or anywhere !
$target_directory = mkdir("uploads/".$yearfolder."/".$project_number);

*// project number wise folder also created.
// If project number is not unique do check like year folder. I think every project number is unique.* 

$target_dir = "uploads/$yearfolder/$project_number/";
*//my target dir has created where my document or pic whatever will be uploaded.*

Now Upload ! Woo 
$target_file = $target_dir.($_FILES["file"]["name"]);

Answer 5

<?php
$dirname = "small";
$filename = "upload/".$dirname."/";

if (!is_dir($filename )) {
    mkdir("upload/" . $dirname, 0777, true);
    echo "The directory $dirname was successfully created.";
    exit;
} else {
    echo "The directory $dirname exists.";
}
 ?>

检查目录是否存在，make目录

5 个答案: