Question

我试图检查ftp服务器上是否已经存在目录，如果它存在，那么它应该只创建并保存一个json文件到这个目录和return true，如果它不存在，然后它应该首先创建目录并将json文件保存到目录，并return true，否则它应该return false。

我的代码目前如下所示：

<?php

    // Function to create the outfit xml file
    function create_outfit_json(){
        if (!file_exists('../user/' . $this->Username)) {
            mkdir('../user/' . $this->username, 0777, true);

            $json['outfits'] = [];
            $json['outfits']['0'] = [
                'outfit'   => [
                    'url' => 'placeholder',
                    'default' => 1,
                    'name' => 'New outfit',
                    'c' => '#bb9977',
                    'mood' => 3,
                    'species' => 'male'
                ]
            ];

            $fp = fopen('../user/' . $this->Username . '/outfits.json', 'w');
            fwrite($fp, json_encode($json));
            fclose($fp);

            return true;
        }else if(file_exists('../user/' . $this->Username)){
            $json['outfits'] = [];
            $json['outfits']['0'] = [
                'outfit'   => [
                    'url' => 'placeholder',
                    'default' => 1,
                    'name' => 'New outfit',
                    'c' => '#bb9977',
                    'mood' => 3,
                    'species' => 'male'
                ]
            ];

            $fp = fopen('../user/' . $this->Username . '/outfits.json', 'w');
            fwrite($fp, json_encode($json));
            fclose($fp);

            return true;
        }else{
            return false;
        }
    }

?>

有没有办法让这段代码看起来更干净简洁？

Answer 1

<?php

    // Function to create the outfit xml file
    function create_outfit_json(){
        if (!file_exists('../user/' . $this->Username)) {
            mkdir('../user/' . $this->Username, 0777, true);
        }

            $json['outfits']['0'] = [
                'outfit'   => [
                    'url' => 'placeholder',
                    'default' => 1,
                    'name' => 'New outfit',
                    'c' => '#bb9977',
                    'mood' => 3,
                    'species' => 'male'
                ]
            ];

            $fp = fopen('../user/' . $this->Username . '/outfits.json', 'w');
            fwrite($fp, json_encode($json));
            fclose($fp);
        }
    }

?>

您不需要冗余代码 - 只需将mkdir包装在一个if。

中

Answer 2

阅读评论，总是用重用能力编写代码，

    // Seperate all data and saving
    function create_outfit_json(){
        $json['outfits'] = [];
        $json['outfits']['0'] = [
            'outfit'   => [
                'url' => 'placeholder',
                'default' => 1,
                'name' => 'New outfit',
                'c' => '#bb9977',
                'mood' => 3,
                'species' => 'male'
            ]
        ];
        return $this->saveToDirectory(json_encode($json), 'outfits.json');
    }

    // passing data and filename will give you flexibility 
    function saveToDirectory($data, $filename){
        if (!file_exists('../user/' . $this->Username)) {
            mkdir('../user/' . $this->username, 0777, true);
        }
        // rather than opening connection let php do it for you
        return file_put_contents('../user/'.$this->Username.'/'.$filename, $data);
    }

将json保存到ftp目录

2 个答案: