How to rename files into a list. CSV in your output with PHP and CURL?

Question

How are you all? This is my first post because I am a beginner with PHP and CURL and I needed to create a script to download several images in repositories through a file .csv. It reads the url where the file is, and saves in a pre-established directory, I use the "basename" function to recover the name of the image that is in the URL. Until then the script does everything smoothly saves the right images everything correctly, but I can’t change this output to rename the files as needed. I wonder if I can use the same. csv add one more column, or insert in the script the specific names for each image, that is, I already know what this name would be and the order to rename would be the one in the . CSV, look at the script:

<?php

$csvFile = file('imagensuno.csv'); // Aqui ele pega o arquivo .csv com as URL das imagens 

foreach ($csvFile as $line) {

  $url = str_getcsv($line);

  $ch = curl_init($url[0]);

  $name = basename($url[0]); //Aqui ele usa o nome da imagens na URL para salvar ela com o nome que ela tem na URL


if (!file_exists('G:\cca_imagens/' . $name)) {
    $fp = fopen('G:\cca_imagens/' . $name, 'wb');

  }

  curl_setopt($ch, CURLOPT_FILE, $fp);
  curl_setopt($ch, CURLOPT_HEADER, 0);
  curl_exec($ch);
  curl_close($ch);
  fclose($fp);

  //echo str_getcsv($line);

}

the function str_getcsv, I will scroll down the line of the csv file "imagensuno.csv" which is located in the same directory as the script.

to file_exists which will check whether the image has been saved before, very useful if script execution is stopped.

Also note the use of the function basename to recover the name of the image that is in the URL.

And this is where my doubt is, and I can’t remember if I can or can make you save the name I need that would be:

Ex of file . csv

Coluna 1                                        Coluna 2
https://enderecodaimagens.com.br/img/foto.png,  foto.png => sku435.png, 
https://enderecodaimagens.com.br/img/foto2.png, foto2.png => sku444.png, 
https://enderecodaimagens.com.br/img/foto3.png  foto.png => sku865.png

How would they all be in the same. csv I want to rename these not by URL name but by Column 2 action, but I don’t know if it’s possible, I can’t think of how to do this output.

People need this Help and if someone wants to use the script this the layout and logico want to improve it also think very good ok?

abs

Answer 1

Renamed file: '.$Rename.'
-
'; //Tests if the file exists if (!file_exists('C: wamp www projects dw photo cca/' . $name)) { //Test to rename the file if(Rename("C: wamp www projects dw photo cca $name", "C: wamp www projects dw photo cca $Rename"){ echo 'Original file: '.$name. '
Renamed file: '.$Rename.'
Status: Successfully renamed, line '.$Count.'
'; }Else{ echo 'Original file: '.$name. '
Renamed file: '.$Rename.'
Status: Error renaming, line '.$Count.'
'; } } $Count +=1; } ?>

This above edited code!

Answer 2

Assuming you have one .csv in the following format:

http://127.0.0.1/stack/fopencsv/imagem1.jpg, novaimagem1.jpg
http://127.0.0.1/stack/fopencsv/imagem2.jpg, novaimagem2.jpg
http://127.0.0.1/stack/fopencsv/imagem3.jpg, novaimagem3.jpg
http://127.0.0.1/stack/fopencsv/imagem4.jpg, novaimagem4.jpg
http://127.0.0.1/stack/fopencsv/imagem5.jpg, novaimagem5.jpg

You can rename using the function name php, and fgescsv with the following script:

<?php

//Abrindo o manipulador
$handle = fopen('csv.csv', 'r');

$count = 1;
$file = fgetcsv($handle);
//Abrindo o arquivo e lendo as linhas
while ($row = fgetcsv($handle)) {
    //removendo a url (url não é aceita no 'rename')
    $name = str_replace('http://127.0.0.1/stack/fopencsv/img/','',$row[0]);
    $rename = $row[1];

    echo 'Arquivo original: ' . $name . '<br>Arquivo renomeado: ' . $rename . '<br>-<br>'.PHP_EOL;
    //Testa se o arquivo existe
    //use __dir__ , ou outra coisa, o importante é apontar o local correto do arquivo, dentro do mesmo servidor.
    if (!file_exists(__DIR__ . '\\' . $name)) {
        //Teste para renomear o arquivo
        if (rename(__DIR__ . '\img\\' . $name, __DIR__ . '\img\\' . $rename)) {
            echo 'Arquivo original: ' . $name . '<br>Arquivo renomeado: ' . $rename . '<br>Status: Renomeado com sucesso, linha ' . $count . '<br>';
        } else {
            echo 'Arquivo original: ' . $name . '<br>Arquivo renomeado: ' . $rename . '<br>Status: Erro ao renomear, linha ' . $count . '<br>';
        }
    }
    $count += 1;
}

Rules:

1 - Files must be on the same server from where the script is executed.

2 - Replace the url where the file is by the local path.

3 - My files are in the folder: Script: c: xampp htdocs stack fgetscsv Images: c: xampp htdocs stack fgetscsv img

4 - Give permission to the file (probably already ok)