Question

目前：我的代码保存了csv文件中给出的URL中的图像。

目标：根据同一csv文件中的值重命名已保存的图像，格式为：csv位置3,4,5,6,7,8
（例如100A1_https：//www.instagram.com/p/BBzUXUFLrGH/_48.8486557_2.3481125）。

我试图将this stackoverflow示例合并到我的代码中。但无法，因为我绝望地失去了。

＆＃13;

<?php

$destination = 'images/';
$dom = new DOMDocument;

$dir = "dir to photos";

$row = 0;

if (($handle = fopen("data/testcsvfile.csv", "r")) !== FALSE) {
  while (($data = fgetcsv($handle, 100, ",")) !== FALSE /*&& $row < 1*/) {
    $row++;

    // number of fields in each row
    $num = count($data);

    // get url from position in csv
    $url = $data[6];
    echo $row . " " . $url. PHP_EOL;

    if( strlen($url) > 0 ){
        // if we have a url, get the contents
        $page = file_get_contents($url);
        //echo $page;

        $dom->loadHTML($page);

        // find url using meta tag
        $my_tags = $dom->getElementsByTagName('meta');

        // find which one is the image and grab and save
        foreach ($my_tags as $tag) {
            if($tag->getAttribute("property") == "og:image"){
                $image_url = $tag->getAttribute('content');
                echo  $image_url . PHP_EOL;
                $the_image = file_get_contents($image_url);

                // rename file based on csv
                // $newname = "$dir"."$names[3] $names[4] $names[5]".";
                rename($newname);


                file_put_contents( $destination . "img_" . $row . ".jpg", $the_image);

            }

        }

    }

  }
  fclose($handle);
}

?>

＆＃13;

// test csv file

Mon,1,0,100,A,1,https://www.instagram.com/p/BBzUXUFLrGH/,48.8486557,2.3481125
Mon,1,0,100,A,1,https://www.instagram.com/p/BAe0tGULrC1/,48.85272468,2.347259349
Mon,1,0,100,A,1,https://www.instagram.com/p/_zik5YLrMf/,48.85356691,2.345645975

＆＃13;

Answer 1

首先，您需要了解使用URL作为文件名是一件非常奇怪/糟糕的事情。网址使用特殊字符，例如＆＃39; /＆＃39;这是某些操作系统中的目录分隔符。如果你真的需要这样做，首先需要在文件名中转义该字符（或者更好地用非特殊字符替换它们，如下划线）：

$newname = str_replace('/', '\/', $newname);

脚本的更新版本，可以安全地替换带下划线的斜杠：

<?php

$destination = 'images';
$dom = new DOMDocument;

if (($handle = fopen("data/testcsvfile.csv", "r")) !== FALSE) {
  while (($data = fgetcsv($handle, 100, ",")) !== FALSE) {
    $num = count($data);

    // get url from position in csv
    $url = $data[6];
    echo $row . " " . $url. PHP_EOL;

    if( strlen($url) > 0 ){
        // if we have a url, get the contents
        $page = file_get_contents($url);
        //echo $page;

        $dom->loadHTML($page);

        // find url using meta tag
        $my_tags = $dom->getElementsByTagName('meta');

        // find which one is the image and grab and save
        foreach ($my_tags as $tag) {
            if($tag->getAttribute("property") == "og:image"){
                $image_url = $tag->getAttribute('content');
                echo  $image_url . PHP_EOL;
                $the_image = file_get_contents($image_url);

                $destinationName = str_replace('/', '_', sprintf(
                  "%s%s%s_%s_%s_%s",
                  $data[3],
                  $data[4],
                  $data[5],
                  $data[6],
                  $data[7],
                  $data[8]
                ));

                $destinationPath = sprintf(
                  "%s/%s",
                  $destination,
                  $destinationName
                );


                file_put_contents($destinationPath, $the_image);

            }

        }

    }

  }
  fclose($handle);
}

?>

php重命名基于csv值从{url}保存的图像

1 个答案: