使用 PHP shell_exec() 调用 sh 脚本时未设置变量

发布于 2024-10-16 15:24:53 字数 1288 浏览 10 评论 0原文

我目前有一个 sh 脚本：

    location=$1
    imageNumber=$2
    keyword=$3
    page=$4
    imagesInPage=$5
    imagesToGet=$6
    imageUrls=`curl -s "http://www.flickr.com/services/rest/?method=flickr.photos.search&api_key=[MYAPIKEY]&text=$keyword&per_page=$imagesInPage&media=photos&license=1,2,4,5,7&page=$page" | sed '1,3d' | sed 'N;$!P;$!D;$d' | ./xml_to_urls 2> /dev/nul`

    currentImage=0

    for url in $imageUrls
    do
            if [ $currentImage -ne $imagesToGet ]
            then
                    curl -s "$url" > "$location/$imageNumber.jpg"
            else
                    break
            fi
            currentImage=`expr $currentImage + 1`
            imageNumber=`expr $imageNumber + 1`
    done

...它基本上请求 Flickr API 提供与参数对应的图像列表。然后，我循环遍历图像 URL，将每个图像保存在指定目录中。

调用它的 PHP 脚本相当长，但这是调用它的行：

shell_exec("./get_images \"" . $dir . "\" " . $currentImageNumber . " \"" . $keyword . "\" " . $page . " 500 500");

当我在命令行上运行命令（例如：./get_images "temp/0.75456300 1297381201" 0 "moose" 1 500 500）时，没有问题发生。但是，当由 PHP 脚本运行时，变量 $imageUrls 永远不会设置。我可以回显我设置 imageUrls 的命令，并在 PHP 创建的页面上显示 URL，但据我所知，变量 self 是空白的。

我能得到的任何帮助都会很棒！请告诉我是否应该添加其他内容或者我的问题是否具有误导性，这是我的第一篇文章：D！谢谢！

原文

I currently have a sh script:

    location=$1
    imageNumber=$2
    keyword=$3
    page=$4
    imagesInPage=$5
    imagesToGet=$6
    imageUrls=`curl -s "http://www.flickr.com/services/rest/?method=flickr.photos.search&api_key=[MYAPIKEY]&text=$keyword&per_page=$imagesInPage&media=photos&license=1,2,4,5,7&page=$page" | sed '1,3d' | sed 'N;$!P;$!D;$d' | ./xml_to_urls 2> /dev/nul`

    currentImage=0

    for url in $imageUrls
    do
            if [ $currentImage -ne $imagesToGet ]
            then
                    curl -s "$url" > "$location/$imageNumber.jpg"
            else
                    break
            fi
            currentImage=`expr $currentImage + 1`
            imageNumber=`expr $imageNumber + 1`
    done

... which basically requests the Flickr API for a list of images corresponding to the arguments. Then, I loop through the image URLs, saving each one in a specified directory.

The PHP script that calls this is quite long, but here's the line that calls it:

shell_exec("./get_images \"" . $dir . "\" " . $currentImageNumber . " \"" . $keyword . "\" " . $page . " 500 500");

When I run the command (ex: ./get_images "temp/0.75456300 1297381201" 0 "moose" 1 500 500) on the command line, no problems occur. However, when run by the PHP script, the variable $imageUrls is never set. I can echo out the command I am setting imageUrls to, and that displays the URLs on the page PHP creates, but as far as I can tell, the variable its self is blank.

Any help I could get on this would be great! Please tell me if I should add anything else or if my question is misleading, this is my first post :D! Thanks!

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╭ゆ眷念 2024-10-23 15:24:53

尝试在 imageList 的变量赋值的输出周围添加双引号，

imageUrls="`curl -s "http://www.flickr.com/services/rest/....."

我希望这会有所帮助。

PS，由于您似乎是新用户，如果您得到的答案对您有帮助，请记住将其标记为已接受，或给它一个+（或-）作为有用的答案

try adding dbl-quotes around the output of your variable assignment for imageList

imageUrls="`curl -s "http://www.flickr.com/services/rest/....."

I hope this helps.

P.S. as you appear to be a new user, if you get an answer that helps you please remember to mark it as accepted, or give it a + (or -) as a useful answer

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