Linux上的无损jpeg批处理作物 [英] Lossless jpeg batch crop on Linux

问题描述

在Linux上，我需要在右侧无损地裁剪许多jpeg格式的图像20像素.

I need to crop a number of images in jpeg format by 20 pixels on the right side losslessly on Linux.

我检查了jpegtran，但是在裁切之前它需要文件大小(以像素为单位)，而且我不知道如何用该文件构建批处理文件.

I checked jpegtran, but it needs the file size in pixels before cropping, and I don't know how to build a batch file with that.

如何以编程方式从图像的右侧无损地裁剪20个像素?

How can I losslessly crop 20 pixels from the right side of images programmatically?

推荐答案

我的shell脚本有点生锈，因此在尝试使用此脚本之前，请先备份图像.

My shell scripting is a little rusty so please make a backup of your images before trying this script.

#!/bin/bash
FILES=/path/to/*.jpg

for f in $FILES
do
    identify $f | awk '{ split($3, f, "x"); f[1] -= 20; cl = sprintf("jpegtran -crop %dx%d+0+0 %s > new_%s", f[1], f[2], $1, $1); system(cl); }'
done

注意事项:

• 将路径调整为正确的值
• 您需要* .jpeg吗?
• identify是ImageMagick命令
• awk将从identify抓取像素尺寸，以用作jpegtran裁剪图像的参数(宽度减小20px)
• 新图像另存为new_[old_name].jpg
• jpegtran可能会调整裁切区域，以便可以无损执行.检查生成的图像尺寸正确且不大.

• Adjust the path to the correct value
• Do you need *.jpeg?
• identify is an ImageMagick command
• awk will grab the pixel dimensions from identify to use as a parameter (with the width reduced by 20px) for jpegtran to crop the image
• The new image is saved as new_[old_name].jpg
• jpegtran might adjust the cropping region so that it can perform losslessly. Check that the resulting images are the correct size and not slightly larger.

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