// This works
convert ${path}${dst} -crop ${crop} ${path}${dst}
// but when changed to this, it fails
convert ${path}${src} -trim ${path}${dst}
convert ${path}${dst} -crop ${crop} ${path}"pdf_"${dst}
我在做什么错?ImageMagick bash脚本问题
// This works
convert ${path}${dst} -crop ${crop} ${path}${dst}
// but when changed to this, it fails
convert ${path}${src} -trim ${path}${dst}
convert ${path}${dst} -crop ${crop} ${path}"pdf_"${dst}
我在做什么错?ImageMagick bash脚本问题
是否改变这一点:
convert ${path}${dst} -crop ${crop} ${path}"pdf_"${dst}
要这样:
convert ${path}${dst} -crop ${crop} "${path}pdf_${dst}"
帮助?
在Bash中,除非'$ path'或'$ dst'中有空格,否则不应有任何区别。 – 2010-05-14 13:17:00
适合我。
$ path=./
$ src=test.jpg
$ dst=test2.jpg
$ crop=100x100+10+10
$ convert ${path}${src} -trim ${path}${dst}
$ convert ${path}${dst} -crop ${crop} ${path}pdf_${dst}
$ identify pdf_test2.jpg
pdf_test2.jpg JPEG 100x100 100x100+0+0 DirectClass 8-bit 3.30078kb
这有点神秘 - 它是如何失败的? – 2010-05-14 12:58:46