我有2个文件
$file = "songOne.wav";
$file = "song Two.wav";
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当我执行
system("command songOne.wav");
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它成功执行命令.
但是当我执行时
system("command song Two.wav");
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它给出了错误,因为它们之间有空格,因此命令将其视为2个参数,
output :
can't open file song
can't open file Two.wav
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如何使命令将它们视为一个文件?
谢谢
您可以传递system参数列表:
my @args = ("command", "song Two.wav");
system (@args);
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或者,如果您不想创建数组,请执行以下操作:
system ("command", "song Two.wav");
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