I'm trying to replace first, second and third %f with a b and c respectively using sed and awk.
我试图用sed和awk分别用b和c替换第一,第二和第三%f。
(51.7296373326*%f)+(41.7319764456*%f)+(-193.993966414*%f)
This is what i've tried
这就是我尝试过的
sed "s|(51.7296373326*%f)|(51.7296373326*a)|g" dump1.txt
The values are not constant..
值不是常数..
1 个解决方案
#1
1
I guess you meant sed or awk.
我想你的意思是sed或awk。
awk 'BEGIN{a[++i]="a";a[++i]="b";a[++i]="c"}
{for(x=1;x<=i;x++)sub(/%f/,a[x])}7' file
if the replacement list is big, like a-z, you can put them in a file, first load it into array a instead of loading them in BEGIN block.
如果替换列表很大,比如a-z,你可以将它们放在一个文件中,首先将它加载到数组a中,而不是将它们加载到BEGIN块中。
Note, the codes could be optimized by checking the return value of sub, if 0, break.
注意,可以通过检查sub的返回值来优化代码,如果为0,则为break。
#1
1
I guess you meant sed or awk.
我想你的意思是sed或awk。
awk 'BEGIN{a[++i]="a";a[++i]="b";a[++i]="c"}
{for(x=1;x<=i;x++)sub(/%f/,a[x])}7' file
if the replacement list is big, like a-z, you can put them in a file, first load it into array a instead of loading them in BEGIN block.
如果替换列表很大,比如a-z,你可以将它们放在一个文件中,首先将它加载到数组a中,而不是将它们加载到BEGIN块中。
Note, the codes could be optimized by checking the return value of sub, if 0, break.
注意,可以通过检查sub的返回值来优化代码,如果为0,则为break。