写一个bash脚本，统计一个文本文件nowcoder.txt中每一行出现的1~5数字的个数，并且计算一下整个文档中一共出现了几个1~5数字的总数。
示例:
假设 nowcoder.txt 内容如下：
a12b8
10ccc
2521abc
9asf
你的脚本应当输出：
line1 number: 2
line2 number: 1
line3 number: 4
line4 number: 0
sum is 7
说明:
不要担心你输出的空格以及换行的问题
[编程题]打印每一行出现的数字个数
coutline=0
count=0
sum=0
while read line
do
for((i=0;i<${#line};i++));do
if [[ ${line:$i:1} =~ [1-5] ]];then
let count=count+1
fi
done
let coutline=coutline+1
echo "line${coutline} number: ${count}"
let sum=sum+count
count=0
done<./nowcoder.txt
echo "sum is ${sum}"
count=0
sum=0
while read line
do
for((i=0;i<${#line};i++));do
if [[ ${line:$i:1} =~ [1-5] ]];then
let count=count+1
fi
done
let coutline=coutline+1
echo "line${coutline} number: ${count}"
let sum=sum+count
count=0
done<./nowcoder.txt
echo "sum is ${sum}"
发表于 2021-05-08 13:44:09
回复(1)
cnt=1
sum=0
while read line
do
r=`echo $line | grep -oE "[12345]" | wc -l`
echo "line${cnt} number: ${r}"
let "cnt++"
let "sum+=$r"
done
echo "sum is ${sum}" < nowcoder.txt awk '{
gsub(/[^1-5]/,"",$0);
print "line"NR" number: "length($0);
sum+=length($0);
} END{ print "sum is "sum }'
编辑于 2021-01-12 19:46:43
回复(1)
idx=1
sum=0
while read line
do
num=$(echo $line |grep -o [1-5]|wc -l)
echo "line$idx number:" $num
let idx=$idx+1 # 这行等同于 $idx++
let sum=$sum+$num
done
echo "sum is"$sum
### 对上面的命令解读：
## grep -o 的用法匹配满足条件的字符，举例如下，你一看就懂
# echo a123kk |grep -o [1-9]
1
2
3
# echo a123kk |grep -o [a-z]
a
k
k
# echo a123kk |grep -o a1
a1
## let 是shell命令运算符，用于整数运算，举例如下
# let a=1+3;echo $a
4
发表于 2022-07-17 22:37:01
回复(0)
awk -F '' '{count=0;{for(i=1;i<=NF;i++)if($i~/[1-5]/)count++;sum+=count}print "line"NR,"number:",count}END{print "sum is ",sum}'
发表于 2022-04-20 22:40:14
回复(0)
awk -F "[1,2,3,4,5]" 'BEGIN{sum=0}{print "line"NR" number:"(NF-1);sum+=(NF-1)}END{print "sum is "sum}' nowcoder.txt
发表于 2021-09-26 16:20:56
回复(5)
#!/bin/bash
sum=0
i=1
while read line;do
#echo $line
a=$(echo "$line" | grep -o -E "[1-5]"| wc -l)
echo line${i} number:$a
i=$[i+1]
sum=$[$a+$sum]
done < nowcoder.txt
echo "sum is $sum"
sum=0
i=1
while read line;do
#echo $line
a=$(echo "$line" | grep -o -E "[1-5]"| wc -l)
echo line${i} number:$a
i=$[i+1]
sum=$[$a+$sum]
done < nowcoder.txt
echo "sum is $sum"
发表于 2021-03-03 22:41:45
回复(0)
awk -F '' '{
cnt[NR]=0 #注意cnt[NR]初始化为0；否则不会记录没有1~5数字的行
for(i=1;i<=NF;i++){
if($i~/[1-5]/){ #注意这里需要加上//表示正则
sum++
cnt[NR]++ #注意这里使用了行号NR
}
}
}END{
for(j in cnt){
printf("line%d number: %d\n",j,cnt[j]) #可以使用print来打印更方便简洁
}
printf("sum is %d",sum)
}' nowcoder.txt
发表于 2024-10-26 21:18:48
回复(0)
awk '{gsub("[^1-5]", ""); print "line" NR " number: " length; sum+=length} END {print "sum is " sum}' nowcoder.txt
# print "line" NR " number: " length; 输出打印 此时 [每一行长度==每一行1-5的字符数]
# sum+=length 做一个变量累加每一行 1-5 的字符数量
# END {print "sum is " sum} 最后一行打印总数
发表于 2024-03-25 01:04:59
回复(0)
awk '{count=0; for(j=1;j<=NF;j++){for(i=1;i<=length($j);i++){char=substr($j,i,1); if(char ~ /^[1-5]$/){count++}}}printf("line%d number:%d\n",NR,count);total+=count} END{printf("sum is %d",total)}' $1
发表于 2025-01-17 11:34:06
回复(0)
sum=0
i=0
while read line; do
let i+=1
cnt=$(echo $line | sed 's/[1-5]/\n/g' | wc -l)
let cnt-=1
let sum+=$cnt
echo line$i number: $cnt
done
echo sum is $sum
发表于 2024-12-07 16:34:57
回复(0)
#!/bin/bash
line_num=0
sum=0
line_number=1
while IFS= read -r line;do
line_count=0
for (( i=0; i<${#line}; i++ )); do
char="${line:$i:1}"
if [[ $char =~ [1-5] ]]; then
((line_count++))
fi
done
echo "line${line_number} number: ${line_count}"
sum=$((sum + line_count))
((line_number++))
done <nowcoder.txt
echo "sum is ${sum}"
发表于 2024-12-05 15:35:33
回复(0)
gawk 'BEGIN {
sum = 0
}
{
count_line = 0
for( i=1; i<=length($0); i++ ) { # 遍历每行的每个字符
char = substr($0, i, 1) # 逐个截取字符
if( char ~ /[1-5]/ ){
count_line++
}
}
print "line" NR " number: " count_line
sum += count_line
}
END {
print "sum is " sum
}' "$1"
sum = 0
}
{
count_line = 0
for( i=1; i<=length($0); i++ ) { # 遍历每行的每个字符
char = substr($0, i, 1) # 逐个截取字符
if( char ~ /[1-5]/ ){
count_line++
}
}
print "line" NR " number: " count_line
sum += count_line
}
END {
print "sum is " sum
}' "$1"
发表于 2024-08-29 14:01:09
回复(0)
file="nowcoder.txt"
cat $file | sed 's/[^1-5]//g' | awk -v sum=0 '{
sum=sum + length($1)
print "line"NR " number:"length($1)
}
END{
print "sum is "sum
}'
发表于 2024-07-03 14:47:08
回复(0)
#!/bin/bash
line_number=0
while IFS= read -r line; do
((line_number++))
count=$(echo "$line" | tr -dc "1-5" | wc -m)
((sum += count))
echo "line${line_number} number: ${count}"
done
echo "sum is ${sum:-0}"
发表于 2024-07-02 15:38:47
回复(0)
#!/bin/bash
col_num=`cat nowcoder.txt| wc -l`
sum=0
for i in `seq 1 $col_num`;do
n=`cat nowcoder.txt|head -n${i}|tail -n1|grep -o '[1-5]'|wc -l`
sum=$((sum+n))
printf "line%d number: %d\n" $i $n
done
printf "sum is %d" $sum
编辑于 2024-02-16 19:35:14
回复(0)
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