Question

sed/awk截取字符串

<table><tr><td>Password:</td><td>123456</td></tr><tr><td>Hash:</td><td>44efce164ab921caaad3b435b5140</td></tr></table> </div>
，想截取44efce164ab921caaad3b435b5140，求命令

Answer 1

[flycat]$cat file
<table><tr><td>Password:</td><td>123456</td></tr><tr><td>Hash:</td><td>44efce164ab921caaad3b435b5140</td></tr></table></div>

[flycat]$cat file | sed 's/^<table>.*<td>//' | awk -F"<" '{print $1}'
44efce164ab921caaad3b435b5140

附送你2个：
cat file | sed 's/^<table>.*<td>//' | sed -r 's/<\/td.*$//'
cat file | awk -F"</td>" '{print $4}'|cut -c 5-

Answer 2

我给你个思路吧，试一下下边的shell脚本

#!/bin/bash

echo "<table><tr><td>Password:</td><td>123456</td></tr><tr><td>Hash:</td><td>44efce164ab921caaad3b435b5140</td></tr></table>" | awk '
BEGIN{
FS="</*td>"
RS="</*tr>"
}

{
print "1 - " $1
print "2 - " $2
print "3 - " $3
print "4 - " $4
}'

追问

$3结果是这个44efce164ab921caaad3b435b5140，怎么去掉后面的...

追答

不可能，我这里执行后是这个结果：
1 -
2 -
3 -
4 -
1 -
2 - Password:
3 -
4 - 123456
1 -
2 -
3 -
4 -
1 -
2 - Hash:
3 -
4 - 44efce164ab921caaad3b435b5140
1 -

2 -
3 -
4 -