printing lines contain double quote

Shell Programming and Scripting
1

Hello ,
I got html file , these file are normal html as we can see .

what i would like to do is in this html file , i want to print only string start with double quote and end with double quote by line by line.

<tr><td valign=top>25.</td><td><A href="../XMLL/CatalogL.htm">XML</a></td></tr>
<tr><td valign=top>26.</td><td><A href="../XMLNQ/CatogXML-LINQ.htm">

the out put should look like
"../XMLL/CatalogL.htm"
"../XMLNQ/CatogXML-LINQ.htm"

is that simple or a lot work to do or can anybody share code.

Many thanks

2 
nawk -F'"' '{for(i=2;i<=NF;i+=2) print FS $i FS}' myFile.html
3

In perl:

perl -ne '/".+"/; print $& . "\n"' file

Shorter perl:

perl -nle 'print /".+"/g' file
4

Using sed

sed -n 's/^.*href=\(".[^"]*"\).*$/\1/p' file
5

Another sed:

sed 's/.*=//;s/>.*//' file
6

awk 'BEGIN {FS="\""} {print "\""$2"\""}' myFile

7

Many thanks for all reply , it work great, but want to ask a question , if i want to modify

if string startwith <div class="codedom"> and endwith <code> , i mean

stringstart = <div class="codedom">
stringend = <code>

i want to print whole thing like

<div class="codedom"> ......................................
......................................................................
.......................................................................
......................................................................
.......................................................................
......................................................................
.............................................................<code>

how do i modify this because i don't see any start and end from above code

thanks

8 
perl -nl -0777 -e '$,="\n";print /<div class="codedom">.*?<code>/sg' file
9 
grep -o '"[^"]*"' infile