Cerrado

DHTML Question

Question: How do you get the links (anchor tag) behind the text selected by a user - in IE?

To be clear, I have no control over the actual webpage, the user (the person making the selection) can be on any website.

For example: Suppose the user has selected "Arrays" on <[url removed, login to view]>.

How do you get just the link from the anchor tag without manually parsing it?

For Example:

if .[url removed, login to view](0).htmltext gives the following result:

<A href="index_software_engineering.html#application_development">Application Development</A>&nbsp;

<A href="index_software_engineering.html#arrays">Arrays</A>&nbsp;

How do i get just: (the destination URL)

<[url removed, login to view]>

<[url removed, login to view]>

The solution should look like this:

.[url removed, login to view]("A");

but this is not a valid expression.

Thank you.

## Deliverables

Just the answer

## Platform

IE

Habilidades: Ingeniería, Javascript, MySQL, PHP, Arquitectura de software, Verificación de software, Visual Basic, Web Hosting, Gestión de páginas web, Verificación de páginas web

Ver más: any question, question, Question Answer, application parsing text, php parsing text, php html parsing, helper php, html parsing php, question text, javascript getelementsbytagname example, application question, htmltext, parsing javascript html, question html, php text parsing, php question answer website, website parsing php, question answer development, development question answer, question answer application, getelementsbytagname, question answer website php, question answer javascript php, php parsing webpage, parsing html document

Información del empleador:
( 6 comentarios ) United States

ID de proyecto: #3476740

3 los freelancers están ofertando un promedio de $11 para este trabajo.

nitinbatravw

See private message.

$17 USD en 1 día
(4 comentarios)
2.6
varisoftpakistan

See private message.

$11.05 USD en 1 día
(9 comentarios)
2.5
sssss123vw

See private message.

$4.25 USD en 1 día
(2 comentarios)
0.0