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how to make echo get_i18n_link($slug) work ?

To have a multilanguage menu in a site is by using:

<?php get_i18n_navigation(); ?>

However, I need to make a customized menu.

One way to make your own customized menu is by using:


<li><?php get_i18n_link(index); ?></li>
<li><?php get_i18n_link(about); ?></li>
<li><?php get_i18n_link(contact); ?></li>

But in my case what I really need is this:

echo '<ul>';
echo '<li>' . get_i18n_link(about) . '</li>';
echo '</ul>';

The problem here is that the part: echo get_i18n_link(about) does not work correctly
and outputs a 1

Is there a parameter to make echo get_i18n_link(about) work correctly

like echo get_i18n_link(about, false) or something like that ?
Quotes are missing in your examples. It should be like this:

<?php get_i18n_link('about'); ?>
Thanks Carlos for your reply,

But no change,
somehow this still adds a 1 directly after the link
echo '<li>' . get_i18n_link('about') . '</li>';

On the front it looks like this

I think with echo the 1 will always be added

Instead, will try to use <?php get_i18n_link('about'); ?>
Do not use echo, the return value is already echoes.

Instead, just:

<li><?php get_i18n_link('about') ?></li>

Thanks for the replies. Yes in this case I have to change
my design using <?php ... instead of echo ...

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