Each tag page contains a list of pages that contain that tag:
\n{{#- _includes/layouts/tagPage.vto -#}}\n\n<section>\n {{ for page of search.pages(tag, \"date=desc\") }}\n {{#- page stuff here -#}}\n <hr class=\"sep\">\n {{ /for }}\n</section>Is it possible to give each tag page pagination?
","upvoteCount":1,"answerCount":1,"acceptedAnswer":{"@type":"Answer","text":"You can use a different generator to paginate the first pages using renderOrder to make sure the render order is correct.
","upvoteCount":2,"url":"https://github.com/lumeland/lume/discussions/648#discussioncomment-10323248"}}}Generator inside a generated page? #648
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I used a generator to make a page for each tag in my blog: // tags/index.page.js
export const layout = "layouts/tagList.vto";
export const title = "Tags";
export const category = "list";
export default function* ({ search }) {
// config
const baseURL = "/tags"
// set args
const tags = search.values("tags");
// yield the index
yield {
url: `${baseURL}/`,
layout: layout,
title: title,
category: category,
tagArr: tags
}
for (const tag of tags) {
const page = {
url: `${baseURL}/${tag}/`,
title: `#${tag}`,
layout: "layouts/tagPage.vto",
tag: tag
};
yield page;
}
}Each tag page contains a list of pages that contain that tag: {{#- _includes/layouts/tagPage.vto -#}}
<section>
{{ for page of search.pages(tag, "date=desc") }}
{{#- page stuff here -#}}
<hr class="sep">
{{ /for }}
</section>Is it possible to give each tag page pagination? |
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Replies: 1 comment 3 replies
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You can use a different generator to paginate the first pages using renderOrder to make sure the render order is correct. |
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thanks! I was able to make tag pages (e.g. |
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After review your code again, I think this can be simplified. You don't need to create two generators. // tags/index.page.js
export const layout = "layouts/tagList.vto";
export const title = "Tags";
export const category = "list";
export default function* ({ search, paginate }) {
// config
const baseURL = "/tags"
// set args
const tags = search.values("tags");
// yield the index
yield {
url: `${baseURL}/`,
layout: layout,
title: title,
category: category,
tagArr: tags
}
for (const tag of tags) {
const pages = search.pages(tag, "date=desc"); // All pages with this tag
const paginates = paginate(pages, {
size: 10,
url: (n) => n === 1 ? `${baseURL}/${tag}/` : `${baseURL}/${tag}/${n}/`,
});
for (const p of paginates) {
yield {
...p,
title: `#${tag}`,
layout: "layouts/tagPage.vto",
tag: tag
}
}
}
}Note: I didn't run this code and I'm not 100% sure if it works, but I think it gives you an idea. |
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Thank you! The code works, with a few modifications to fit my templates. |
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You can use a different generator to paginate the first pages using renderOrder to make sure the render order is correct.