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Fix social URL host matching #1174

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Merged
MaheshtheDev merged 2 commits into from
Jun 28, 2026
+35 −8
Merged

Fix social URL host matching #1174

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7e37463
Fix social URL host matching
anirudh5harma Jun 27, 2026
c978fd5
Remove URL helper test file
MaheshtheDev Jun 28, 2026
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43 changes: 35 additions & 8 deletions apps/web/lib/url-helpers.ts
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Original file line number Diff line number Diff line change
Expand Up @@ -77,7 +77,7 @@ export const isValidUrl = (url: string): boolean => {
*/
export const normalizeUrl = (url: string): string => {
if (!url.trim()) return ""
if (url.startsWith("http://") || url.startsWith("https://")) {
if (/^https?:\/\//i.test(url)) {
return url
}
return `https://${url}`
Expand Down Expand Up @@ -120,25 +120,52 @@ export const extractUrls = (
return { urls, duplicates }
}

const parseWebUrl = (url: string): URL | null => {
const trimmed = url.trim()
if (!trimmed) return null

try {
const parsed = new URL(trimmed)
return parsed.protocol === "http:" || parsed.protocol === "https:"
? parsed
: null
} catch {
try {
return new URL(`https://${trimmed}`)
} catch {
return null
}
}
}

const hostnameMatches = (hostname: string, domain: string): boolean => {
const normalizedHostname = hostname.toLowerCase()
return (
normalizedHostname === domain || normalizedHostname.endsWith(`.${domain}`)
)
}

/**
* Checks if a URL is a Twitter/X URL.
*/
export const isTwitterUrl = (url: string): boolean => {
const normalizedUrl = url.toLowerCase()
const parsed = parseWebUrl(url)
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anirudh5harma marked this conversation as resolved.
if (!parsed) return false
return (
normalizedUrl.includes("twitter.com") || normalizedUrl.includes("x.com")
hostnameMatches(parsed.hostname, "twitter.com") ||
hostnameMatches(parsed.hostname, "x.com")
)
}

/**
* Checks if a URL is a LinkedIn profile URL (not a company page).
*/
export const isLinkedInProfileUrl = (url: string): boolean => {
const normalizedUrl = url.toLowerCase()
return (
normalizedUrl.includes("linkedin.com/in/") &&
!normalizedUrl.includes("linkedin.com/company/")
)
const parsed = parseWebUrl(url)
if (!parsed || !hostnameMatches(parsed.hostname, "linkedin.com")) return false

const [section, handle] = parsed.pathname.split("/").filter(Boolean)
return section?.toLowerCase() === "in" && Boolean(handle)
}

/**
Expand Down