Cut Your SERP API Bill in Half

The Problem

Legacy SERP vendors price like it is still 2019. Entry plans start in the hundreds of dollars for tens of thousands of searches, overage rates punish success, and anything premium like AI Overviews or location-specific results sits behind a higher tier. Teams that built around a legacy vendor often find that the per-search cost of an agent workflow eats most of the gross margin. Switching is risky because the response schemas are nearly, but not quite, compatible, so a migration becomes a multi-sprint project that no one has time to champion.

The Scavio Solution

Scavio ships a flat per-search price that covers every feature on day one. There is no Pro tier gate on AI Overviews, no separate add-on for local results, no premium multiplier for mobile versus desktop. Volume discounts kick in automatically, and the free tier is generous enough to test at real production traffic before you commit. Most teams report cutting their effective per-thousand-search cost by forty to sixty percent after moving over, with no drop in coverage or quality.

Who It Is For

Engineering leads and finance-aware founders watching the SERP line item creep. If your current vendor just raised prices or gated AI Overviews behind a higher plan, a migration pays for itself inside one billing cycle.

Key Benefits

• Flat per-search pricing with no feature-tier upsells
• AI Overviews, Knowledge Graph, and local packs included by default
• Automatic volume discounts without annual contracts
• Free tier sized for real pilot traffic, not just hello-world testing
• Transparent usage dashboard with per-key spend breakdown

Python Example

import requests
from collections import Counter

API_KEY = "your_scavio_api_key"

def cheap_batch(queries: list[str]):
    hits = Counter()
    for q in queries:
        r = requests.post(
            "https://api.scavio.dev/api/v1/search",
            headers={"x-api-key": API_KEY},
            json={"platform": "google", "query": q},
            timeout=10,
        )
        data = r.json()
        for org in data.get("organic", []):
            hits[org["link"].split("/")[2]] += 1
    return hits.most_common(10)

print(cheap_batch(["running shoes", "trail shoes", "marathon shoes"]))

JavaScript Example

const API_KEY = "your_scavio_api_key";

async function cheapBatch(queries) {
  const hits = new Map();
  for (const q of queries) {
    const r = await fetch("https://api.scavio.dev/api/v1/search", {
      method: "POST",
      headers: { "x-api-key": API_KEY, "content-type": "application/json" },
      body: JSON.stringify({ platform: "google", query: q }),
    });
    const data = await r.json();
    for (const org of data.organic ?? []) {
      const host = new URL(org.link).host;
      hits.set(host, (hits.get(host) ?? 0) + 1);
    }
  }
  return [...hits.entries()].sort((a, b) => b[1] - a[1]).slice(0, 10);
}

console.log(await cheapBatch(["running shoes", "trail shoes", "marathon shoes"]));

Platforms Used