{"id":43114,"date":"2026-07-06T09:03:00","date_gmt":"2026-07-06T06:03:00","guid":{"rendered":"https:\/\/fatihsoysal.com\/blog\/robust-webhook-retry-stratejisi-nasil-kurulur\/"},"modified":"2026-07-06T09:03:32","modified_gmt":"2026-07-06T06:03:32","slug":"robust-webhook-retry-stratejisi-nasil-kurulur","status":"publish","type":"post","link":"https:\/\/fatihsoysal.com\/blog\/robust-webhook-retry-stratejisi-nasil-kurulur\/","title":{"rendered":"Robust Webhook Retry Stratejisi Nas\u0131l Kurulur?"},"content":{"rendered":"<h2>Robust Webhook Retry Stratejisi Nas\u0131l Kurulur?<\/h2>\n<p>Web uygulamalar\u0131nda veri ak\u0131\u015f\u0131n\u0131 kesintisiz sa\u011flamak i\u00e7in robust (dayan\u0131kl\u0131) bir webhook retry (yeniden deneme) stratejisi kurmak hayati \u00f6nem ta\u015f\u0131r.<\/p>\n<h2>Webhook Nedir ve Neden Ba\u015far\u0131s\u0131z Olur?<\/h2>\n<p>Webhooks (web kancalar\u0131), modern web mimarisinde uygulamalar\u0131n birbiriyle ger\u00e7ek zamanl\u0131 olarak haberle\u015fmesini sa\u011flayan hafif y\u00f6ntemlerdir. Temel olarak, bir sistemde belirli bir olay (event) ger\u00e7ekle\u015fti\u011finde, di\u011fer sisteme HTTP POST iste\u011fi g\u00f6nderilmesi mant\u0131\u011f\u0131na dayan\u0131r. \u00d6rne\u011fin, bir e-ticaret platformunda \u00f6deme tamamland\u0131\u011f\u0131nda, kargo sistemine otomatik bilgi g\u00f6nderilmesi bu y\u00f6ntemle ger\u00e7ekle\u015fir. Ancak, internet d\u00fcnyas\u0131 her zaman kararl\u0131 ve kesintisiz de\u011fildir. Bu nedenle, webhook entegrasyonu (uyumla\u015ft\u0131rma) s\u00fcre\u00e7lerinde s\u0131k s\u0131k kesintiler ya\u015fan\u0131r.<\/p>\n<p>\u00d6ncelikle, hedef sunucuda anl\u0131k bir \u00e7\u00f6kme meydana gelebilir. Bununla birlikte, a\u011f ba\u011flant\u0131s\u0131ndaki ge\u00e7ici kopmalar veya sunucunun a\u015f\u0131r\u0131 y\u00fck alt\u0131nda kalmas\u0131 (rate limiting &#8211; istek s\u0131n\u0131rlama) gibi durumlar webhook g\u00f6nderiminin ba\u015far\u0131s\u0131z olmas\u0131na yol a\u00e7ar. E\u011fer g\u00f6nderici sistem, al\u0131c\u0131n\u0131n iste\u011fi alamad\u0131\u011f\u0131n\u0131 fark edip i\u015flemi \u00f6ylece b\u0131rak\u0131rsa, veri kayb\u0131 ka\u00e7\u0131n\u0131lmaz olur. \u00d6rne\u011fin, iyzico gibi bir \u00f6deme ge\u00e7idi \u00fczerinden al\u0131nan bir \u00f6demenin ba\u015far\u0131l\u0131 bilgisi e-ticaret sitenize ula\u015fmazsa, m\u00fc\u015fterinin sipari\u015fi onaylanmaz. Bu durum hem m\u00fc\u015fteri memnuniyetsizli\u011fine hem de operasyonel karma\u015faya neden olur. \u0130\u015fte tam bu noktada, ak\u0131ll\u0131 ve hata tolerans\u0131 (fault tolerance) y\u00fcksek bir yeniden deneme mekanizmas\u0131 kurmak zorunlu hale gelir.<\/p>\n<h2>Exponential Backoff (\u00dcstel Geri \u00c7ekilme) Nedir ve Nas\u0131l \u00c7al\u0131\u015f\u0131r?<\/h2>\n<p>Bir webhook ba\u015far\u0131s\u0131z oldu\u011funda, en s\u0131k yap\u0131lan hata sistemi hemen ard\u0131ndan tekrar tekrar aramakt\u0131r. \u00d6rne\u011fin, her 5 saniyede bir ayn\u0131 iste\u011fi g\u00f6ndermek mant\u0131kl\u0131 g\u00f6r\u00fcnebilir. Ancak, hedef sunucu zaten a\u015f\u0131r\u0131 y\u00fck alt\u0131ndaysa veya ge\u00e7ici bir veritaban\u0131 kilitlemesi ya\u015f\u0131yorsa, bu s\u0131k aral\u0131klarla yap\u0131lan istekler sunucunun daha da \u00e7\u00f6kmesine yol a\u00e7ar. Yaz\u0131l\u0131m d\u00fcnyas\u0131nda bu duruma &#8220;Thundering Herd Problem&#8221; (G\u00fcr\u00fclt\u00fcl\u00fc S\u00fcr\u00fc Problemi) ad\u0131 verilir.<\/p>\n<p>Bu problemi \u00e7\u00f6zmek i\u00e7in Exponential Backoff (\u00fcstel geri \u00e7ekilme) algoritmas\u0131 kullanmal\u0131s\u0131n\u0131z. Bu algoritma, her ba\u015far\u0131s\u0131z denemeden sonra bekleme s\u00fcresini katlayarak art\u0131r\u0131r. B\u00f6ylece, kar\u015f\u0131 sunucuya kendini toparlamas\u0131 i\u00e7in zaman tan\u0131m\u0131\u015f olursunuz. Matematiksel olarak form\u00fcl \u015fu \u015fekildedir:<\/p>\n<blockquote><p><code>Bekleme S\u00fcresi = Ba\u015flang\u0131\u00e7 Gecikmesi * (\u00c7arpan ^ Deneme Say\u0131s\u0131)<\/code><\/p><\/blockquote>\n<p>\u00d6rne\u011fin, ba\u015flang\u0131\u00e7 gecikmesini 2 saniye, \u00e7arpan\u0131 ise 2 olarak belirleyelim. Bu durumda deneme s\u00fcre\u00e7leri \u015fu \u015fekilde ilerler:<\/p>\n<ul>\n<li><strong>1. Deneme:<\/strong> Ba\u015far\u0131s\u0131z. 2 saniye bekle.<\/li>\n<li><strong>2. Deneme:<\/strong> Ba\u015far\u0131s\u0131z. 4 saniye bekle.<\/li>\n<li><strong>3. Deneme:<\/strong> Ba\u015far\u0131s\u0131z. 8 saniye bekle.<\/li>\n<li><strong>4. Deneme:<\/strong> Ba\u015far\u0131s\u0131z. 16 saniye bekle.<\/li>\n<\/ul>\n<p>G\u00f6r\u00fcld\u00fc\u011f\u00fc \u00fczere, her ad\u0131mda s\u00fcre iki kat\u0131na \u00e7\u0131kar. Bu sayede, hedef sunucunun \u00fczerindeki bask\u0131 kademeli olarak azalt\u0131l\u0131r. Ancak, t\u00fcm ba\u015far\u0131s\u0131z isteklerin ayn\u0131 anda tekrar denemeye ba\u015flamas\u0131n\u0131 engellemek i\u00e7in bu form\u00fcle &#8220;Jitter&#8221; (rastgele sapma) eklemek gerekir.<\/p>\n<h3>Jitter (Rastgele Sapma) Neden Eklenmelidir?<\/h3>\n<p>Sistemlerimizde sadece \u00fcstel geri \u00e7ekilme kullanmak bazen yeterli olmayabilir. \u00c7\u00fcnk\u00fc ayn\u0131 anda \u00e7\u00f6ken y\u00fczlerce webhook iste\u011fi, ayn\u0131 gecikme s\u00fcrelerini hesaplayarak hedef sunucuya tam olarak ayn\u0131 saniyede tekrar h\u00fccum edebilir. Bu senaryo, hedef sunucunun aya\u011fa kalkt\u0131\u011f\u0131 anda tekrar \u00e7\u00f6kmesine sebep olur.<\/p>\n<p>Jitter (rastgele sapma), bekleme s\u00fcresine rastgele bir g\u00fcr\u00fclt\u00fc veya zaman dilimi ekleme i\u015flemidir. \u00d6rne\u011fin, hesaplanan 8 saniyelik bekleme s\u00fcresine rastgele +\/- 1.5 saniye ekleyerek s\u00fcreyi 6.5 ile 9.5 saniye aras\u0131na \u00e7ekebilirsiniz. B\u00f6ylelikle, kuyruktaki y\u00fczlerce webhook iste\u011fi farkl\u0131 milisaniyelerde tetiklenir. Bu durum, hedef sunucunun gelen trafi\u011fi daha dengeli ve p\u00fcr\u00fczs\u00fcz bir \u015fekilde i\u015flemesini sa\u011flar. Sonu\u00e7 olarak, a\u011f trafi\u011findeki tepe noktalar\u0131 (peaks) d\u00fczle\u015ftirilir ve sistem genelinde y\u00fcksek bir hata tolerans\u0131 elde edilir.<\/p>\n<h2>Ad\u0131m Ad\u0131m Webhook Retry Mekanizmas\u0131 Nas\u0131l Tasarlan\u0131r?<\/h2>\n<p>G\u00fcvenilir bir webhook mimarisi tasarlarken, i\u015flemleri kesinlikle ana uygulama ak\u0131\u015f\u0131nda (senkron olarak) yapmamal\u0131s\u0131n\u0131z. Bunun yerine, asenkron (e\u015fzamans\u0131z) bir arka plan i\u015fleyicisi kullanmal\u0131s\u0131n\u0131z. Bir webhook tetiklendi\u011finde, bu istek hemen bir mesaj kuyru\u011funa (message broker) aktar\u0131lmal\u0131d\u0131r.<\/p>\n<p>Tasar\u0131m s\u00fcrecinde \u015fu ad\u0131mlar\u0131 izleyebilirsiniz:<\/p>\n<ol>\n<li><strong>Veritaban\u0131 Kayd\u0131:<\/strong> G\u00f6nderilecek her webhook iste\u011fini benzersiz bir kimlik (ID) ile veritaban\u0131na kaydedin ve durumunu &#8220;PENDING&#8221; (Beklemede) olarak i\u015faretleyin.<\/li>\n<li><strong>Kuyru\u011fa Ekleme:<\/strong> \u0130ste\u011fi Redis, RabbitMQ veya Amazon SQS gibi bir kuyruk sistemine g\u00f6nderin.<\/li>\n<li><strong>\u0130\u015fleme ve \u0130zleme:<\/strong> Kuyruk t\u00fcketicisi (consumer) iste\u011fi hedef sunucuya g\u00f6nderir. E\u011fer HTTP 200 OK d\u00f6nerse, durumu &#8220;SUCCESS&#8221; (Ba\u015far\u0131l\u0131) yap\u0131n.<\/li>\n<li><strong>Hata Yakalama ve Hesaplama:<\/strong> E\u011fer sunucu 5xx hatas\u0131 veya zaman a\u015f\u0131m\u0131 (timeout) verirse, deneme say\u0131s\u0131n\u0131 1 art\u0131r\u0131n. Exponential backoff ve jitter form\u00fcl\u00fcn\u00fc kullanarak yeni bir g\u00f6nderim zaman\u0131 hesaplay\u0131n.<\/li>\n<li><strong>Gecikmeli Yeniden Deneme:<\/strong> \u0130ste\u011fi, hesaplanan s\u00fcre kadar geciktirerek kuyru\u011fa yeniden ekleyin.<\/li>\n<li><strong>DLQ (Dead Letter Queue) Aktar\u0131m\u0131:<\/strong> Maksimum deneme s\u0131n\u0131r\u0131na (\u00f6rne\u011fin 5 veya 10 kez) ula\u015f\u0131ld\u0131\u011f\u0131nda, iste\u011fi &#8220;FAILED&#8221; (Ba\u015far\u0131s\u0131z) olarak i\u015faretleyin ve manuel inceleme i\u00e7in Dead Letter Queue (Al\u0131namayan Mesaj Kuyru\u011fu) havuzuna aktar\u0131n.<\/li>\n<\/ol>\n<p>A\u015fa\u011f\u0131daki tablo, tipik bir 5 ad\u0131ml\u0131 retry s\u00fcrecinin zaman ak\u0131\u015f\u0131n\u0131 g\u00f6stermektedir:<\/p>\n<table>\n<thead>\n<tr>\n<th>Deneme No<\/th>\n<th>Temel S\u00fcre (sn)<\/th>\n<th>Jitter Aral\u0131\u011f\u0131 (sn)<\/th>\n<th>Toplam Gecikme (Yakla\u015f\u0131k)<\/th>\n<th>Durum<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>1<\/td>\n<td>5<\/td>\n<td>+\/- 1<\/td>\n<td>4 &#8211; 6 saniye<\/td>\n<td>Yeniden Deneniyor<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>10<\/td>\n<td>+\/- 2<\/td>\n<td>8 &#8211; 12 saniye<\/td>\n<td>Yeniden Deneniyor<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>20<\/td>\n<td>+\/- 4<\/td>\n<td>16 &#8211; 24 saniye<\/td>\n<td>Yeniden Deneniyor<\/td>\n<\/tr>\n<tr>\n<td>4<\/td>\n<td>40<\/td>\n<td>+\/- 8<\/td>\n<td>32 &#8211; 48 saniye<\/td>\n<td>Yeniden Deneniyor<\/td>\n<\/tr>\n<tr>\n<td>5<\/td>\n<td>80<\/td>\n<td>+\/- 16<\/td>\n<td>64 &#8211; 96 saniye<\/td>\n<td>Son Deneme \/ DLQ<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Node.js ve Redis ile \u00d6rnek Bir Uygulama Nas\u0131l Yap\u0131l\u0131r?<\/h2>\n<p>Bu b\u00f6l\u00fcmde, pop\u00fcler bir JavaScript \u00e7al\u0131\u015fma ortam\u0131 (runtime) olan Node.js kullanarak basit ama etkili bir \u00fcstel geri \u00e7ekilme algoritmas\u0131 yazaca\u011f\u0131z. Bu kod \u00f6rne\u011finde, hata durumunda bekleme s\u00fcresini hesaplayan fonksiyonu ve kuyruk y\u00f6netim mant\u0131\u011f\u0131n\u0131 g\u00f6rebilirsiniz.<\/p>\n<div class=\"code-container\">\n<pre><code>\/\/ Exponential Backoff ve Jitter Hesaplama Fonksiyonu\nfunction calculateDelay(attempt, baseDelay = 1000, maxDelay = 30000) {\n    \/\/ \u00dcstel gecikmeyi hesapla: base * 2^attempt\n    const tempDelay = baseDelay * Math.pow(2, attempt);\n    \n    \/\/ Maksimum gecikme s\u0131n\u0131r\u0131n\u0131 a\u015fmamas\u0131n\u0131 sa\u011fla\n    const backoffLimit = Math.min(tempDelay, maxDelay);\n    \n    \/\/ Jitter (Rastgele Sapma) ekle (%20 oran\u0131nda)\n    const jitter = backoffLimit * 0.2 * Math.random();\n    \n    \/\/ Nihai bekleme s\u00fcresi\n    return Math.floor(backoffLimit + jitter);\n}\n\n\/\/ \u00d6rnek Webhook G\u00f6nderim Fonksiyonu\nasync function sendWebhookWithRetry(webhookUrl, payload, attempt = 0, maxAttempts = 5) {\n    try {\n        console.log(<code>Deneme ${attempt + 1}: Webhook g\u00f6nderiliyor...<\/code>);\n        \n        const response = await fetch(webhookUrl, {\n            method: 'POST',\n            headers: { 'Content-Type': 'application\/json' },\n            body: JSON.stringify(payload)\n        });\n\n        if (!response.ok) {\n            throw new Error(<code>Sunucu Hatas\u0131: ${response.status}<\/code>);\n        }\n\n        console.log(\"Webhook ba\u015far\u0131yla iletildi!\");\n        return true;\n    } catch (error) {\n        console.error(<code>Hata Olu\u015ftu: ${error.message}<\/code>);\n\n        if (attempt &lt; maxAttempts) {\n            const delay = calculateDelay(attempt);\n            console.log(<code>${delay} ms sonra tekrar denenecek...\\n<\/code>);\n            \n            \/\/ Belirlenen s\u00fcre kadar asenkron olarak bekle ve yeniden dene\n            await new Promise(resolve =&gt; setTimeout(resolve, delay));\n            return sendWebhookWithRetry(webhookUrl, payload, attempt + 1, maxAttempts);\n        } else {\n            console.error(\"Maksimum deneme s\u0131n\u0131r\u0131na ula\u015f\u0131ld\u0131. \u0130stek DLQ kuyru\u011funa aktar\u0131l\u0131yor.\");\n            return false;\n        }\n    }\n}\n\n\/\/ Fonksiyonu Test Etme (S\u00fcrekli 503 hatas\u0131 veren test servisi)\nconst targetUrl = \"https:\/\/httpstat.us\/503\"; \nconst data = { orderId: 12345, status: \"completed\" };\nsendWebhookWithRetry(targetUrl, data);<\/code><\/pre>\n<\/div>\n<p>Yukar\u0131daki kod blo\u011funda g\u00f6rd\u00fc\u011f\u00fcn\u00fcz \u00fczere, <code>calculateDelay<\/code> fonksiyonu her ba\u015far\u0131s\u0131z denemede s\u00fcreyi katlayarak art\u0131r\u0131r ve \u00fczerine rastgele bir sapma ekler. Bu sayede, hedef sunucunun anl\u0131k kesintilerinden etkilenmeden, sistem kaynaklar\u0131n\u0131 en verimli \u015fekilde kullanm\u0131\u015f oluruz.<\/p>\n<h2>Ger\u00e7ek D\u00fcnya Senaryosu: Trendyol Entegrasyonunda \u00d6deme Kayb\u0131<\/h2>\n<p>Gelin konuyu daha net anlamak i\u00e7in yerel e-ticaret pazar\u0131ndan bir \u00f6rnek \u00fczerinden ilerleyelim. Farz edelim ki, Trendyol veya Hepsiburada \u00fczerinde sat\u0131\u015f yapan bir ma\u011fazan\u0131z var ve kendi ERP (Kurumsal Kaynak Planlamas\u0131) sisteminizi bu platformlara entegre ettiniz. M\u00fc\u015fteri sipari\u015f verdi\u011finde, Trendyol sunucular\u0131 sizin ERP sisteminize bir webhook g\u00f6nderir.<\/p>\n<p>Bir g\u00fcn, ERP sisteminizin kurulu oldu\u011fu sunucu, yo\u011fun bir veritaban\u0131 yedekleme i\u015flemi nedeniyle 5 dakika boyunca yan\u0131t veremez hale geldi. Bu s\u0131rada Trendyol&#8217;dan gelen 50 adet sipari\u015f webhook iste\u011fi, sunucunuzdan &#8220;504 Gateway Timeout&#8221; hatas\u0131 ald\u0131. E\u011fer Trendyol taraf\u0131nda ak\u0131ll\u0131 bir retry mekanizmas\u0131 olmasayd\u0131, bu 50 sipari\u015f sizin sisteminize asla d\u00fc\u015fmeyecekti. Sonu\u00e7 olarak, m\u00fc\u015fterilerin kargolar\u0131 haz\u0131rlanamayacak, ma\u011faza puan\u0131n\u0131z d\u00fc\u015fecek ve ciddi bir ciro kayb\u0131 ya\u015fayacakt\u0131n\u0131z.<\/p>\n<p>Ancak, \u00fcstel geri \u00e7ekilme kullanan bir webhook sistemi sayesinde, ilk ba\u015far\u0131s\u0131z denemeden sonra istekler hemen silinmez. Sistem \u00f6nce 10 saniye, sonra 30 saniye, ard\u0131ndan 2 dakika bekleyerek istekleri yeniden g\u00f6nderir. Sunucunuz yedekleme i\u015flemini tamamlay\u0131p tekrar aya\u011fa kalkt\u0131\u011f\u0131nda, biriken t\u00fcm webhook istekleri g\u00fcvenli bir \u015fekilde ERP sisteminize ula\u015f\u0131r ve sipari\u015fleriniz kesintisiz \u015fekilde i\u015flenmeye devam eder.<\/p>\n<h2>Webhook G\u00fcvenli\u011fi ve \u0130leri D\u00fczey \u0130pu\u00e7lar\u0131 Nelerdir?<\/h2>\n<p>Dayan\u0131kl\u0131 bir webhook retry stratejisi kurarken, g\u00fcvenli\u011fi ve veri tutarl\u0131l\u0131\u011f\u0131n\u0131 asla g\u00f6z ard\u0131 etmemelisiniz. Yeniden deneme mekanizmalar\u0131 do\u011fas\u0131 gere\u011fi ayn\u0131 iste\u011fin birden fazla kez g\u00f6nderilmesine yol a\u00e7abilir. Bu durum, sistemlerinizde m\u00fckerrer (tekrarlanan) i\u015flemlere sebep olabilir.<\/p>\n<p>Bu sorunlar\u0131 \u00f6nlemek i\u00e7in \u015fu ileri d\u00fczey teknikleri uygulayabilirsiniz:<\/p>\n<ul>\n<li><strong>Idempotency (Ayn\u0131 G\u00fc\u00e7l\u00fc \u0130\u015flem \/ Tekrarlanabilirlik):<\/strong> Al\u0131c\u0131 taraf, gelen her webhook iste\u011findeki benzersiz i\u015flem ID&#8217;sini (\u00f6rne\u011fin sipari\u015f numaras\u0131) kontrol etmelidir. E\u011fer bu ID daha \u00f6nce ba\u015far\u0131yla i\u015flendiyse, i\u015flemi tekrar yapmadan do\u011frudan HTTP 200 OK d\u00f6nmelidir. Bu, sisteminizin m\u00fckerrer \u00f6deme almas\u0131n\u0131 veya ayn\u0131 kargoyu iki kez \u00e7\u0131karmas\u0131n\u0131 engeller.<\/li>\n<li><strong>Webhook \u0130mzalar\u0131 (Signature Verification):<\/strong> G\u00f6nderdi\u011finiz webhook isteklerinin g\u00f6vdesini (payload) gizli bir anahtar (secret key) ile imzalay\u0131n. Al\u0131c\u0131, gelen iste\u011fin ger\u00e7ekten sizden geldi\u011fini do\u011frulamak i\u00e7in bu imzay\u0131 kontrol etmelidir.<\/li>\n<li><strong>Zaman A\u015f\u0131m\u0131 S\u0131n\u0131rlar\u0131 (Timeout Limits):<\/strong> \u0130stek g\u00f6nderirken ba\u011flant\u0131 zaman a\u015f\u0131m\u0131n\u0131 (connection timeout) k\u0131sa tutun (\u00f6rne\u011fin 5-10 saniye). Yan\u0131t vermeyen bir sunucuyu dakikalarca beklemek, sizin kendi kuyruk sisteminizi kilitleyebilir.<\/li>\n<li><strong>Kuyruk \u0130zolasyonu (Queue Isolation):<\/strong> \u0130lk denemeler i\u00e7in h\u0131zl\u0131 kuyruklar kullan\u0131rken, \u00e7ok kez ba\u015far\u0131s\u0131z olmu\u015f ve saatler sonra denenecek istekleri farkl\u0131 bir yava\u015f kuyru\u011fa (slow queue) aktar\u0131n. B\u00f6ylece yeni ve sa\u011fl\u0131kl\u0131 webhook istekleri, eski ve sorunlu isteklerin arkas\u0131nda beklemek zorunda kalmaz.<\/li>\n<\/ul>\n<h2>S\u0131k\u00e7a Sorulan Sorular (FAQ)<\/h2>\n<h3>Webhook istekleri i\u00e7in ideal maksimum deneme say\u0131s\u0131 ka\u00e7t\u0131r?<\/h3>\n<p>Genellikle end\u00fcstri standartlar\u0131nda maksimum deneme say\u0131s\u0131 5 ile 10 aras\u0131nda de\u011fi\u015fir. \u00d6nemli olan deneme say\u0131s\u0131ndan ziyade, toplam deneme s\u00fcresidir. Exponential backoff kullanarak bu denemeleri 24 saate yaymak en sa\u011fl\u0131kl\u0131 yakla\u015f\u0131md\u0131r.<\/p>\n<h3>Ba\u015far\u0131s\u0131z olan webhook&#8217;lar\u0131 ne kadar s\u00fcreyle saklamal\u0131y\u0131z?<\/h3>\n<p>Maksimum deneme s\u0131n\u0131r\u0131na ula\u015fan ve DLQ (Dead Letter Queue) sistemine d\u00fc\u015fen webhook&#8217;lar, manuel analiz ve hata ay\u0131klama s\u00fcre\u00e7leri i\u00e7in en az 7 ila 14 g\u00fcn boyunca veritaban\u0131nda saklanmal\u0131d\u0131r.<\/p>\n<h3>Al\u0131c\u0131 sunucu 404 hatas\u0131 veriyorsa yeniden denemeli miyiz?<\/h3>\n<p>Hay\u0131r. HTTP 404 (Bulunamad\u0131) veya 401 (Yetkisiz) gibi 4xx serisi istemci hatalar\u0131, genellikle kal\u0131c\u0131 yap\u0131land\u0131rma hatalar\u0131na i\u015faret eder. Bu durumlarda yeniden deneme yapmak a\u011f trafi\u011fini bo\u015funa me\u015fgul edecektir. Yeniden deneme mekanizmas\u0131 sadece 5xx (Sunucu Hatalar\u0131) ve a\u011f zaman a\u015f\u0131m\u0131 durumlar\u0131nda tetiklenmelidir.<\/p>\n<h3>Kuyruk sistemi olarak ne tercih edilmelidir?<\/h3>\n<p>K\u00fc\u00e7\u00fck ve orta \u00f6l\u00e7ekli projeler i\u00e7in Redis tabanl\u0131 BullMQ olduk\u00e7a ba\u015far\u0131l\u0131d\u0131r. B\u00fcy\u00fck \u00f6l\u00e7ekli, y\u00fcksek trafikli sistemlerde ise RabbitMQ, Apache Kafka veya AWS SQS gibi geli\u015fmi\u015f mesaj kuyru\u011fu sistemleri tercih edilmelidir.<\/p>\n<p>#Teknoloji #WebGeli\u015ftirme #Backend #SoftwareArchitecture #API<\/p>\n<div class=\"github-example-link\"><strong>\u00d6rnek kod:<\/strong> <a href=\"https:\/\/github.com\/fatihsoysalcom\/robust-webhook-retry-with-exponential-backoff\" target=\"_blank\" rel=\"noopener noreferrer\">github.com\/fatihsoysalcom\/robust-webhook-retry-with-exponential-backoff<\/a><\/div>\n","protected":false},"excerpt":{"rendered":"Web uygulamalar\u0131nda veri ak\u0131\u015f\u0131n\u0131 kesintisiz sa\u011flamak i\u00e7in robust (dayan\u0131kl\u0131) bir webhook retry (yeniden deneme) stratejisi kurmak hayati \u00f6nem ta\u015f\u0131r.","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"csco_page_header_type":"","csco_page_load_nextpost":"","csco_page_subscribe_form":"","csco_page_contact_form":"","footnotes":""},"categories":[1],"tags":[],"class_list":{"0":"post-43114","1":"post","2":"type-post","3":"status-publish","4":"format-standard","6":"category-genel","7":"cs-entry","8":"cs-video-wrap"},"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v20.5 (Yoast SEO v25.3.1) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Robust Webhook Retry Stratejisi Nas\u0131l Kurulur? 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