Résolution d'un bug sur le calcul des quadrigrammes.

src/js/class/Event.js

@@ -33,20 +33,42 @@ export class Events {
    * Takes a name and returns its tetragraph.
    * "Pierre Martin" will give "PMAR"
    * "Jean-Paul Dupont" will give "JPDU"
+   * If the last name is short, we try to use elements from the previous names :
+   * "Pierre O" => "PIEO", "Pierre O-LY" => "POLY", "Pierre-Jean O" => "PJEO", "Pierre Martin-O" => "PMAO"
    * @param {string} name of the user, separated by spaces or '-' for composed names.
    */
   static compute_tetragraph(name) {
-    const words = name.replace("-", " ").split(" ");  // split by space or -
-    let tetragraph = ""
-    for (let i = 0; i < words.length; i++) {
-      if (i >= words.length - 1) {
-        // takes all first letters from the last word to finish the tetragraph
-        let lettersMissing = Math.max(0, 4 - tetragraph.length)
-        tetragraph += words[i].substring(0, lettersMissing).toUpperCase();
-      } else {
-        tetragraph += words[i][0].toUpperCase();
-      }
+    const words = name.replace("-", " ").replace(/\s+/, " ").trim().split(" ");  // split by space or -, removing duplicate blanks
+    let tetragraph = "";
+    let n = Math.min(3, words.length - 1);
+    // take first letters of the first words, up to the before-last one, or 3 (consider the last one as the family name)
+    for (let i = 0; i < n; i++) {
+        tetragraph += words[i][0];
     }
+
+    // Add the last first letter (family name)
+    tetragraph += words[words.length-1][0]
+
+    // Now, fix the tetragraph to have 4 letters, using the first letters of the last words
+    let lettersMissingCount = Math.max(0, 4 - tetragraph.length);
+    let suffix = "";
+    let j = words.length - 1;
+    // Handle the case where the last word is too short (e.g we need 3 letters, but the name is LY)
+    // In this case, we just add letters from the previous word (e.g. Mary Ly --> MALY)
+    while (lettersMissingCount > 0 && j >= 0) {
+      // Remove the last letter from the tetragraph, it will be added back by the suffix at the right place
+      tetragraph = tetragraph.slice(0, -1)
+
+      // substring treats everything > length as if it were length, so it's fine to do that
+      let toAdd = words[j].substring(0, lettersMissingCount + 1);
+      lettersMissingCount -= toAdd.length - 1; // -1 since we remove a letter from the tetragraph
+      suffix = toAdd + suffix;
+      j--;
+    }
+
+    tetragraph += suffix;
+    tetragraph = tetragraph.toUpperCase();
+
     return tetragraph;
   }
 }