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js-assignments's Issues

getRectangleString

js-assignments/task/01-strings-tasks.js

Lines 205 to 210 in 03703cc

let str = `┌${'─'.repeat(width-2)}┐\n`;
for (let i=0; i< height-2; i++){
str+=`│${' '.repeat(width-2)}│\n`;
}
str+= `└${'─'.repeat(width-2)}┘\n`
return str;

Шаблонное мышление. Можно без цикла сделать, еще одним string.repeat

extractNameFromTemplate

js-assignments/task/01-strings-tasks.js

Line 72 in 03703cc

return value.slice(7,value.length-1);

Лаконичнее будет

return value.slice(7,-1);

slice отличный инструмент с довольно емкими аргументами, надо учиться им пользоваться. Причем для массивов все будет точно также.

И в задаче unbracketTag тоже надо поправить.

getCardId

js-assignments/task/01-strings-tasks.js

Lines 277 to 319 in 03703cc

if (+(value.charAt(0))>=2 && +(value.charAt(0))<=9){
switch(value.charAt(1)){
case '♣':return +(value.charAt(0))-1 ;break;
case '♦':return +(value.charAt(0))-1 + 13;break;
case '♥':return +(value.charAt(0))-1 + 13*2;break;
case '♠':return +(value.charAt(0))-1 + 13*3;break;
}
}else if (+(value.charAt(1))==0){
switch(value.charAt(2)){
case '♣':return 9 ;break;
case '♦':return 9 + 13;break;
case '♥':return 9 + 13*2;break;
case '♠':return 9 + 13*3;break;
}
}else if (value.charAt(0)=='A'){
switch(value.charAt(1)){
case '♣':return 0 ;break;
case '♦':return 13;break;
case '♥':return 13*2;break;
case '♠':return 13*3;break;
}
}else if (value.charAt(0)=='J'){
switch(value.charAt(1)){
case '♣':return 10 ;break;
case '♦':return 10+13;break;
case '♥':return 10+13*2;break;
case '♠':return 10+13*3;break;
}
}else if (value.charAt(0)=='Q'){
switch(value.charAt(1)){
case '♣':return 11 ;break;
case '♦':return 11+13;break;
case '♥':return 11+13*2;break;
case '♠':return 11+13*3;break;
}
}else if (value.charAt(0)=='K'){
switch(value.charAt(1)){
case '♣':return 12 ;break;
case '♦':return 12+13;break;
case '♥':return 12+13*2;break;
case '♠':return 12+13*3;break;
}
}

Приз за самое длинное решение. Я в таких случаях запрещаю на полгода пользоваться Ctrl+C и Ctrl+V.
Можно решить в одну строку, если немного подумать:

return '♣♦♥♠'.indexOf(value.slice(-1)) * 13 + ...

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