如何在插入之前重命名元素?

Question

我想将新列表(newList)的元素添加到我已经存在的列表(oldList)中，但是在插入之前，我必须验证我的newList元素是否已存在于我的旧列表中，以防万一，必须添加一个额外的数字。

let newList = [{name: "abc.jpeg"}, {name: "abc.jpeg"}]
let oldList = [{name:"fs%2FPanel%8SmgM7rhJ4T9j%2Fpictures%2Fa.jpeg"},
{name:"fs%2FPanel%8SmgM7rhJ4T9j%2Fpictures%2Fa1.jpeg"},
{name:"fs%2FPanel%8SmgM7rhJ4T9j%2Fpictures%2Fa2.jpeg"}]
const addElement = (colection, new_item, tally) => {
const a = new_item.name.split('.')
const reps = colection.filter(elem => {
var reg = /.*2Fpictures%2F(.*)/
const name = elem.name.match(reg)
if(name)
return name[1].includes(a[0])
}).length
new_item.sname = a[0] + (reps > 0 ? reps : '') + '.' + a[1]
colection.push(new_item)
console.log(colection)
return colection
}

let tally = {}
for (let index = 0; index < newList.length; index++) {
oldList = addElement(oldList, newList[index], tally)
}

我需要的预期结果是这样的：

[{name:"fs%2FPanel%8SmgM7rhJ4T9j%2Fpictures%2Fa.jpeg"},
{name:"fs%2FPanel%8SmgM7rhJ4T9j%2Fpictures%2Fa1.jpeg"},
{name:"fs%2FPanel%8SmgM7rhJ4T9j%2Fpictures%2Fa2.jpeg"},
{name: "abc.jpeg"}, 
{name: "abc1.jpeg"}]

因为三个元素包含"fs%2FPanel%8SmgM7rhJ4T9j%2Fpictures%2Fabc.jpeg"abc.jpeg

Answer 1

另一种方法使用您从 Maptally开始的方法，并且比使用多个filter()实例更有效

的查找

let newList = [{name: "abc.jpeg"}, {name: "abc.jpeg"},{name:"fs%2FPanel%8SmgM7rhJ4T9j%2Fpictures%2Fa.jpeg"}]
let oldList = [{name:"fs%2FPanel%8SmgM7rhJ4T9j%2Fpictures%2Fa.jpeg"},
{name:"fs%2FPanel%8SmgM7rhJ4T9j%2Fpictures%2Fa1.jpeg"},
{name:"fs%2FPanel%8SmgM7rhJ4T9j%2Fpictures%2Fa2.jpeg"}, {name: "abc.jpeg"}];
const baseName = (str) => str.replace(/d+./, '.');
const getNamesCount = arr => arr.reduce((a,{name})=>{
const base = baseName(name);
return a.set(base, (a.get(base)||0)+1)
},new Map())

const mergeLists = (oList, nList) => {
const tally = getNamesCount(oList);
nList.forEach(({name,...rest})=>{
const base = baseName(name);    
if(tally.has(base)){
let count = tally.get(base)
name = name.split('.').join(count + '.');
tally.set(base, ++count)
}else{
tally.set(base, 1)
}
oList.push({name,...rest})
})
return oList;
}

console.log(mergeLists(oldList,newList))

Answer 2

//I added some element for testing
let newList = [
{name: "abc.jpeg"}, 
{name: "abc.jpeg"},
{name:"xxx123.jpg"},
{name:"xxx123.jpg"},
{name:"xxx123.jpg"},
{name:"xxx123.jpg"},
{name:"yyy143.jpg"},
{name:"yyy143.jpg"},
{name:"yyy143.jpg"},
{name:"yyy144.jpg"},
{name:"yyy145.jpg"},
{name:"yyy146.jpg"}
];
let oldList = [
{name:"fs%2FPanel%8SmgM7rhJ4T9j%2Fpictures%2Fa.jpeg"},
{name:"fs%2FPanel%8SmgM7rhJ4T9j%2Fpictures%2Fa1.jpeg"},
{name:"fs%2FPanel%8SmgM7rhJ4T9j%2Fpictures%2Fa2.jpeg"}];
//Creating a Set 
const mySet = new Set()
//Fill mySet with files names
for (i=0;i<oldList.length;i++){
mySet.add(oldList[i].name);
}
//Looping newList
for (i=0;i<newList.length;i++){  
if (mySet.has(newList[i].name)){  
//  The point "." position in filename string
pointpos = newList[i].name.lastIndexOf(".");    

//  File name and file extension 
name = newList[i].name.substring(0,pointpos);    
ext  = newList[i].name.substring(pointpos);

// Getting numbers in file name string from right to left 
// initializing cursor pos at the end of the file name string 
pos=name.length-1;
//Where to store the numbers of the file name , if any .

numbers=[];

// String numbers 0,1,2,3,4,5,6,7,8,9 
// have char codes from 48 to 57
while ( 48 <= name.charCodeAt(pos) && 57 >= name.charCodeAt(pos) && pos >= 0){

//Add values at top of the array , not at bottom.             
numbers.unshift(name.charAt(pos));
pos--;
}

//if file name has numbers...
if (numbers.length > 0){
//join numbers and sum 1 
num  = parseInt(numbers.join(""))+1;                    
snum = String(num); 

// name without numbers part
name = name.substring(0,name.length-numbers.length);

//Sum 1 again while we have a filename in the Set
while (mySet.has(name+snum+ext)){               
snum=String(num++);
}
//Here we have a new file name not present in the Set 
//So add it to the Set with number and extension 
mySet.add( name+snum+ext);

}else{
//The file name is duplicate but 
//there are no others with numbers
//so add "1" string to the name and extension 
//and add it to the Set          
name+="1";
mySet.add(name+ext);
}

}else{
// The filename is not present in oldlist so we can add it in the Set
mySet.add(newList[i].name);
}
}
//Reset oldList array
oldList=[];
//Fill it with objects like it was but with new set items as values 
for (let item of mySet) {      
oldList.push({name:item})
}
//Show it 
console.log(oldList);

没有评论：

const mySet = new Set();
for (i=0;i<oldList.length;i++)mySet.add(oldList[i].name);
for (i=0;i<newList.length;i++){  
if (mySet.has(newList[i].name)){          
pointpos = newList[i].name.lastIndexOf(".");    
name = newList[i].name.substring(0,pointpos);    
ext  = newList[i].name.substring(pointpos);
pos=name.length-1;
numbers=[];
while ( 48 <= name.charCodeAt(pos) && 57 >= name.charCodeAt(pos) && pos >= 0){        
numbers.unshift(name.charAt(pos));
pos--;
}            
if (numbers.length > 0){            
num  = parseInt(numbers.join(""))+1;                    
snum = String(num);         
name = name.substring(0,name.length-numbers.length);
while (mySet.has(name+snum+ext)){               
snum=String(num++);
}
mySet.add( name+snum+ext);                             
}else{         
name+="1";
mySet.add(name+ext);
}       
}else{
mySet.add(newList[i].name);
}
}
oldList=[];
for (let item of mySet) {      
oldList.push({name:item})
}
console.log(oldList);

Answer 3

您可以找到字符串的出现次数，然后将该数字附加到字符串中。

PP_7