[英]how can i fix this: Uncaught TypeError: Cannot read property 'toString' of undefined in javascript
trying to write a function that takes a string, and returns the string back with first letter of each word in capital. 尝试编写一个函数,该函数需要一个字符串,然后将字符串返回,并以大写形式包含每个单词的第一个字母。 i am able to do so.
我能够做到。 But i keep on getting undefined value at my first index value in the iteration, tho the i have items in my list.
但是我一直在迭代中获得第一个索引值的不确定值,因为我的列表中有项目。
i wrote my function that takes a string parameter. 我写了一个带有字符串参数的函数。 i pass my parameter in to split using split(" ") method it so i can get an array of words.
我使用split(“”)方法将参数传递给split,这样我就可以得到一个单词数组。 and when i created my forloop to go over all of them to be able to make each of their first letters Capital.
当我创建我的forloop遍历所有这些对象时,可以使每个字母的首字母大写。 now in the iteration, i create a variable to store the first index element converted into a string using the toString() method so i can later apply the toUpperCase() method on it
现在在迭代中,我创建了一个变量来存储使用toString()方法转换为字符串的第一个索引元素,以便稍后可以在其上应用toUpperCase()方法
When i do a console.log() in the for loop i can see the elements converted all listed out (when i execute the function and pass a string to it) 当我在for循环中执行console.log()时,我可以看到所有列出的元素都已转换(当我执行该函数并将一个字符串传递给它时)
But i do not understand why i get: Uncaught TypeError: Cannot read property 'toString' of undefined...and this error points to the variable holding the element at index [i] when i am converting it to string in the forloop: var pickString = newStringList[i].toString() 但是我不明白为什么会得到:Uncaught TypeError:无法读取未定义的属性'toString'...并且当我在forloop中将其转换为字符串时,此错误指向变量将索引[i]上的元素保留pickString = newStringList [i] .toString()
function capitalizeLertters (letter) {
var newStringList = letter.split(" ");
// console.log(newStringList[0])
var addAll = []
for (var i = 0; i <= newStringList.length; i++) {
console.log(newStringList[i])
var pickString = newStringList[i].toString()
// console.log(pickString)
var finalString = pickString.charAt(0).toUpperCase() + pickString.slice(1)
// console.log(finalString)
}
addAll.push(finalString)
// console.log(addAll)
return addAll
}
console.log(capitalizeLertters("js string exercises"))
This is my expected result: "Js String Exercises" 这是我的预期结果:“ Js弦乐练习”
And these are the different console.log results 这些是不同的console.log结果
""js Js string String exercises Exercises"" “” js Js字符串字符串练习练习“
But my function does not execute in the end...it throws this: 但是我的函数最终没有执行...它抛出了这个:
"""learnJs.js:100 Uncaught TypeError: Cannot read property 'toString' of undefined at capitalizeLertters (learnJs.js:100) at learnJs.js:110""" “”“ learnJs.js:100未捕获的TypeError:无法在learningJs.js:110处的capitalizeLertters(learnJs.js:100)读取未定义的属性'toString'
So 所以
You don't need toString()
on a String 您不需要在字符串上使用
toString()
The for loop crashes at the last element because you should stop at length -1
or simply replace <=
with <
. for循环在最后一个元素处崩溃,因为您应该在
length -1
处停止,或者只是将<=
替换为<
。
You are calling addAll
after the loop, so it only adds the last element. 您在循环后调用
addAll
,因此它仅添加最后一个元素。
You are returning an array and I assume you want the String, so you can use join()
on the array. 您正在返回一个数组,并且我假设您要使用String,因此可以在数组上使用
join()
。
function capitalizeLertters (letter) { var newStringList = letter.split(" "); var addAll = []; for (var i = 0; i < newStringList.length; i++) { var pickString = newStringList[i]; var finalString = pickString.charAt(0).toUpperCase() + pickString.slice(1); addAll.push(finalString); } return addAll.join(" "); } console.log(capitalizeLertters("js string exercises"))
please check the modifications 请检查修改
Solution : 解决方案 :
function capitalizeLertters (letter) {
var newStringList = letter.split(" ");
// console.log(newStringList[0])
var addAll = [];
for (var i = 0; i < newStringList.length; i++) {
console.log(newStringList[i]);
var pickString = newStringList[i];
// console.log(pickString)
var finalString = pickString.charAt(0).toUpperCase() + pickString.slice(1);
// console.log(finalString)
addAll.push(finalString);
}
// console.log(addAll)
return addAll.join(" ");
}
capitalizeLertters("js string exercises");
An array starts with the index number 0. The length of newStringList is 3. If you give a less than (<) then 'i' will go like 0("js"), 1("string"), 2("exercises"). 数组以索引号0开头。newStringList的长度为3。如果给出的长度小于(<),则'i'会像0(“ js”),1(“ string”),2(“ exercises ”)。 If you give a less than equal (<=) it will go like 0("js"), 1("string"), 2("exercises"), 3(undefined).
如果给出的值小于等于(<=),它将像0(“ js”),1(“ string”),2(“ exercises”),3(undefined)一样。
function capitalizeLertters (letter) {
var newStringList = letter.split(" ");
var addAll = ""
for (var i = 0; i < newStringList.length; i++) {
var pickString = newStringList[i].toString()
var finalString = pickString.charAt(0).toUpperCase() + pickString.slice(1)
addAll += finalString
if(i != newStringList.length-1){
addAll +=" "
}
}
return addAll
}
console.log(capitalizeLertters("js string exercises"))
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