const name = document.getElementById('fname');
const submit = document.getElementById('submit');
const out = document.getElementById('out');
const gname = name.value;
// greet function
const greet = (n) => {
return `hello ${n}`;
};
// output
const pout = (fname) => {
out.innerHTML = greet(fname);
};
// call event
submit.addEventListener("click", () => { pout(gname) });<input type="text" id="fname" placeholder="Please enter your name"><br>
<button id="submit" >Submit</button>
<br><br>
OUTPUT: <span id="out"></span>
请帮我找到代码中的错误,我正在学习es6 JavaScript,找不到错误
发布于 2019-07-04 02:59:06
我假设您希望在output上显示输入的值。在这种情况下,您应该将name.value传递给greet()函数!
我已经删除了gname常量,因为这里不需要它们。但是,您可以在output函数中移动gname,因为它只需要在函数中,而不是在其他任何地方。
const name = document.getElementById('fname');
const out = document.getElementById('out');
const submit = document.getElementById('submit');
// greet function
const greet = (n) => {
return `hello ${n}`;
};
// output
const pout = () => {
out.innerHTML = greet(name.value);
};
// call event
submit.addEventListener("click", (event) => {
pout();
});<input type="text" id="fname" ><br>
<button id="submit" >Submit</button>
<br><br>
OUTPUT: <span id="out"></span>
发布于 2019-07-04 03:41:47
gname被分配为输入元素值- const gname = name.value;,您的input元素最初是空的。因此,每次调用pout(gname)时,gname都会有空字符串。如果要在输入中传递输入的值,请传递name.value而不是gname。
submit.addEventListener("click", () => { pout(name.value) });
const name = document.getElementById('fname');
const submit = document.getElementById('submit');
const out = document.getElementById('out');
// greet function
const greet = (n) => {
return `hello ${n}`;
};
// output
const pout = (fname) => {
out.innerHTML = greet(fname);
};
// call event
submit.addEventListener("click", () => {
pout(name.value)
});<input type="text" id="fname" placeholder="Please enter your name"><br>
<button id="submit">Submit</button>
<br><br> OUTPUT: <span id="out"></span>
https://stackoverflow.com/questions/56880058
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