我正在使用代码学院自学JavaScript,我正在尝试制作一些简单的代码,以便在提示提出问题时,用户回复给出了回答。
示例。
prompt says "what's your favourite colour?"
user says "blue"
response "that's the same colour as the sky!"
但是,当我尝试添加不同的选项时,我会收到语法错误:意外令牌。
我尝试做到这一点,以便如果我问一个问题,答复会得到答复,但其他任何回答。
这是代码。
prompt("what do you want?");
if ("coke");
{console.log ("no coke, pepsi.")};
else
console.log ("pepsi only.")};
如果有人有任何想法,我会非常感谢!
免责声明:我不适用于可口可乐
如果您以后要使用prompt
的返回值。另外,您有一些应该纠正的语法错误:
var answer = prompt('what do you want?');
if (answer === 'coke') {
console.log('you said coke!');
} else {
console.log('why didn't you say coke!?');
}
您也可以在获得更多情况时使用开关:
var answer = prompt('what do you want?');
switch (answer) {
case 'coke':
console.log('you said coke!');
break;
default:
console.log('why didn't you say coke!?');
break;
}
或一个对象,因为大多数人更喜欢这样切换:
var answer = prompt('what do you want?');
var responses = {
coke: 'you said coke!',
defaultResponse: 'why didn't you say coke!?'
};
console.log(responses[answer] || responses.defaultResponse);
if最后不需要半隆。而是这样做:
if ("coke") {
console.log ("no coke, pepsi.");
} else {
console.log ("pepsi only.");
}
删除尾随的分号:
prompt("what do you want?");
if ("coke") {
console.log ("no coke, pepsi.");
} else {
console.log ("pepsi only.");
}
您在近距离后有半柱。尝试:
var ans = prompt("what do you want?");
if (ans == "coke") {
console.log ("no coke, pepsi.");
} else {
console.log ("pepsi only.");
}
var name = prompt("what do you want?");
if (name == "coke")
{
console.log ("no coke, pepsi.")
}
else
{
console.log ("pepsi only.")
}
喜欢上面的
实际上不做
if (ans == "whatever") {
console.log ("whatever");
} else {
console.log ("whatever.");
}
做
if (ans == "whatever") {
confirm ("whatever");
} else {
confirm ("whatever.");
}
需要确定variable
。同样,"if" "else"
语句之间的括号和半结肠也有问题。我不确定console log
,但是如果您想要弹出警报请尝试以下操作:
var brand = prompt ('what do you want?');
if (brand="coke") {
alert ("no coke, pepsi.")
}else {
alert ("pepsi only.")
};
dicaimer:我充其量是新手,贾特(Jut)碰巧调试了一个类似的问题。希望它有帮助。