spb*_*spb 6 javascript logic eval calculator
所以,我搜索了高低,我找不到答案.我已经尝试了大约三次,并通过基本上将输入作为字符串存储在数组中,解析数字,然后打开运算符,以便评估整数,得到一个基本的一个,但我有一个找出链接逻辑真的很难.有没有人有什么建议?甚至可能只是伪造的代码?我真的不想使用eval.非常感谢
对于只有5个运算符(^,*,/,+, - )且没有括号的简单计算器,您可以执行类似的操作.首先,将字符串转换为数字和运算符数组很方便.然后,我们按照优先顺序遍历数组查找每个运算符,并将运算符应用于它之前和之后的数字.
function parseCalculationString(s) {
// --- Parse a calculation string into an array of numbers and operators
var calculation = [],
current = '';
for (var i = 0, ch; ch = s.charAt(i); i++) {
if ('^*/+-'.indexOf(ch) > -1) {
if (current == '' && ch == '-') {
current = '-';
} else {
calculation.push(parseFloat(current), ch);
current = '';
}
} else {
current += s.charAt(i);
}
}
if (current != '') {
calculation.push(parseFloat(current));
}
return calculation;
}
function calculate(calc) {
// --- Perform a calculation expressed as an array of operators and numbers
var ops = [{'^': (a, b) => Math.pow(a, b)},
{'*': (a, b) => a * b, '/': (a, b) => a / b},
{'+': (a, b) => a + b, '-': (a, b) => a - b}],
newCalc = [],
currentOp;
for (var i = 0; i < ops.length; i++) {
for (var j = 0; j < calc.length; j++) {
if (ops[i][calc[j]]) {
currentOp = ops[i][calc[j]];
} else if (currentOp) {
newCalc[newCalc.length - 1] =
currentOp(newCalc[newCalc.length - 1], calc[j]);
currentOp = null;
} else {
newCalc.push(calc[j]);
}
console.log(newCalc);
}
calc = newCalc;
newCalc = [];
}
if (calc.length > 1) {
console.log('Error: unable to resolve calculation');
return calc;
} else {
return calc[0];
}
}
var calculateButton = document.getElementById('calculate'),
userInput = document.getElementById('userInput'),
result = document.getElementById('result');
calculateButton.addEventListener('click', function() {
result.innerHTML = "The answer is " + calculate(parseCalculationString(userInput.value));
});Run Code Online (Sandbox Code Playgroud)
<input type="text" id="userInput" />
<input type="button" value="Calculate" id="calculate" />
<div id="result"></div>Run Code Online (Sandbox Code Playgroud)
(jsfiddle).要允许括号,您可以告诉calculate函数在开始查找任何其他运算符之前检查括号,然后在每组括号内的表达式上递归调用自身.还可以改进解析功能,例如移除任何空白区域并处理错误.
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