有没有办法从正则表达式中匹配和删除逻辑运算符

Question

我不想在正则表达式中使用 Match 函数时包含逻辑运算符（and、or、not）。我在下面尝试过，但它没有按预期工作。有人可以帮帮我吗。

我用于解析的字符串类型：示例：

1. Input -->'(Value1==6) and (Value2==0)?1:0'

Output --> ["Value1", "Value2"]

2. Input : 'Value_1'

Output -->["Value_1"]

3. Input : '(Value_1 * Value_2)'

Output : ["Value1", "Value2"]

4. Input : 'Value_Machine_Outcome==4?1:0'

Output : Value_Machine_Outcome

嵌套条件：无

条件是否始终在括号中：否， 我将在下一步中使用Math.evaluate评估它们

请举例如下：

const paragraph = '(Value1==6) and (Value2==0)?1:0';

const regex = /([a-zA-Z]+[a-zA-Z0-9_]+)(?<!and|or|not)/g;

const found = paragraph.match(regex);

console.log(found);

// expected output: Array ["Value1", "Value2"]

Answer 1

试试这个正则表达式：([a-zA-Z]+[a-zA-Z0-9_]+)(?<![and|or|not]). 我刚刚更新了您的代码中的正则表达式，请测试并让我知道您是否有任何疑问。const paragraph = '(Value1==6) and (Value2==0)?1:0';const regex = /([a-zA-Z]+[a-zA-Z0-9_]+)(?<![and|or|not])/g;const found = paragraph.match(regex);console.log(found);

Answer 2

您更新的问题完全改变了输入的性质。如果输入如此不同，您将需要匹配任何不以 、 或 以外的数字开头的“单词”&nbsp;and（or但这not符合您最初的尝试，所以我想这是有道理的） :const&nbsp;regex&nbsp;=&nbsp;/(?!and|or|not)\b[A-Z]\w*/gi;实例：const tests = [&nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; str: "(Value1==6) and or not (Value2==0)?1:0",&nbsp; &nbsp; &nbsp; &nbsp; expect: ["Value1", "Value2"]&nbsp; &nbsp; },&nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; str: "Value_1",&nbsp; &nbsp; &nbsp; &nbsp; expect: ["Value_1"]&nbsp; &nbsp; },&nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; str: "(Value_1 * Value_2)",&nbsp; &nbsp; &nbsp; &nbsp; expect: ["Value_1", "Value_2"]&nbsp; &nbsp; },&nbsp; &nbsp; {&nbsp; &nbsp; &nbsp; &nbsp; str: "Value_Machine_Outcome==4?1:0",&nbsp; &nbsp; &nbsp; &nbsp; expect: ["Value_Machine_Outcome"] // Note I put this in an array&nbsp; &nbsp; }];const regex = /(?!and|or|not)\b[A-Z]\w*/gi;for (const {str, expect} of tests) {&nbsp; &nbsp; const result = str.match(regex);&nbsp; &nbsp; const good = result.length === expect.length && result.every((v, i) => v === expect[i]);&nbsp; &nbsp; console.log(JSON.stringify(result), good ? "Ok" : "<== ERROR");}其工作原理是不允许and、or、 和not，并要求在单词边界 (&nbsp;\b) 处进行匹配。请注意，在测试中，我将Value_Machine_Outcome==4?1:0字符串的预期结果更改为数组，而不仅仅是字符串，就像所有其他结果一样。问题完全改变输入之前的原始答案：如果你想使用String.prototype.match，你可以对 a 使用正向后视（自 ES2018 起）(并匹配 a 之前的所有内容=：const&nbsp;regex&nbsp;=&nbsp;/(?<=\()[^=]+/g;实例：const paragraph = '(Value1==6) and (Value2==0)?1:0';const regex = /(?<=\()[^=]+/g;const found = paragraph.match(regex);console.log(found);// expected output: Array ["Value1", "Value2"]如果您同意循环，则可以通过使用捕获组来避免后向查找（因为它们仅在 ES2018 中添加）：const regex = /\(([^=]+)/g;const found = [];let match;while (!!(match = regex.exec(paragraph))) {&nbsp; &nbsp; found.push(match[1]);}实例：const paragraph = '(Value1==6) and (Value2==0)?1:0';const regex = /\(([^=]+)/g;const found = [];let match;while (!!(match = regex.exec(paragraph))) {&nbsp; &nbsp; found.push(match[1]);}console.log(found);// expected output: Array ["Value1", "Value2"]在评论中你问：我的表达式也可以包含下划线。就像它可以是 value_1、value_2 一样。那里能行得通吗？我说会是因为上面的两个都匹配除了=.后来你说：当我的结构包含“Value_1”时它会忽略同样，以上两者都可以与Value_1和配合使用Value_2：第一的：const paragraph = '(Value_1==6) and (Value_2==0)?1:0';const regex = /(?<=\()[^=]+/g;const found = paragraph.match(regex);console.log(found);// expected output: Array ["Value1", "Value2"]第二：const paragraph = '(Value_1==6) and (Value_2==0)?1:0';const regex = /\(([^=]+)/g;const found = [];let match;while (!!(match = regex.exec(paragraph))) {&nbsp; &nbsp; found.push(match[1]);}console.log(found);// expected output: Array ["Value1", "Value2"]