替换所有用“或”括起来的字符串

Question

text = abc "def" g'i' jklm'no' p"qrs uv w"

应该成为abc !string! g !string! jklm!string! p !string!

到目前为止：

if(/(?:'[^']+'|"[^"]+")/.test(text){
   text = text.replace(/(?:'[^']+'|"[^"]+")/, "!string!");
}   

（杰瑞的回答）

它只替换用''或'“括起来的第一个文本。这实际上与我之前的问题有关，但我认为这更复杂：How to search for the second occurence of ' or " in regex?

Answer 1

g修饰符执行全局匹配

text = text.replace(/(?:'[^']+'|"[^"]+")/g, "!string!");

Answer 2

试试这个

str = "abc \"def\" g'i' jklm'no' p\"qrs uv w\"";
mystring = str.replace(/("|')+[a-z ]+("|')/g, '!sring!');
alert(mystring);

<强>输出：

abc !string! g !string! jklm!string! p !string!

Answer 3

使用全局参数“g”

str.replace(/blue/g,"red");

Answer 4

这个怎么样

var str = "abc \"def\" g'i' jklm'no' p\"qrs uv w\"";
var regexp = /("|')[a-z ]+("|')/g;
console.log(str.replace(regexp,'!string!'));

结果：abc !string! g!string! jklm!string! p!string!