Question

所以我想取一个字符串并删除它中的所有[]以及那些[] s中的所有文本/字符，但只删除名称后面的那些字符/字符。我不确定是否应该使用正则表达式或其他方法示例：

"Name []"` to `"Name"
"Another name [text123]"` to `"Another name"
"Yet another name [a] [] [c]"` to `"Yet another name"
"[So many names] [test] []"` to `"[So many names]" 
"[TITLE] name []"` -> `"name"
"[txt2] [namez] [] a"` -> `"a"

Answer 1

它会运行一些正则表达式。

示例如下

var originalStrings1 = '[So many names] [test] []';
var originalStrings2 = '[TITLE] name []';

console.log(getName(originalStrings1));
console.log(getName(originalStrings2));

function getName(str)
{
	var hasName = str.replace(/\[(.*?)\]/gm, '');
	var noName = str.match(/\[(.*?)\]/gm);
	if(isEmpty(hasName))
	{
		return 'Name is: '+noName[0];
	}
	else
	{
		return 'Name is: '+hasName.replace(/^\s+|\s+$/g,'');
	}
}

function isEmpty(str)
{
	return str.replace(/^\s+|\s+$/g, '').length == 0;
}

Answer 2

我当然使用REGEX，更快，更强，更容易。

以下是一个工作片段：

＆＃13;

var str = "Yet another name [a] [] [c]";
var res = str.replace(/\[.*\]/i, " ");
document.getElementById('res').innerHTML = res;

＆＃13;

<div id="res"></div>

＆＃13;

从字符串中删除＆＃34; [<content>]＆＃34; s

2 个答案: