js 零碎笔记

叁歲伎倆 2022-06-11 01:52 150阅读 0赞

# ![aHR0cHM6Ly9pbWcyMDIwLmNuYmxvZ3MuY29tL2Jsb2cvODA2OTU2LzIwMjAwNy84MDY5NTYtMjAyMDA3MzEyMzQ1MDc3MjItMTExNjgwOTkyMi5naWY][] #

# js 零碎笔记 #

1. 集合，字符串包含  
var array = \[1,2,3\]  
array.indexOf(5) = -1  
  
2. 选择器匹配  
$("input\[id^='code'\]");//id属性以code开始的所有input标签  
$("input\[id$='code'\]");//id属性以code结束的所有input标签  
$("input\[id\*='code'\]");//id属性包含code的所有input标签  
可以联想到正则表达式，也有^开头，$结尾，^就是以什么开头的意思，$以什么结尾。  
  
3. 全部替换   
"abcdabcd".replace(/b/g,"1234");  
或者   
"abcdabcd".replace(new RegExp("b","gm"),"1234");

4. 转义特殊字符 \# | " " . \[ \] ===> \\\\\# ....

function escapSpecialCharacters (str) {
    	if (null == str) {
    		return "";
    	}
    	str = str.replace(/\s/g,"\\ ");
    	str = str.replace(/\|/g,"\\|");
    	str = str.replace(/\[/g,"\\[");
    	str = str.replace(/\]/g,"\\]");
    	str = str.replace(/\./g,"\\.");
    	return str;
    }

5. 场景：实体类字段A是List集合。requestJson传给后台的是String。需要将String转为List集合

// prototype :
    var countrys = $A();
    countryObj.each(function(e){
    	if (e.checked) {
    		countrys.push(e.value);
    	}
    });
    var params = $H({});
    var requestJson = $H({});
    requestJson.set("countrys",countrys.join(",")); //join()方法用于把数组中的所有元素放入一个字符串 默认是","
    // java :
    Gson gson = new Gson();
    Map<String, String> request = gson.fromJson(requestJson, Map.class);
    String[] countrys = request.get("countrys").split(",");
    Arrays.asList(countrys);

6. 返回内容是：Object \{ price=55.51,  selectedImage=\[0\],  upc="111111",  更多...\} 情况

取值：  
for(key in args)\{  
args\[key\];  
\}

7. prototype each循环，break; continue; 使用分别是 throw $break; return $continue;

未完待更新......

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