2013-12-10 59 views

回答

4

JavaScript並不是PHP和字符串連接在這裏正在通過+而不是.

var names = ["andy","amy","randy","ronaldo","nani"]; 

for(var i = 0;i<names.length;i++){ 
    console.log("My name is " + names[i]); 
} 
+0

我不好,可怕的錯誤,你的循環笑 – user3057928

1

的錯誤是因爲.

console.log("My name is " . names[i]); 

將其更改爲逗號(,)或加號(+)的這樣

console.log("My name is ", names[i]); 
console.log("My name is " + names[i]); 
0

我想你混淆javascript拼接與php

變種名稱=「安迪」,「艾米」,「好色」,「羅」,「納尼」];

for(var i in names){ 
    console.log("My name is " + names[i]); // change . with + here 
} 

還可以最大限度地減少與var i in names

檢查這個http://jsfiddle.net/g96T8/

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