如何遍歷數組並檢查數組值

Question

所以我有一個數組，並且我想遍歷數組並檢查它中的值是否與某些字符串匹配，如果是這樣，請執行一些操作。

這是我到目前爲止有：

var rgt_count = a_related_genres_leon.length; // <- 3 items 

for (var i = 0; i < rgt_count; i++) { 

    if inArray("Music", a_related_genres_leon) { 
     console.log('Music'); 
    } else if inArray("TV", a_related_genres_leon) { 
     console.log('TV'); 
    } else if inArray("Comedy", a_related_genres_leon) { 
     console.log('Comedy'); 
    } 
} 

我得到一個未捕獲的SyntaxError：意外的標記（錯誤現在

我也試過這個

if text($.inArray("Music", a_related_genres_leon)) { 
     console.log('Music'); 
    } else if text($.inArray("TV", a_related_genres_leon)) { 
     console.log('TV'); 
    } else if text($.inArray("Comedy", a_related_genres_leon)) { 
     console.log('Comedy'); 
    } 

Answer 1

無論是搜索陣列中的特定元素：

if ($.inArray("Music", a_related_genres_leon) > -1) { 
    console.log('Music'); 
} 
if ($.inArray("TV", a_related_genres_leon) > -1) { 
    console.log('TV'); 
} 
if ($.inArray("Comedy", a_related_genres_leon) > -1) { 
    console.log('Comedy'); 
} 

或迭代陣列

var rgt_count = a_related_genres_leon.length; // <- 3 items 
for (var i = 0; i < rgt_count; i++) { 
    console.log(a_related_genres_leon[i]); 
} 

中的所有元素，但不這樣做兩者的混合物。

Answer 2

試試這個：

if ($.inArray("Music", a_related_genres_leon) > -1) { 
    console.log('Music'); 
} else if ($.inArray("TV", a_related_genres_leon) > -1) { 
    console.log('TV'); 
} else if ($.inArray("Comedy", a_related_genres_leon) > -1) { 
    console.log('Comedy'); 
} 

---

From the jQuery.inArray() API Docs

The $.inArray() method is similar to JavaScript's native .indexOf() method in that it returns -1 when it doesn't find a match. If the first element within the array matches value, $.inArray() returns 0.

Because JavaScript treats 0 as loosely equal to false (i.e. 0 == false, but 0 !== false), if we're checking for the presence of value within array, we need to check if it's not equal to (or greater than) -1.

---

Answer 3

請參閱本作正確的語法和用法

https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Statements/if...else http://api.jquery.com/jQuery.inArray/

if($.inArray("Music", a_related_genres_leon) > -1) { 
    console.log('Music'); 
} else if($.inArray("TV", a_related_genres_leon) > -1) { 
    console.log('TV'); 
} else if($.inArray("Comedy", a_related_genres_leon) > -1) { 
    console.log('Comedy'); 
}