比较串来串元素的数组

Question

所以，我想从字符串比较字符串

String [] string = new String [10]; 
string [1] = "pencil"; 
string [2] = "pen"; 
string [3] = "eraser"; 

的数组如何比较字符串，字符串从上面的阵列？

Answer 1

更快的方法是使用此代码

if(Arrays.asList(string).contains(search_string)) 

Answer 2

如果我理解正确，写一个循环，其中迭代多个字符串（示例数组）和等于src字符串。

String [] string = new String [10]; 
... 
String src = ..;//src string 

for(String string : string){ 
    if(src.equals(string)){ 
     //Equal 
    } 
} 

Answer 3

尝试使用这样的：

for (int i = 0; i < string.length; i++) { 
    if (ch.equals(string[i])) { // ch is the string to compare 
    System.out.println("equal");//or whatever your action is 
    } 
} 

编辑：

如果你正在寻找做的是验证搜索到的字符串是否出现在字符串[i]中，您可以尝试：

string[0] = "pentagon"; 
string[1] = "pencil"; 
string[2] = "pen"; 
string[3] = "eraser"; 
string[4] = "penny"; 
string[5] = "pen"; 
string[6] = "penguin"; 
string[7] = "charp"; 
string[8] = "charpen"; 
string[9] = ""; 
String ch = "pen"; 

然后测试，如果数组元素包含搜索字符串：

for (int i = 0; i < string.length; i++) { 
    if (string[i].indexOf(ch) != -1) { 
    System.out.println(string[i]+" contains "+ch); 
    } else 
    System.out.println(string[i]+" doesn't contain "+ch); 
} 

编辑2：

int i=0; 
Boolean found=false; 
while(i < string.length && !found) { 
    if (string[i].indexOf(ch) != -1) { 
    found=true; 
    } 
i++; 
} 
if(found){ 
system.out.println(ch+" is found"); 
}