java 比较两个字符串数组以检查输入的值是否存在于字符串数组中
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comparing two string arrays to check if the entered value is there in the string array
提问by sahana
I have two string arrays
我有两个字符串数组
String[] Mids contains
MSI
MSU
MSV
String[] sl contains
MSU
MSV
MSS
Actual output should be
实际输出应该是
Action
Action
Action cannot be set
for(int i=0;i<sl.length;i++){
for(int j=0;j<Mids.length;j++){
if((sl[i].equals(Mids[j]))){
System.out.println("Action");
}else{
System.out.println("Action cannot be set");
}
}
}
OUTPUT which i get
我得到的输出
Action cannot be set
Action cannot be set
Action cannot be set
Action cannot be set
Action
Action cannot be set
Action
Action cannot be set
Action cannot be set
回答by RoflcoptrException
The problem is you're iterating over both arrays and always printing if you have found the same value. But you should do this only in the first loop. I changed the for loop:
问题是您要遍历两个数组，并且如果找到相同的值，则总是打印。但是您应该只在第一个循环中执行此操作。我改变了 for 循环：
for(int i=0;i<sl.length;i++){
boolean found = false;
for(int j=0;j<Mids.length;j++){
if((sl[i].equals(Mids[j]))){
found = true;
break;
}
}
if (found) {
stdOut.println("Action");
} else {
stdOut.println("Action cannot be set");
}
}
回答by codaddict
To say if an element is not found in an array you need to compare it with all the elements. Just because one comparison fails you cannot conclude that it is not found in the array.
要说是否在数组中找不到元素，您需要将它与所有元素进行比较。不能仅仅因为一个比较失败就断定它没有在数组中找到。
Try something like:
尝试类似：
for(int i=0;i<sl.length;i++){
boolean found = false;
for(int j=0;j<Mids.length;j++){
if((sl[i].equals(Mids[j]))){
found = true;
break;
}
}
if(found) {
// print found.
} else {
// print not found.
}
}
回答by Andrew A.
Another way to do it, with fewer lines of code and fewer iterations would be:
使用更少的代码行和更少的迭代来做到这一点的另一种方法是：
List<String> midsList = new ArrayList<String>(Arrays.asList(Mids));
for (String string : sl) {
if (midsList.contains(string)) {
System.out.println("Action");
} else {
System.out.println("Action cannot be set");
}
}
回答by tamarintech
Why don't you add another printline under the for loop(i) that displays s1 and mids so that you can better understand the execution?
为什么不在显示 s1 和 mids 的 for loop(i) 下添加另一个打印行，以便更好地理解执行？
