Java 为什么我的字符串到字符串比较失败？
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Why is my String to String comparison failing?
提问by Chuck Hriczko
I have an Android app where I want to check to see if an app name that is installed matches a string passed to the function containing this code. The code and example is below:
我有一个 Android 应用程序，我想在其中检查已安装的应用程序名称是否与传递给包含此代码的函数的字符串匹配。代码和示例如下：
private Boolean checkInstalledApp(String appName){
PackageManager pm = this.getPackageManager();
Intent mainIntent = new Intent(Intent.ACTION_MAIN, null);
mainIntent.addCategory(Intent.CATEGORY_LAUNCHER);
List<ResolveInfo> list = pm.queryIntentActivities(mainIntent, 0);
Boolean isInstalled = false;
for(ResolveInfo info: list) {
if (info.activityInfo.applicationInfo.loadLabel( pm ).toString()==appName){
isInstalled = true;
break;
}
}
return isInstalled;
}
Assuming you called checkInstalledApp("SetCPU");and the app name on the phone is called the same thing it should return true. However, it never does. I logged the results and it should match up but it does not. Can anyone please enlighten me as to why this doesn't work?
假设您checkInstalledApp("SetCPU");拨打电话并且手机上的应用程序名称与它应该返回的名称相同true。然而，它从来没有。我记录了结果，它应该匹配，但它不匹配。任何人都可以请教我为什么这不起作用？
采纳答案by Eton B.
Use the String's equals() method instead of the == operator for comparing strings:
使用 String 的 equals() 方法而不是 == 运算符来比较字符串：
info.activityInfo.applicationInfo.loadLabel( pm ).toString().equals(appName)
In Java, one of the most common mistakes newcomers meet is using ==to compare Strings. You have to remember, ==compares the object references, not the content.
在 Java 中，新手遇到的最常见错误之一是使用==比较字符串。您必须记住，==比较的是对象引用，而不是内容。
回答by Blundell
Check out number 7: Top 10 - New Java Developer Errors
查看第 7 项：前 10 名 - 新的 Java 开发人员错误
回答by Bouchehboun Saad
public static boolean compaireString (String string, String string2)
{
// string == null && String2 == null or they reference the same object
if (string == string2) return true;
//we have to be sure that string is not null before calling a methode on it
if (string != null && string.equals(string2)) return true;
return false;
}
