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I have two textbox one is named adminname and other is adminpassword. I want to get input from these two text box and use in if condition but i am not getting the result
here is my Code
String name=adminname.getText();
        String password=adminpassword.getText();
        if(name="anoop" && password="sharma")
        {
            JOptionPane.showMessageDialog(rootPane, "Login successfull");
        }
Please Help me in finding my mistake/error.
Thanx in advance
Posted 26-May-13 3:00am
Comments
ThePhantomUpvoter 26-May-13 8:05am
   
You need to use equals() to test strings for equality in java.
Anoop Kr Sharma 26-May-13 8:09am
   
@ThePhantomUpvoter can you help me in creating the code using equals()
ThePhantomUpvoter 26-May-13 8:15am
   
You are kidding right? You are seriously unable to to write if(name.equals("anoop")?
Anoop Kr Sharma 26-May-13 8:22am
   
Thanx ThePhantomUpvoter for solving my problem
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Solution 3

if(new String("test").equals("test")) 
{
  // Expression is true
}
 
if(new String("test") == "test")
{
  // Expresson is false since you are not comparing 
  // the same instance of the object to itself
}

Best regards
Espen Harlinn
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Comments
   
Sure, a 5.
—SA
Espen Harlinn 6-Jun-13 10:21am
   
Thank you, Sergey :-D
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Solution 1

In Java, '=' is not a comparison operator but assignment. For comparison, you need to replace it with '=='.

—SA
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Solution 4

For assiging we use single = sign, but for comparing we use double = sign.
String name=adminname.getText();
        String password=adminpassword.getText();
        if(name=="anoop" && password=="sharma")
        {
            JOptionPane.showMessageDialog(rootPane, "Login successfull");
        }
Ref :
http://www.pointbaba.com/faq/65/username-and-password-checking-in-java[^]
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Comments
Richard MacCutchan 12-Sep-15 4:03am
   
This question is more that two years old and has already been answered. Also your "solution" is incorrect, which you could have discovered by reading the others.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

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