在我的自定义ListView项上，为什么（string.length（）<= 0）始终为true，而（string ==“””）始终为false？

Question

I've created a custom ListView adapter/item. The item contains a CheckedTextView and a TextView.

In the adapter's getView() method, the custom item is created and its setTask(Task) method is called right before it's returned.

The setTask(Task) method gets 2 Strings and a boolean from the Task object passed in as a parameter. (I do recognize that its address is actually what gets passed.)

One of the Strings is assigned to the custom item's TextView's text property IF that String contains text. Otherwise, the TextView's visible property is set to "gone."

To make that determination, I check to see if the String's length is less than or equal to 0... it produces unexpected behavior - it ALWAYS evaluates true.

If I change it to check if the String is equal to "", it always returns false.

This leads me to believe that the String I'm checking is null. Why would that be?

protected void onFinishInflate() {
    super.onFinishInflate();
    checkbox = (CheckedTextView)findViewById(android.R.id.text1);   
    description = (TextView)findViewById(R.id.description);
}

public void setTask(Task t) {
    task = t;
    checkbox.setText(t.getName());
    checkbox.setChecked(t.isComplete());
    if (t.getDescription().length() <= 0)
        description.setVisibility(GONE);
    else
        description.setText(t.getDescription());
}

And here's the getView() method in the adapter:

public View getView(int position, View convertView, ViewGroup parent) {

    TaskListItem tli;
    if (convertView == null)
        tli = (TaskListItem)View.inflate(context, R.layout.task_list_item, null);
    else
        tli = (TaskListItem)convertView;

    tli.setTask(currentTasks.get(position));
    return tli; 
}

Answer 1

myString == ""

checks the reference of the string object rather than the content. You should use:

"".equals(myString)

which checks the content of the string, rather then the reference. You could use:

myString.equals("")

but that has a chance of a NullPointerException if myString is null.

Answer 2

In Java you can't use == to compare the contents of a String with another String. The == operator always compares the reference or memory address of the variables. In other languages == can be made to compare the contents of objects. In Java it cannot. Use .equals() or .equalsIgnoreCase().

Answer 3

You're comparing strings using == which is always a bad idea with Objects. Your empty string "" points to a different String object than t.getDescription(), though the values may be the same. With Strings sometimes == will work because the String class maintains a pool of Strings and will it use for string literals and constants. A demonstration of how String works is better than words, however.

public class StringTest
{
  public static void main(String[] args)
  {
    String s0="";  // literal
    System.out.println("Length of s0 = " + s0.length());
    System.out.println("Does s0 point to same object as empty string? " 
      + (s0==""));  // since both are literals the same object is used for both
    System.out.println("Does s0 equal an empty string? "
      + s0.equals("")); // comparing values
    System.out.println("Is s0 empty? " + s0.isEmpty());
    System.out.println("");

    StringBuilder sb = new StringBuilder("");
    String s1 = sb.toString(); // don't use a literal for s1
    System.out.println("Length of s1 = " + s1.length());
    System.out.println("Does s1 point to same object as empty string? " 
      + (s1==""));  // s1 is a different object
    System.out.println("Does s1 equal an empty string? " + s1.equals("")); 
    System.out.println("Is s1 empty? " + s1.isEmpty());
    System.out.println("");

    // Do this like s1 but with a twist
    sb = new StringBuilder("");
    String s2 = sb.toString(); // again not a literal
    System.out.println("Does s2 point to same object as empty string? " 
      + (s2=="")); 
    System.out.println("Length of s2 = " + s0.length());
    s2 = s2.intern(); // returns the object from the pool that equals() s2
    System.out.println("Now does s2 point to same object as empty string? " 
      + (s2==""));  // s2 now has the reference to the empty string in the pool
    System.out.println("Does s2 equal an empty string? " + s2.equals(""));
    System.out.println("Is s2 empty? " + s2.isEmpty());
  }
}

I hope this helps.

Answer 4

Additionally, you should check t.getDescription() before using .length() on it.

if(t.getDescription()!=null){
(t.getDescription().length()>0)?description.setText(t.getDescription()):description.setVisibility(GONE);}
else { description.setVisibility(GONE);}

To avoid playing with String, insure that t->description always start as null instead of empty String. This can be done in Class constructor.

Answer 5

As it turns out, other than comparing String object directly, there is another serious bug which JUST occurred to me. (Please correct me if I'm wrong.)

Because views in a ListView are sometimes recycled, the "tli" reference sometimes references a view who's visibility property is already set to "gone." That property never gets changed back to "visible" even if the String ISN'T empty. I just had to explicitly set that property under both conditions. Here's the fixed code:

public void setTask(Task t) {
    task = t;

    checkbox.setText(task.getName());
    checkbox.setChecked(task.isComplete());
    if (task.getDescription().equals("")) 
        description.setVisibility(GONE);
    else
        description.setVisibility(VISIBLE);
        description.setText(task.getDescription());
}

It now functions whether or not I check the content or the length of the String from task.getDescription().

Answer 6

The == operator checks if both sides refer to the same object. Two strings that have the same contents don't (usually) refer to the same object, see below.

String s = "";
String t = "";
if (s == t){
  //false! s and t refer to different objects
}
if (s.equals(t) && t.equals(s)){ //these two are equivalent
  //true!
}
s = t;
if (s == t){
  //now they refer to the same object, so true!
}