Question

Hy guys！

我需要一点帮助，因为我无法弄清楚我做错了什么。我正在写一个信封写作程序。该程序有4个JTextField。姓名，城市，地址，邮政编码。我在Name JTextField上使用keylistener来识别Enter按钮，并添加一个新的Name JTextField。（基本上，它设置第二个名称JTextField可见）。总共可以有4个名称JTextField和其他（城市，地址，邮政编码）共7个。还有打印jbutton，jtable用于查看已写入的信封等待打印等。当我使用g.drawString（）时，我使用一个预定义的x，y值，它根据Font Metrics的StringWidth（）方法测量的Names字段的文本长度动态变化，并用右侧边缘值计算..无论如何。我的主要问题在于：

if (name.length() > address.length()) {

            g.drawString(name,
                (250 - nameCord_X) + 46, 214);       //this is working
            g.drawString(city, (250 - nameCord_X) + 46, 226);
            g.drawString(address, (250 - nameCord_X) + 46, 238);
            g.drawString(postal, (250 - nameCord_X + 46, 250);
        } else {
            if (address.length() > name.length()
                && name2.isEmpty())
            { // this is working
                g.drawString(name, (250 - addressCord_X) + 46, 214);
                g.drawString(city, (250 - addressCord_X) + 46, 226);
                g.drawString(address, (250 - addressCord_X) + 46, 238);
                g.drawString(postal, (250 - addressCord_X + 46, 250);

            }
}


                if (name2.length() > name.length()
                    && name2.length() > address.length()
                    && name2.isVisible())
                {// this is working

                    g.drawString(name, (250 - name2Cord_X) + 46, 202);
                    g.drawString(name2, (250 - name2Cord_X) + 46, 214);
                    g.drawString(city, (250 - name2Cord_X) + 46, 226);
                    g.drawString(address, (250 - name2Cord_X) + 46, 238);
                    g.drawString(postal, (250 - name2Cord_X) + 46, 250);

                }
      //This is the part that doesnt working. it prints out 5 lines
      // but the 2nd line is the same as first line,
      // and the real 2nd line value is printed over.
// For example Name: Lion Name2:Lion (altough I typed for
// example horse into name2 field, and the program draw
// the horse string over the 2nd line which contains Lion.)  
//Altough it prints out the all 5 row, it messing only with
//the second row :-/       
                if( (name.length() > name2.length()
                    && name.length() > address.length()
                    && name2.isVisible()))
                {
                    g.drawString(name, (250 - nameCord_X) + 46, 202);
                    g.drawString(name2, (250 - nameCord_X) + 46, 214);
                    g.drawString(city, (250 - nameCord_X) + 46, 226);
                    g.drawString(address, (250 - nameCord_X) + 46, 238);
                    g.drawString(postal, (250 - nameCord_X) + 46, 250);
                   }

Answer 1

那里有一些复杂的if块。还不清楚变量的类型是什么。它们似乎是String，但是您在isVisible()上致电name2。

我们了解您使用nameCord_X，addressCord_X，name2Cord_X进行对齐。您希望所有字符串都保持对齐并在该点开始绘制，以允许最大的字符串正确匹配，因此它不会超出信封的右侧。这是对的吗？

如果是，那么我建议您首先找到上述长度的最大值。只有当它具有有意义的值时，才应考虑每一个。然后，您可以执行以下操作：

int max = ... // maximum of the above values - only not empty values should be taken into account
int dy = 12;  // it seems you always leave 12 pixels between the lines
int y = ...;  // the value with the topmost y. It seems you use 202 when both names are present 214 when name2 is not given
int x = 250 - max + 46;

String[] labels = new String[] {name, name2, city, address, postal};
for (String label : labels) {
    if (label == null || label.length() == 0) {
        continue; // ignore null or empty values
    }
    g.drawString(name, x, y);
    y += dy;
}

我希望它有所帮助

在信封上绘图文本

1 个答案: