我正在尝试将"玩家名称"添加到数据库中的表中。它完美地插入了第一个名字,然后在第二个名字上崩溃。
public void addTeam1Members(){//TODO自动生成的方法存根
EditText playername, playerinit;
DBAccessMatch dbAccess = new DBAccessMatch(this);
String pname, pinit;
dbAccess.open();
for (int x = 0; x < team1Players; x++) {
switch (x) {
case 0:
playername = (EditText) findViewById(R.id.etTeam1Player1);
GlobalVars.setTeam1PlayerNames(playername.getText().toString(),
x);
GlobalVars.sT1P1 = playername.getText().toString();
playerinit = (EditText) findViewById(R.id.etTeam1Player1Init);
GlobalVars.setTeam1PlayerInit(playerinit.getText().toString(),
x);
if (playername.getText().toString().trim().isEmpty() == true) {
pname = " ";
} else {
pname = playername.getText().toString().trim();
}
if (playerinit.getText().toString().trim().isEmpty() == true) {
pinit = " ";
} else {
pinit = playerinit.getText().toString().trim();
}
Toast.makeText(getApplicationContext(), pname + " - " + pinit, Toast.LENGTH_SHORT).show();
dbAccess.createPlayer(pname, pinit);
break;
case 1:
playername = (EditText) findViewById(R.id.etTeam1Player2);
GlobalVars.setTeam1PlayerNames(playername.getText().toString(),
x);
GlobalVars.sT1P2 = playername.getText().toString();
playerinit = (EditText) findViewById(R.id.etTeam1Player2Init);
GlobalVars.setTeam1PlayerInit(playerinit.getText().toString(),
x);
if (playername.getText().toString().trim().isEmpty() == true) {
pname = " ";
} else {
pname = playername.getText().toString().trim();
}
if (playerinit.getText().toString().trim().isEmpty() == true) {
pinit = " ";
} else {
pinit = playerinit.getText().toString().trim();
}
Toast.makeText(getApplicationContext(), pname + " - " + pinit, Toast.LENGTH_SHORT).show();
dbAccess.createPlayer(pname, pinit);
break;
这是显示播放器1和播放器2的一小段代码。除了它们所在的编辑文本框之外,它们最终是完全相同的。
当createPlayer(name,initial)第一次完美插入时运行,第二次则不运行。
public void createPlayer(String playername, String playerinitials) {
try {
ContentValues cv = new ContentValues();
cv.put(KEY_PLAYERNAME, playername);
cv.put(KEY_PLAYERINITIALS, playerinitials);
ourDatabase.insert(DATABASE_TABLE3, null, cv);
} catch (Exception e) {
String error = e.toString();
Dialog d = new Dialog(null);
d.setTitle("Dang it!");
TextView tv = new TextView(null);
tv.setText(error);
d.setContentView(tv);
d.show();
}
}
在对这个问题进行了一些研究后,我无法确定错误是什么(使用logcat)。我把"创建用户"函数移到了其他地方,并在这个类之外进行,一切都很好。