在Android中拆分字符串后,它不会包含一些拆分部分。
代码在for循环中,其中SetServerString每次迭代都在改变,它适用于10个元素中的6个但是由于某种原因,当字符串是Switch它不起作用时,在Switch字符串之前有一个空格所以我有删除但仍然无法工作..任何建议请..
我正在尝试拆分字符串,以便从字符串中获取设备名称和节点编号,然后将其插入到List视图和数据库中。
SetServerString ="开关2:0" //在Switch之前有一个空格
for (int i = 0; i < sizeo; i++) {
...
if(SetServerString.substring(0,1).equals(dashremove)){
SetServerString=SetServerString.substring(1,SetServerString.length()-1);
Log.v("NEW", "Contains setset " +SetServerString + "sixe "+ String.valueOf(SetServerString.length()) );
}
SetServerString = SetServerString.replaceAll("\\W", "-");
Log.v("NEW", "the name of the device is : " +SetServerString ); // its ok here
GWave.Gdevicenameo.add(SetServerString);
String node = SetServerString;
Log.v("NEW", "the name of the node is : " +node ); // still on here
String string = node;
String[] parts = string.split("-"); // disappears here
String part1 = parts[0];
String part2 = parts[1];
String part3 = parts[2];
String part4 = parts[3];
String part5 = parts[4];
String devnamee = part1 + " " + part2 + " " + part3 ;
String devnamee2 = part1 + " " + part2 + " " + part3 + " "+ part4;
Log.v("NEW","node "+ node +" devname1 "+ devnamee2+ " devname2 "+ devnamee2);
.....
else if(devnamee2.contains("Switch")){
Log.v("ZZ","I AM HERE SWITCH");
String name = devnamee2.substring(devnamee2.length()-2,devnamee2.length()-1);
if(name.equals(" "))
{
devnamee2=devnamee2.substring(0, devnamee2.length()-2);
//Log.v("ZZ","NOT " +devnamee2);
}
devicenamelist.add(devnamee);
devicenodelist.add(part3);
Log.v("ZZ","NODE LIST " +part3 +"| Sensor" + part1 + " p2 " + part2 + " p3 " + part3 + " p4 "+ part4 + " p5 "+ part5) ;
GWave.Gdevicename.add(SetServerString);
}