为什么在将字符串转换为整数时会出现NumberFormatException?

时间:2012-11-09 15:56:53

标签: java parsing

这是我的代码:

 public void setAreaAccessPoints(){
    String mac = "",essid = "",status = "";
    int strength = 0,kanali = 0;
    List<String> AccessPoints = new ArrayList<String>(); //i lista me ta access points
    String temp;
    try{
        String[] command = {"/bin/sh", "-c", "sudo iwlist " + wirelessName + " scanning | grep -A5 \"Cell\" "};
        Process child = Runtime.getRuntime().exec(command);
        BufferedReader r = new BufferedReader(new InputStreamReader(child.getInputStream()));
        while((temp = r.readLine()) != null){
            if(temp.contains("Cell")){
                String[] info = temp.split(" ");
                mac = info[3];
                System.out.println(mac);
                do{
                    temp = r.readLine();
                    if(temp.contains("ESSID:")){
                        essid = temp.replace("ESSID:","");
                    }
                    if(temp.contains("Frequency:")){
                        String[] info1 = temp.split(" ");
                        info1[3] = info1[3].replace(")","");
                        kanali = Integer.parseInt(info1[3]);
                    }
                    if(temp.contains("Mode:")){
                        status = temp.replace("Mode:","");
                    }
                    if(temp.contains("Quality=")){
                        String[] info2 = temp.split(" ");
                        info2[3] = info2[3].replace("level=","");
                        strength = Integer.parseInt(info2[3]);
                    }
                    if(temp.contains("Protocol:")){
                        temp = r.readLine();
                    }
                }while(!(temp.contains("Cell")));
                AccessPoint newAP = new AccessPoint(mac,essid,kanali,status,strength); 
                AccessPoints.add(newAP.toString());  //vazoume ta access points sti lista san strings
            }
        }
        r.close();
        for(String s : AccessPoints)
            System.out.println(s);
    }catch(IOException e){e.printStackTrace();}
}

我正在解析的输出如下所示:

      Cell 04 - Address: 00:05:59:30:C1:7C
                Protocol:802.11b/g
                ESSID:"NA home"
                Mode:Managed
                Frequency:2.437 GHz (Channel 6)
                Quality=2/100  Signal level=-89 dBm  Noise level=-92 dBm
  --
      Cell 05 - Address: 00:05:59:43:AE:C9
                Protocol:802.11b/g
                ESSID:"NetFasteR IAD 2 (PSTN)"
                Mode:Managed
                Frequency:2.437 GHz (Channel 6)
                Quality=0/100  Signal level=-91 dBm  Noise level=-94 dBm
  --
      Cell 06 - Address: 00:05:59:3B:C1:FA
                Protocol:802.11b/g
                ESSID:"Kpanagiotou"
                Mode:Managed
                Frequency:2.437 GHz (Channel 6)
                Quality=0/100  Signal level=-91 dBm  Noise level=-94 dBm
  --

错误在2行"strength = Integer.parseInt(info2[2]);""kanali = Integer.parseInt(info1[3]);"中......我似乎无法弄清问题在哪里。当我分割字符串时,我想要的信息是根据输出在第二个和第三个字段中。那么为什么它会尝试传递null字符串进行整数解析?

StackTrace:

 Exception in thread "Thread-1" java.lang.NumberFormatException: For input string: ""
    at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
    at java.lang.Integer.parseInt(Integer.java:504)
    at java.lang.Integer.parseInt(Integer.java:527)
    at askisi1.Wireless.setAreaAccessPoints(Wireless.java:213)
    at askisi1.Wireless.run(Wireless.java:43)
    at java.lang.Thread.run(Thread.java:722)

4 个答案:

答案 0 :(得分:1)

你正在检查某些东西并替换其他东西......

 if(temp.contains("Quality=")){
                    String[] info2 = temp.split(" ");
                    info2[1] = info2[1].replace("level=","");
                    strength = Integer.parseInt(info2[2]);
                }

答案 1 :(得分:1)

该行

Quality=2/100  Signal level=-89 dBm  Noise level=-92 dBm

包含双重空格,因此在split结果中,您有空String s,非空String s不在您认为的索引处。

打印出带有指数的"Quality=2/100 Signal level=-89 dBm Noise level=-92 dBm".split(" ");的结果

0: Quality=2/100
1: 
2: Signal
3: level=-89
4: dBm
5: 
6: Noise
7: level=-92
8: dBm

您的文件中似乎有前导空格,因此非空String s的索引将会更晚。

答案 2 :(得分:0)

由于行前面有空格,因此格式错误 没有任何空格,你的逻辑正在运作

你可以修剪()并检查你的结果

 temp = r.readLine().trim();

例如,这是有效的

String str ="Frequency:2.437 GHz (Channel 6)";
String [] str1 = str.split(" ");
System.out.println(Integer.parseInt(str1[3].replace(")", "")));

答案 3 :(得分:0)

您从info2 [1]中删除“level =”,然后再对其执行任何操作。你应该从info2 [2]中删除它。另请注意,它们之间用双倍空格分隔,如此处的另一个答案中所述。