本文介绍了ParseLong引发NumberFormatException的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
java.lang.NumberFormatException: For input string: "10"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Long.parseLong(Long.java:441)
相关代码段:
public static class NodeWritable implements Writable {
public double msg;
public double rank;
public String others;
public NodeWritable(double msg, double rank, String others) {
this.msg = msg;
this.rank = rank;
this.others = others;
}
public NodeWritable() {
this.msg = 0.0;
this.rank = 0.0;
this.others = "";
}
@Override
public void write(DataOutput out) throws IOException {
out.writeDouble(msg);
out.writeDouble(rank);
out.writeChars(others + "\n");
}
@Override
public void readFields(DataInput in) throws IOException {
msg = in.readDouble();
rank = in.readDouble();
others = in.readLine();
}
@Override
public String toString() {
return "" + rank;
}
}
ArrayList<Long> incoming_vids = new ArrayList<Long>();
for (NodeWritable msg : messages) {
String in_vid = msg.others.trim();
incoming_vids.add(Long.parseLong(in_vid));
}
这怎么会发生?我已经与Google进行了一些研究.有时 NumberFormatException
似乎是由大数字引起的.但是我只是无法为我的情况做一个解释.
How can this happen? I've done some research with Google. Sometime NumberFormatException
seems to be caused by big numbers. But I just can't find a possible explanation for my case.
推荐答案
您可以使用此
for(int i=0;i<in_vid.length();i++) {
char ch = in_vid.charAt(i);
if( Character.isDigit(ch)) {
// do something
}
}
如果不是数字,则可以在循环中将其消除,并仅传递具有数字的字符串.
if its other than digit then you can eliminate it in the loop and pass only the string which has digits.
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