BigInteger 抛出 numberFormatException

Question

编辑：解决了以下代码的问题：

    String tokenId="0x1800000000001289000000000000000000000000000000000000000000000000";

    BigInteger token1;
    if (tokenId.startsWith("0x")){
        token1=new BigInteger(tokenId.substring(2),16);

我有一个很长的字符串，我需要将其分配为 BigInteger 并将其传递给 Web3j 库的另一个方法。但是，我不断收到数字格式异常。有什么帮助吗？

下面是抛出异常的方法：

public void getBalance1155(String walletAddress) throws ExecutionException, InterruptedException {

    //define constant values

    Web3j web3j=Web3j.build(new HttpService("https://mainnet.infura.io/v3/<apiKey>>"));
    String contractAddress = "0xfaaFDc07907ff5120a76b34b731b278c38d6043C";
    BigInteger tokenId=new BigInteger("0x1800000000001289000000000000000000000000000000000000000000000000",16);
    NoOpProcessor processor = new NoOpProcessor(web3j);
    Credentials credentials = Credentials.create("privatekey");
    TransactionManager txManager = new FastRawTransactionManager(web3j, credentials, processor);

    //Query Blockchain to get balance of WALLETADDRESS from Contract for given TokenID

    ERC1155 token = ERC1155.load(contractAddress, web3j, txManager, DefaultGasProvider.GAS_PRICE, DefaultGasProvider.GAS_LIMIT);
    RemoteCall<BigInteger> sendCall = token.balanceOf(walletAddress, tokenId);
    BigInteger balance=sendCall.sendAsync().get();
    log.info("balance >>>>>> " +balance);
}

例外情况：

java.lang.NumberFormatException: For input string: "0x1800000000001289000000000000000000000000000000000000000000000000" at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) at java.base/java.lang.Long.parseLong(Long.java:692) at java.base/java.lang.Long.parseLong(Long.java:817)

Answer 1

从字符串中删除多余的 0x。

BigInteger's constructor 的文档说

The String representation consists of an optional minus or plus sign followed by a sequence of one or more digits in the specified radix.

[...] The String may not contain any extraneous characters

没有提到像 0x（或八进制的 0）这样的前缀。

Answer 2

您需要删除 0x，您可以通过在 BigInteger

上使用 toString(16) 来检索六进制值

BigInteger tokenId=new BigInteger("1800000000001289000000000000000000000000000000000000000000000000",16);
System.out.println("tokenId.toString(16) = " + tokenId.toString(16));
System.out.println("tokenId.toString(10) = " + tokenId.toString(10));

String originalString = "0x" + tokenId.toString(16);
System.out.println("originalString = " + originalString);

输出：

tokenId.toString(16) = 1800000000001289000000000000000000000000000000000000000000000000
tokenId.toString(10) = 10855508365998423105807514254364715762064874182780947284375732482585619595264
originalString = 0x1800000000001289000000000000000000000000000000000000000000000000

Answer 3

由于您已经在使用 web3j，因此您可以使用 Numeric.decodeQuantity("0x1800000000001289000000000000000000000000000000000000000000000000") 将十六进制解码为 BigInteger

BigInteger 抛出 numberFormatException

BigInteger throws numberFormatException

java

biginteger

web3-java