java - BigInteger 抛出 numberFormatException

Question

编辑:使用以下代码解决了问题:

    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。对此有什么帮助吗？

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);
}

这是异常(exception):

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

从字符串中删除无关的 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

没有提及像0x(或八进制的0)这样的前缀。