BigInteger引发numberFormatException

debugcn 发表于 Dev

角斗士9120

编辑：使用以下代码解决了该问题：

    String tokenId="0x1800000000001289000000000000000000000000000000000000000000000000";

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

我有一个长字符串，需要将其分配为BigInteger并将其传递给Web3j库的另一种方法。但是，我一直收到数字格式异常。有什么帮助吗？

下面是方法抛出异常：

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);
}

这是例外：

java.lang.NumberFormatException：对于输入字符串：java.base / java.lang.NumberFormatException.forInputString（NumberFormatException.java:65）处的java.base / java.lang.Long.parseLong（Long.java:692）为“ 0x1800000000001289000000000000000000000000000000000000000000000000000000” ），位于java.base / java.lang.Long.parseLong（Long.java:817）

安东尼·丹尼尔

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

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