How to return string over 32 characters from bytes32[] array?

Question

I have a bytes32[] array. I convert strings to bytes32 before they are added to the array as follows:

function stringToBytes32(string memory source) returns (bytes32 result) {
    assembly {
    result := mload(add(source, 32))
    }
}

which will turn the string into a hex like 0x5468697320697320612074657374206d65737361676520746861742077696c6c, for example.

However, if the string converted into bytes32 is over 32 characters, I don't know how to turn it back into its original string over 32 characters. Using web3.toAscii only brings back the first 32 characters.

Answer 1

struct DocumentStruct{bytes32 string1; bytes32 string2; bytes32 string3;}
mapping(bytes32 => DocumentStruct) public documentStructs;

function StoreDocument(bytes32 key,bytes32 string1,bytes32 string2,bytes32 string3) onlyOwner returns (bool success) {
    documentStructs[key].string1 = value;
    documentStructs[key].string2 = name;
    documentStructs[key].string3 = name;
    return true;
}

function RetrieveData(bytes32 key) public constant returns(string,string,string) {
    var d = bytes32ToString(documentStructs[key].string1);
    var e = bytes32ToString(documentStructs[key].string2);
    var f = bytes32ToString(documentStructs[key].string3);
    return (d,e,f);
}
function bytes32ToString(bytes32 x) constant returns (string) {
bytes memory bytesString = new bytes(32);
uint charCount = 0;
for (uint j = 0; j < 32; j++) {
    byte char = byte(bytes32(uint(x) * 2 ** (8 * j)));
    if (char != 0) {
        bytesString[charCount] = char;
        charCount++;
    }
}
bytes memory bytesStringTrimmed = new bytes(charCount);
for (j = 0; j < charCount; j++) {
    bytesStringTrimmed[j] = bytesString[j];
}
return string(bytesStringTrimmed);
}

This works pretty well for me. I would agree that its a little complex to encode and decode, but just run your 32 byte array through it to get a string back

Edit - Here's some added stuff. A rather hacky way, but it might work for you. I'm sure Rob will yell at this answer though