Question

我试图将csv文件中给出的数据解析为“数据”文件中的ID，AGE和GPA字段，但是我认为我做的不正确（当我尝试打印数据时，打印奇怪的数字）。我在做什么错了？

char data[1000];
FILE *x = fopen("database.csv","rt");
char NAME[300];
int ID[300],AGE[300],GPA[300];
int i,j;
i = 0;

while(!feof(x)) {

        fgets(data,999,x);

        for (j = 0; j < 300 && data[i] != ','; j++, i++) {
                ID[j] = data[i];
                i++;
        }

        for (j = 0; j < 300 && data[i] != ','; j++, i++) {
                NAME[j] = data[i];
                i++;
        }

        for (j = 0; j < 300 && ( data[i] != '\0' || data[i] != '\r' || data[i] != data[i] != '\n'); j++, i++) {

                GPA[j] = data[i];

        }

}

Answer 1

首先：对于您正在执行的操作，您可能需要仔细查看函数strtok和atoi宏。但是考虑到您发布的代码，这也许仍然有点太先进了，所以我在这里花了更长的时间。

假设该行类似于

172,924,1182

然后您需要解析这些数字。数字172实际上由内存中的两个或四个字节表示，格式非常不同，而字节“ 0”与数字0完全不同。您将看到ASCII码，十进制为48，或者十六进制为0x30。

如果将一个数字的ASCII值减去48，您将得到一个数字，因为幸运的是数字是按数字顺序存储的，因此“ 0”是48，“ 1”是49，依此类推。 / p>

但是您仍然存在将三个数字1 7 2转换为172的问题。

因此，一旦有了“数据”：（我添加了注释代码来处理CSV内未加引号，未转义的文本字段，因为在您的问题中提到了AGE字段，但随后您似乎想使用NAME字段。在文本字段加引号或完全逃脱了另一罐蠕虫）

size_t i   = 0;
int number = 0;
int c;
int field = 0; // Fields start at 0 (ID).
// size_t x = 0;

// A for loop that never ends until we issue a "break"
for(;;) {
    c = data[i++];
    // What character did we just read?
    if ((',' == c) || (0x0c == c) || (0x0a == c) || (0x00 == c)) {
        // We have completed read of a number field. Which field was it?
        switch(field) {
            case 0: ID[j] = number; break;
            case 1: AGE[j] = number; break;
            // case 1: NAME[j][x] = 0; break; // we have already read in NAME, but we need the ASCIIZ string terminator.
            case 2: GPA[j] = number; break;
        }
        // Are we at the end of line?
        if ((0x0a == c) || (0x0c == c)) {
            // Yes, break the cycle and read the next line
            break;
        }
        // Read the next field. Reinitialize number.
        field++;
        number = 0;
        // x = 0; // if we had another text field
        continue;
    }
    // Each time we get a digit, the old value of number is shifted one order of magnitude, and c gets added. This is called Horner's algorithm:
    // Number   Read    You get
    // 0        "1"     0*10+1 = 1
    // 1        "7"     1*10+7 = 17
    // 17       "2"     17*10+2 = 172
    // 172      ","     Finished. Store 172 in the appropriate place.
    if (c >= '0' && c <= '9') {
        number = number * 10 + (c - '0');
    }
    /*
       switch (field) {
           case 1:
               NAME[j][x++] = c; 
               break;
       }
    */
}

解析CSV文件问题

1 个答案: