Question

我想从我的数据库（MySQL）中获取结果并将它们存储在一个数组中。我尝试过一些东西，但我没有办法解决。

    mysql_real_query(mysql, query, strlen(query));
    mysql_res = mysql_store_result(mysql);
    rows_number = (unsigned long) mysql_num_rows (mysql_res);
    printf ("Number of Rows: %lu\n\n", rows_number);

    while ((row = mysql_fetch_row (mysql_res)) != NULL) {
    for (i = 0;  i < mysql_num_fields(mysql_res);  i ++)
    printf ("%s ",row[i]); //this is printed like it should be.
    printf("\n");

    char *val = row[i]; //should save it to an char
    printf("%s",val);

但是如果我启动程序，输出是:(只提取！）这是printf发送的内容（＆＃34;％s＆＃34;，row [i]）; ：

1 1 1 0006

这是打印形式char val：

</alias>
<alias>windows-1256</alias>
<collation name="cp1256_bin"        id="67" order="Binary"  flag="binary"/>
<collation name="cp1256_general_ci" id="57" order="Arabic"  flag="primary">
<order>Arabic</order>
<order>Persian</order>
<order>Pakistani</order>
<order>Urdu</order>
</collation>
</charset>

那有什么不对？谢谢你的帮助!!

Answer 1

当您正确缩进代码时，您可以看到问题

while ((row = mysql_fetch_row (mysql_res)) != NULL) {
    for (i = 0;  i < mysql_num_fields(mysql_res);  i ++)
        printf ("%s ",row[i]); //this is printed like it should be.
    printf("\n");

    char *val = row[i]; //should save it to an char
    printf("%s",val);

分配

时，for循环已经完成

char *val = row[i];

所以，这是一个经典的错误。

Answer 2

i在for循环的边界之外使用：

而不是：

 for (i = 0;  i < mysql_num_fields(mysql_res);  i ++)
     printf ("%s ",row[i]); //this is printed like it should be.

你可能意味着这个：

 for (i = 0;  i < mysql_num_fields(mysql_res);  i ++)
 {
    printf ("%s ",row[i]); //this is printed like it should be.
    printf("\n");

    char *val = row[i]; //should save it to an char
    printf("%s",val);
 }

C - 将MySQL行取回char

2 个答案: