我有一行数据
211L CRYST1 60.970 60.970 97.140 90.000 90.000 120.000 P 32 2 1 6
我想用 C 进行解析。具体来说,我想将 P 32 2 1
提取为单个字符串。
当我使用 strtok 时,它使用所有空格作为分隔符,返回各个字符串
P
32
2
1
问题的更简洁的措辞:
如果我有可变数量的字符串(在本例中为 4 个),如何将它们连接成单个字符串?
到目前为止我的代码:
while (fgets(line,sizeof line, PDBlist)!=NULL)
{
p=0;
pch=strtok(line,"\t");
sprintf(space[p],"%s",pch);
while(pch!=NULL){
pch=strtok(NULL," ");
p++;
sprintf(space[p],"%s",pch);
}
for(i=8;i<(p-1);i++){
if(i==(p-2))printf("%s\n",space[i]);
else printf("%s ",space[i]);
} }*
最佳答案
如果行的格式始终如发布的示例所示,则使用 strtok()
的替代方法是 sscanf()
。它为行内容提供一定程度的验证,无需额外编码(例如,验证 float
值):
const char* input = "211L CRYST1 ....";
char first_token[32];
char second_token[32];
float float_1, float_2, float_3, float_4, float_5, float_6;
char last_token[32];
/* The '%31s' means read next sequence of non-whitespace characters
but don't read anymore than 31. 31 is used to leave space
for terminating NULL character.
'%f' is for reading a float.
'%31[^\n]' means read next sequence of characters up to newline
but don't read anymore than 31. */
if (9 == sscanf(input, "%31s %31s %f %f %f %f %f %f %31[^\n]",
first_token,
second_token,
&float_1,
&float_2,
&float_3,
&float_4,
&float_5,
&float_6,
last_token))
{
/* Successfully read 9 tokens. */
}
查看在线演示:http://ideone.com/To4ZP .
关于c - 使用strtok解析数据,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11941825/