Question

我用XCode编写了一个相对简单的C程序，我可以将带有词汇表的无格式文本文件输入到其中，然后返回给我一个在漂亮的表中包含它们的HTML文件。在Xcode中运行时，它可以正常工作，它可以从我设置的工作目录中提取文件，然后将html文件放回到那里。不过，我不想每次运行该程序时都必须打开XCode。我尝试通过命令行进行编译，但是完成的exec找不到应该打开的文件，无论它们是在Xcode中设置的工作目录中还是与exec相同的文件中。我尝试输入整个文件路径，但这也无济于事。这是该项目的代码：

#include <stdio.h>
#include <stdlib.h>

int main() {
    char name1[50], name2[50], Eng[15][256], Kanji[15][256], Roma[15][256]; //declare file names and string arrays
    FILE *fptr; //declare file pointer

    printf("Input: ");
    scanf("%s",name2); //input input file, has to be .txt

    printf("\nOutput: ");
    scanf("%s",name1); //input output file, has to be .html

    int i, m;
    printf("\nNumber: ");
    scanf("%d",&m); //input number of kanji in list

    if ((fptr = fopen(name2,"r")) == NULL){
        printf("Error! opening file");
        // error in case input file doesn't exist
        exit(1);
    }

    for (i=0;i<m;i++) {
        fscanf(fptr,"%s%s%s", Eng[i], Kanji[i], Roma[i]); //reads the list into the arrays
    }

    fptr = fopen(name1,"w");

    fprintf(fptr,"<!DOCTYPE html>\n<head>\n<meta charset='utf-8'>\n<script src='/Volumes/PRIVAT/Hobbies/Code/ressources/jquery-3.3.1.min.js'></script>\n\n<link href='https://fonts.googleapis.com/css?family=Kosugi+Maru' rel='stylesheet'>\n\n<link rel='stylesheet' href='style.css'>\n<script src='script.js'></script>\n\n</head>\n<body>\n<table>\n<colgroup> <col id='col1' span='1'> </colgroup>\n<colgroup> <col id='col2' span='2'> </colgroup>\n\n<caption>WEEK 1</caption>\n<tr>\n<th>English</th>\n<th>Kanji</th>\n<th>Reading</th>\n</tr>"); //writes constant part of html file

    for (i=0;i<m;i++) {
        fprintf(fptr,"\n<tr>\n<td>%s</td>\n<td>%s</td>\n<td>%s</td>\n</tr>",Eng[i],Kanji[i],Roma[i]); // writes variable part of html file
    }

    fprintf(fptr,"\n</body>"); //closing html

    fclose(fptr); //closes output 


    return 0;
}

html很好，所以不要在这里尝试阅读。真的，我唯一的问题是，是否有一种方法可以编译此项目以在保留工作目录的终端中运行。感谢您的帮助！

Answer 1

您的代码对我有用。可执行文件的工作目录是运行它的位置，而不是可执行文件所在的位置。

您可以向getcwd添加呼叫，以查看程序的当前工作目录设置为（from this Stack Overflow answer）。如果perror返回fopen，则向NULL添加呼叫也很有帮助，这样您就可以了解失败的原因。

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <limits.h>

int main() {
    char name1[50], name2[50], Eng[15][256], Kanji[15][256], Roma[15][256]; //declare file names and string arrays
    FILE *fptr; //declare file pointer

    // print current working directory
    char cwd[PATH_MAX];
    if (getcwd(cwd, sizeof(cwd)) != NULL) {
        printf("Current working dir: %s\n", cwd);
    } else {
        perror("getcwd() error");
        return 1;
    }

    printf("Input: ");
    scanf("%s",name2); //input input file, has to be .txt

    printf("\nOutput: ");
    scanf("%s",name1); //input output file, has to be .html

    int i, m;
    printf("\nNumber: ");
    scanf("%d",&m); //input number of kanji in list

    if ((fptr = fopen(name2,"r")) == NULL){
        perror("Error! opening file");
        // error in case input file doesn't exist
        exit(1);
    }

    for (i=0;i<m;i++) {
        fscanf(fptr,"%s%s%s", Eng[i], Kanji[i], Roma[i]); //reads the list into the arrays
    }

    fptr = fopen(name1,"w");

    fprintf(fptr,"<!DOCTYPE html>\n<head>\n<meta charset='utf-8'>\n<script src='/Volumes/PRIVAT/Hobbies/Code/ressources/jquery-3.3.1.min.js'></script>\n\n<link href='https://fonts.googleapis.com/css?family=Kosugi+Maru' rel='stylesheet'>\n\n<link rel='stylesheet' href='style.css'>\n<script src='script.js'></script>\n\n</head>\n<body>\n<table>\n<colgroup> <col id='col1' span='1'> </colgroup>\n<colgroup> <col id='col2' span='2'> </colgroup>\n\n<caption>WEEK 1</caption>\n<tr>\n<th>English</th>\n<th>Kanji</th>\n<th>Reading</th>\n</tr>"); //writes constant part of html file

    for (i=0;i<m;i++) {
        fprintf(fptr,"\n<tr>\n<td>%s</td>\n<td>%s</td>\n<td>%s</td>\n</tr>",Eng[i],Kanji[i],Roma[i]); // writes variable part of html file
    }

    fprintf(fptr,"\n</body>"); //closing html

    fclose(fptr); //closes output 

    return 0;
}

编译为exec时如何保存xcode工作目录？

1 个答案: