2015-02-10 74 views
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我建立一個POS應用VeriFone公司(C語言),它應與來自摩洛哥M2M開關通信,但發送應該有這樣(08\00)但反斜線初始化消息時,我被困當發送這個我有08\5c00。 它用十六進制(5c)中的值轉換反斜槓。我正在使用的工具是套接字工作臺來模擬服務器。 如何在不轉換爲\5c的情況下發送反斜槓? 它需要用C語言完成。發送一個反斜槓用C插座ISO8583消息

編輯

這是我要發送到與頭中的服務器上的數據,但嘗試打印\00當我\5C00

sprintf(data,"%s%s%s%s%s%s%s%s%s%s%s%s%s","\x30\x60\x60\x20\x15\x35\x35","\x08",‌"\\00","\x0x00","\x01\x30\x30\x30\x30\xC0\x30\x30\x30\x30","\x97","\\00","\x30\x30"‌,"\x00\x00\x01\x00","\x02",idTerminal,idCommercant,"\x20\x20\x20\xA4\xBC"); 
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向我們展示相關的部分代碼的。 – 2015-02-10 11:32:31

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也許服務器正在接收'\',但將其打印爲'\ 5c'? – 2015-02-10 11:43:35

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這是我想向服務器發送與它的頭,但是當我要打印\ 00我有\ 5C00數據:\t \t \t的sprintf(數據,「%s%s%S%S%s%S% S%S%S%S%S%S%S 「 」\ X30 \ X60 \ X60 \ X20 \ X15 \ X35 \ X35「, 」\ X08「, 」\\ 00「, 」\ x0x00「,」 \ X01 \ X30 \ X30 \ X30 \ X30 \ XC0 \ X30 \ X30 \ X30 \ X30" , 「\ X97」, 「\\ 00」, 「\ X30 \ X30」, 「\ X00 \ X00 \ X01 \ X00」 「\ X02」,idTerminal,idCommercant, 「\ X20 \ X20 \ X20 \ XA4 \ XBC」); – knk 2015-02-10 13:18:34

回答

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如果我理解正確的話,第一部分你的例子:

sprintf(data,"%s%s", 
      "\x30\x60\x60\x20\x15\x35\x35", 
      "\x08"); 

正在做你想要的。問題是,在接下來的%s,你正在使用"\\00",你想服務器接收ASCII \00(這將是0x5c,爲0x30,的0x30),而是服務器報告它接收ASCII \5c00(這將是0x5c, 0x,35,0x43,0x30,0x30)。

我克拉斯Lindbäck同意的,因爲它聽起來像的VeriFone終端正在做正確的事情,但服務器顯示是錯誤的。爲了證明這是正確的(你可以只做一個或另一個,或者你可以一起做),我會考慮做兩件事來排除這種情況。

第一條:您可以使用LOG_PRINTF(或者打印到紙張或屏幕上,如果你喜歡)打印每個字節的值,你把它關閉之前。下面是我寫過的一個快速而骯髒的函數,當我對一次類似的問題進行故障排除時。請注意,我只關心字符串的開頭(如你所見,似乎是這樣),所以如果我用完了緩衝區空間,我不打印結尾。

void LogDump(unsigned char* input, int expectedLength) 
{ 
#ifdef LOGSYS_FLAG 
    char buffer[100]; 
    int idx, bfdx; 
    memset(buffer, 0, sizeof(buffer)); 

    bfdx = 0; 
    for (idx = 0; idx < expectedLength && bfdx < sizeof(buffer); idx++) 
    { 
     //if it is a printable character, print as is 
     if (input[idx] > 31 && input[idx] < 127) 
     { 
      buffer[bfdx++] = (char) input[idx]; 
      continue; 
     } 
     //if we are almost out of buffer space, show that we are truncating 
     // the results with a ~ character and break. Note we are leaving 5 bytes 
     // because we expand non-printable characters like "<121>" 
     if (bfdx + 5 > sizeof(buffer)) 
     { 
      buffer[bfdx++] = '~'; 
      break; 
     } 
     //if we make it here, then we have a non-printable character, so we'll show 
     // the value inside of "<>" to visually denote it is a numeric representation 
     sprintf(&buffer[bfdx], "<%d>", (int) input[idx]); 
     //advance bfdx to the next 0 in buffer. It will be at least 3... 
     bfdx += 3; 
     //... but for 2 and 3 digit numbers, it will be more. 
     while (buffer[bfdx] > 0) 
      bfdx++; 
    } 
    //I like to surround my LOG_PRINTF statements with short waits because if there 
    // is a crash in the program directly after this call, the LOG_PRINTF will not 
    // finish writing to the serial port and that can make it look like this LOG_PRINTF 
    // never executed which can make it look like the problem is elsewhere 
    SVC_WAIT(5); 
    LOG_PRINTF(("%s", buffer)); 
    SVC_WAIT(5); 
#endif 
} 

其次:嘗試爲char數組中的每個位置分配一個顯式值。如果您已經使用了我的LOG_PRINTF建議,並且發現它並未發送您認爲應該的內容,則這將是解決此問題的一種方法,以便它能夠正確發送您想要的內容。這種方法比較繁瑣一些,但因爲你是拼寫出來的每個值,無論如何,它應該不會太大更多的開銷:

data[0] = 0x30; 
//actually, I'd probably use either the decimal value: data[0] = 48; 
// or I'd use the ASCII value: data[0] = '0'; 
// depending on what this data actually represents, either of those is 
// likely to be more clear to whomever has to read the code later. 
// However, that's your call to make. 
data[1] = 0x60; 
data[2] = 0x60; 
data[3] = 0x20; 
data[4] = 0x15; 
data[5] = 0x35; 
data[6] = 0x35; 
data[7] = 0x08; 
data[8] = 0x5C; // This is the '\' 
data[9] = 0x48; // The first '0' 
data[10]= 0x48; // The second '0' 
data[11]= 0; 
//for starters, you may want to stop here and see what you get on the other side 

後你已經證明自己,這ISVeriFone代碼不是導致問題,您將知道您是需要關注終端還是服務器端。

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我意識到這是一個非常古老的職位(我不知道我今天之前沒有看到它...),你可能已經解決了你的問題。如果你有*解決了它,你會分享你的解決方案嗎? @knk – David 2015-07-24 18:14:08