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我有一個在php腳本(給定完全權限)內創建的shell腳本。當我嘗試從終端運行shell腳本時,我得不到任何錯誤,但腳本不運行(沒有任何命令執行)。儘管一旦我複製了shell腳本的內容,將它們粘貼到XCode中的一個新文件中,並覆蓋舊的shell腳本,它就會正常運行。無法運行具有完全權限的shell腳本(UNIX)
有什麼建議嗎?我一直試圖找出很長一段時間,現在沒有進展。
我假設從PHP腳本編寫shell腳本時存在問題,因爲它在XCode或文本編輯器中編寫時工作。
這裏是PHP腳本編寫shell腳本:
<code>
$filePath = "/Applications/MAMP/htdocs/php/Batch/modulator/Release23/Library/irShell.sh";
$script = fopen($filePath, 'w+');
chmod($filePath, 0777);
fwrite($script,"#!/bin/sh");
$irPath = "/Applications/MAMP/htdocs/php/Batch/modulator/Release23/Library"; //path to .ir files
$modPath = "/Applications/MAMP/htdocs/php/Batch/modulator";
if($dir = opendir($irPath)){
while(($file = readdir($dir)) !== false){
$posA = strpos($file, ".IR");
$posB = strpos($file, ".ir");
$posC = strpos($file, ".Ir");
if ($posA == true){
$fileName = trim($file, ".IR");
$noT = substr_replace($fileName, "", 0, 1);
echo "$noT\n";
fwrite($script, "\r" . $modPath . "/mod -o " . $irPath . "/codes/" . $noT . " " . $fileName . ".IR");
}
else if ($posB == true){
$fileName = trim($file, ".ir");
$noT = substr_replace($fileName, "", 0, 1);
echo "$noT\n";
fwrite($script, "\r" . $modPath . "/mod -o " . $irPath . "/codes/" . $noT . " " . $fileName . ".ir");
}
else if ($posC == true){
$fileName = trim($file, ".Ir");
$noT = substr_replace($fileName, "", 0, 1);
echo "$noT\n";
fwrite($script, "\r" . $modPath . "./mod -o " . $irPath . "/codes/" . $noT . " " . $fileName . ".Ir");
}
}
}
</code>
下面是由這個PHP生成的shell腳本的例子:
#!/bin/sh
/Applications/MAMP/htdocs/php/Batch/modulator/mod -o /Applications/MAMP/htdocs/php/Batch/modulator/Release23/Library/codes/1294 T1294.ir
/Applications/MAMP/htdocs/php/Batch/modulator/mod -o /Applications/MAMP/htdocs/php/Batch/modulator/Release23/Library/codes/1295 T1295.ir
/Applications/MAMP/htdocs/php/Batch/modulator/mod -o /Applications/MAMP/htdocs/php/Batch/modulator/Release23/Library/codes/1296 T1296.ir
寫入文件的權限仍爲777嗎? – chrisaycock 2011-02-11 17:03:59
我對那些`\ r`有點懷疑。 – 2011-02-11 17:16:11