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我有一個php代碼,有if語句(導出爲pdf,導出爲excel,最後一個導出爲excel)。在循環內執行一個php內的另一個php文件
前兩個如果上傳pdf和excel工作的很好,當我點擊按鈕來運行第三個if語句時,它顯示了我的空白頁面,即使它的代碼運行正常分別。
我的PHP代碼如下:
<?php
define('FPDF_FONTPATH','/Applications/XAMPP/xamppfiles/lib/php');
if (isset($_POST['pdf'])) {
//code//
}
}
if (isset($_POST['excel'])) {
//code//
}
if (isset($_POST['upload'])) {
exec('imp.php');
}
?>
而我試圖運行的imp.php文件是:
<?php
ini_set('display_errors', 1);
error_reporting(E_ALL);
$db_host = 'localhost';
$db_user = 'root';
$db_pwd = '';
$database = 'mydatabase';
$table = 'demo';
if ([email protected]_connect($db_host, $db_user, $db_pwd))
die("Can't connect to database");
if (!mysql_select_db($database))
die("Can't select database");
if(isset($_POST['submit']))
{
$fname = $_FILES['sel_file']['name'];
echo 'upload file name: '.$fname.' ';
$chk_ext = explode(".",$fname);
if(strtolower(end($chk_ext)) == "csv")
{
$filename = $_FILES['sel_file']['tmp_name'];
$handle = fopen($filename, "r");
while (($data = fgetcsv($handle, 1000, ",")) !== FALSE)
{
$sql = "INSERT into demo(orders,quantity,country,dates,comment) values('$data[0]','$data[1]','$data[2]', '$data[3]','$data[4]')";
mysql_query($sql) or die(mysql_error());
}
fclose($handle);
echo "Successfully Imported";
}
else
{
echo "Invalid File";
}
}
?>
<h1>Import CSV file</h1>
<form action='<?php echo $_SERVER["PHP_SELF"];?>' method='post' enctype="multipart/form-data">
Import File : <input type='file' name='sel_file' size='20'>
<input type='submit' name='submit' value='submit'>
</form
也許至少它應該是'exec('php imp.php');'?順便說一句,當你執行'exec()' –
時,你會在瀏覽器中看到任何東西,它仍然顯示我一個空白的屏幕,而不是顯示imp.php的內容 – compcrk
我應該使用什麼,而不是exec()呢? – compcrk