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我可以將文檔上傳到我的主測試目錄。但是我希望將我的變量叫做staffNo並將其放置在特定的staffNo目錄中。但即使我已經爲我的輸入聲明瞭適當的名稱,我也無法解析變量。PHP調用變量開關
例如:
人員頁
$staffNo = 1;
<form id="Staff" name="Staff" method="post" action="upload.php" enctype="multipart/form-data">
//My staffno variable
<input type="hidden" id="staffNo" name="staffNo" value="<?php echo $staffNo ?>"/>
//Upload document
<input name="upload" id="upload" type="file"/>
<input type="hidden" name="upload" value="upload"/>
//Submit button
<input type="submit" name="submit" value="Submit">
</form>
upload.php的
switch($_POST['submit'])
{
case 'Submit':
if ($_FILES["upload"]["error"] > 0)
{
echo "Error: " . $_FILES["upload"]["error"] . "<br />";
}
else
{
echo "Upload: " . $_FILES["upload"]["name"] . "<br />";
echo "Type: " . $_FILES["upload"]["type"] . "<br />";
echo "Stored in: " . $_FILES["upload"]["tmp_name"];
move_uploaded_file($_FILES["upload"]["tmp_name"],
"documents/"."staffNo"."/".$_FILES["upload"]["name"]);
echo "Stored in: " ."./documents/"."staffNo"."/". $_FILES["upload"]["name"];
}
break;
case 'others':
break;
default;
我staffNo變量無法調出,即使它的形式下已經。我在某處做錯了嗎?我還想爲員工創建一個新文件夾,如果沒有找到它的話。但是現在基本員工變數無法呼叫。好心提醒。
應用差距<?PHP的echo $ staffNo?>到<?PHP的echo $ staffNo?> – swapnesh 2012-07-07 08:48:24
不影響〜:( – JLearner 2012-07-07 08:51:55
@ user976050的var_dump的$ _POST [ '提交'] – swapnesh 2012-07-07 08:52:52