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我已經被分配了修復舊的php網站的任務，因爲它已被移動到一個新的服務器。它現在所在的服務器不允許全球化變量，這幾乎是所有這個網站運行。當試圖上傳圖像時，我的sql語句顯示了所有內容，但是我添加了圖像的列表的id。我希望有人能幫我弄清楚這一點。php文件上傳不能正常工作
這是我上傳功能：
function upload(){
global $imagefolder, $id;
global $tbl_units;
include "globalizePOSTGET.php";
// $uid = uuid();
$minsize = 5000; // 5kb
$maxsize = 3000000; // 3mb
$ext = explode('.',basename($_FILES['userfile']['name']));
$ext = $ext[count($ext)-1];
$ext = strtolower($ext);
if ($ext != "jpg" && $ext != "jpeg" && $ext != "png") {
echo "<script> alert('Image is not a png or jpeg format'); </script>";
return false;
}
$imagename = $_POST['id']."_img".$_FILES['img'].".$ext";
$imagename2 = "X_".$imagename;
$uploadfile = $imagefolder . $imagename;
$uploadfile2 = $imagefolder . $imagename2;
$uploadthumb = $imagefolder . "tn_" . $imagename;
if (file_exists($uploadfile)) unlink($uploadfile);
if (file_exists($uploadthumb)) unlink($uploadthumb);
if (file_exists($uploadfile)) {
echo "<script> alert('Image already exists!'); </script>";
}
else
{
if(is_uploaded_file($_FILES['userfile']['tmp_name'])) {
// check the file is less than the maximum file size
if($_FILES['userfile']['size'] < $maxsize) {
$imgData = addslashes(file_get_contents($_FILES['userfile']['tmp_name'])); // prepare the image for insertion
$size = getimagesize($_FILES['userfile']['tmp_name']); // get the image info..
if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadfile2)) {
$Image = @imagecreatefromjpeg($uploadfile2);
if ($Image) {
$img_height = imagesy($Image);
$img_width = imagesx($Image);
imagedestroy($Image);
}
if ($img_height > $img_width) { // portrait
$tempMultiplier = 150/$img_height;
$tempMultiplierFull = 600/$img_height;
} else {
$tempMultiplier = 150/$img_width;
$tempMultiplierFull = 600/$img_width;
}
$imageHeight = $img_height * $tempMultiplier;
$imageWidth = $img_width * $tempMultiplier;
$fullimageHeight = $img_height * $tempMultiplierFull;
$fullimageWidth = $img_width * $tempMultiplierFull;
createthumb($imagename2,"tn_".$imagename,$imageWidth,$imageHeight);
if($_FILES['userfile']['size'] > $minsize) {
createthumb($imagename2,$imagename,$fullimageWidth,$fullimageHeight);
if (file_exists($uploadfile2)) unlink($uploadfile2);
} else {
rename($uploadfile2, $uploadfile);
}
$sql = "UPDATE $tbl_units SET photo".$_FILES['img']." = \"" . $imagename . "\" WHERE id = " . $_POST['id'];
echo $sql;
if(!mysql_query($sql)) {
echo "<script> alert('Unable to upload file'); </script>";
} else {
?> <script>location.replace('memonly.php?action=edit_record&id=<?php echo $id; ?>');</script> <?php
}
}
} else {
// if the file is not less than the maximum allowed, print an error
$file_n = basename($_FILES['userfile']['name']);
$file_s = $_FILES['userfile']['size'];
?>
<script> alert("File exceeds the maximum limit of <?php echo $maxsize; ?>\nFile <?php echo $file_n; ?> is <?php echo $file_s; ?>");</script>
<?php
}
}
}
}
我呼應上是給我的錯誤就行了sql語句，我想。點擊提交後，頁面告訴我Unable to upload file'. Which is why I echoed the sql there. I end up with a sql statement looking like this: UPDATE member_units SET photo =「_img.jpg」WHERE id =`
有人請幫幫我！我對PHP很缺乏經驗，我不知道該怎麼做。
這裏是做上傳的形式：
<form enctype="multipart/form-data" action="<?php echo $_SERVER['PHP_SELF']?>" method="post">
<input type="hidden" name="_submit_check" value="1" />
<input type="hidden" name="id" value="<?php echo $id; ?>" />
<input type="hidden" name="img" value="<?php echo $img; ?>" />
Image URL: <input type="file" name="userfile" value="" style="font-size: 10px; width: 100%;">
<input type="submit" value="Submit" onClick="return validate();">
<input type="button" value="Cancel" onClick="location.href='/memonly.php?action=edit_record<?php echo "&id=$id&memberid=$memberid"; ?>';">
</form>
這是什麼：$ _FILES ['img']？ – Shehabix 2013-02-27 23:30:37
你讓我了！它曾經獲得'$ _POST'，但另一位程序員告訴我需要將它改爲'$ _FILES'，所以我做了。 – jlg 2013-02-28 14:06:58