它给了我这个错误,有人可以帮我吗? (!)解析错误:语法错误,第59行的C:\ wamp64 \ www \ photogallery \ includes \ gallery-upload.inc.php中出现意外的'else'(T_ELSE)
她是我的代码,希望您能对我有所帮助。我找不到我的错误。 有人可以回答我的考试吗,大约需要2个小时。
$newFileName = $_POST['filename'];
if(empty(empty($newFileName))) {
$newFileName = "gallery";
} else {
$newFileName = strtolower(str_replace(" ", "-", $newFileName));
}
$imageTitle = $_POST['filetitle'];
$imageDesc = $_POST[''];
$file = $_FILES['file'];
$fileName = $file["name"];
$fileType = $file["type"];
$fileTempName = $file["tmp_name"];
$fileError = $file["error"];
$fileSize = $file["size"];
$fileExt = explode($fileName);
$fileActualExt = strtolower(end($fileExt));
$allowed = array("jpg", "jpeg", "png");
if (in_array($fileActualExt, $allowed)) {
if ($fileError === 0) {
if(fileSize < 2000000) {
$imageFullName = $newFileName ."." . uniqid("", true) ."." . $fileActualExt;
$fileDestination = "../img/gallery/" . $imageFullName;
include_once "dbh.inc.php";
if (empty($imageTitle) || empty($imageDesc)) {
header("Location: ../gallery.php?upload=empty");
exit();
} else {
$sql = "SELECT * FROM gallery";
$stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt, $sql)) {
echo "SQL statement failed!";
} else {
mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt);
$rowCount = mysqli_num_rows($result);
$setImageOrder = $rowCount + 1;
$sql = "INSERT INTO gallery (titleGallery, descGallery, imgFullNameGallery, orderGallery) VALUES (?, ?, ?, ?);";
if (!mysqli_stmt_prepare($stmt, $sql)) {
echo "SQL statement failed!";
} else {
mysqli_stmt_bind_param($stmt, "ssss", $imageTitle, $imageDesc, $imageFullName, $setImageOrder);
mysqli_stmt_execute($stmt);
move_uploaded_file($fileTempName, $fileDestination);
header("Location: ../gallery.php?upload=succes");
}
}
} else {
echo "File size is to big";
exit();
} else {
echo "You had an error!";
exit();
} else {
echo "You need to upload a proper file type!";
exit();
}
} ?>
答案 0 :(得分:0)
此错误意味着在ELSE语句之前紧挨着缺少一些内容。它说unexpected 'else'
,这意味着它在else
之前还有其他期望。
代码很可能在前一行缺少分号或右括号。但是,如果您在gallery-upload.inc.php
文件的第59行附近发布代码,我们可以为您提供更明确的答案。您从哪里获得此文件的?
答案 1 :(得分:0)
我认为您不能使用多个其他导致上述错误的信息。
错误代码:
if (condition) {
echo 'test';
} else {
echo "File size is to big";
exit();
} else {
echo "You had an error!";
exit();
} else {
echo "You need to upload a proper file type!";
exit();
}
在条件中使用else-if。
if (condition) {
//something
} else if (condition) {
//something
} else {
//something
}
希望这会有所帮助! Tq