在 PHP 文件中获取解析错误

Question

我用php写了一些代码。示例代码片段如下：

include('config.php');
require_once "variables.php";
global $uploadID = " " ;    //getting error in this line
function uploadImage($wtI,$tbln,$pri,$db){
if(is_array($_FILES)) {
if(is_uploaded_file($_FILES['image']['tmp_name'])) {
$sourcePath = $_FILES['image']['tmp_name'];
$targetFolder = "../upload_images/$wtI/";
if (!file_exists($targetFolder)) {
mkdir($targetFolder, 0777, true);
}
$targetPath = $targetFolder.$_FILES['image']['name'];
while(file_exists($targetPath)){
$targetPath = $targetFolder.uniqid().'-'.$_FILES['image']['name'];
}
if(move_uploaded_file($sourcePath,$targetPath)){
$sql = "UPDATE `$tbln` SET image='".substr($targetPath,3)."' WHERE $pri=$uploadID;";
$result=mysqli_query($db,$sql);
return true;
}
else return false;
}
}
}

问题是我在运行 php 文件时收到以下错误消息：

Parse error:syntax error, unexpected '=', expecting ',' or ';' in C:wamp64wwwprojectphpadditem.php on line 6

这个错误有什么解决方案吗？

Answer 1

global 关键字允许您访问全局变量，而不是创建新变量。只需删除那里的全局关键字即可。 全局关键字必须放在要使用的变量的函数中。 查看 https://www.w3schools.com/php/php_variables.asp 以了解如何使用它。

代码的更正将是：

include('config.php');
require_once "variables.php";
// Changes start here
$uploadID = " ";    //getting error in this line
function uploadImage($wtI,$tbln,$pri,$db){
global $uploadID;
//Changes end here
if(is_array($_FILES)) {
if(is_uploaded_file($_FILES['image']['tmp_name'])) {
$sourcePath = $_FILES['image']['tmp_name'];
$targetFolder = "../upload_images/$wtI/";
if (!file_exists($targetFolder)) {
mkdir($targetFolder, 0777, true);
}
$targetPath = $targetFolder.$_FILES['image']['name'];
while(file_exists($targetPath)){
$targetPath = $targetFolder.uniqid().'-'.$_FILES['image']['name'];
}
if(move_uploaded_file($sourcePath,$targetPath)){
$sql = "UPDATE `$tbln` SET image='".substr($targetPath,3)."' WHERE $pri=$uploadID;";
$result=mysqli_query($db,$sql);
return true;
}
else return false;
}
}

}

我正在用手机接听，所以请原谅格式问题