Question

我有这个HTML表单：

<form action="../functions/createproject.php" method="post">
Name: <input type="text" name="name"><br>
Description: <input type="text" name="description"><br>
<input type="submit">
</form>

这个PHP代码（createproject.php）：

<?php
$projectname = $_POST["name"];
$projectdescription = $_POST["description"];
function createproject() {
    if (!file_exists("../projects/$projectname")) {
        mkdir("../projects/$projectname", 0777, true);
    }
    $fp = fopen($_SERVER['DOCUMENT_ROOT'] . "../projects/$projectname/info.txt","wb");
    fwrite($fp,$content);
    fclose($fp);
}
createproject();
?>

但是我收到以下错误：

Notice: Undefined variable: projectname in /home/wdele/domains/hypah.org/public_html/functions/createproject.php on line 5 Notice: Undefined variable: projectname in /home/wdele/domains/hypah.org/public_html/functions/createproject.php on line 8 Warning: fopen(/home/wdele/domains/hypah.org/public_html../projects//info.txt): failed to open stream: No such file or directory in /home/wdele/domains/hypah.org/public_html/functions/createproject.php on line 8 Notice: Undefined variable: content in /home/wdele/domains/hypah.org/public_html/functions/createproject.php on line 9 Warning: fwrite() expects parameter 1 to be resource, boolean given in /home/wdele/domains/hypah.org/public_html/functions/createproject.php on line 9 Warning: fclose() expects parameter 1 to be resource, boolean given in /home/wdele/domains/hypah.org/public_html/functions/createproject.php on line 10

我想不出有什么理由不起作用。

有什么想法吗？感谢。

Answer 1

您在全局范围内定义$projectname，但在函数范围内使用它。有两种方法可以解决这个问题：

使用global在函数中提供全局变量：

function createproject() {
    global $projectname;
    ...

或者，更改函数以$projectname作为参数：

function createproject($projectname) {
    ...

然后使用createproject($projectname)调用它。

后者是通常的方式。全局变量的使用通常是不受欢迎的，因为它使代码更难理解而且模块化程度更低。

Answer 2

您无法在函数外调用已定义的变量。使用global关键字。像这样：

function createproject() {
    global $projectname;
    if (!file_exists("../projects/$projectname")) {
        mkdir("../projects/$projectname", 0777, true);
    }
    $fp = fopen($_SERVER['DOCUMENT_ROOT'] . "../projects/$projectname/info.txt","wb");
    fwrite($fp,$content);
    fclose($fp);
}

PHP不会处理表单

2 个答案: