Php 表单上传如何显示?当我运行这个文件时
我有一个php脚本,如下所示Php 表单上传如何显示?当我运行这个文件时,php,upload,Php,Upload,我有一个php脚本,如下所示 <?php if($_FILES['img']['type'] != "image/gif") { echo "<center><br>param name: img<br>directory file /challenge/ex"; exit; } $uploaddir = 'ex/'; $uploadfile = $uploaddir . basename($_FI
<?php
if($_FILES['img']['type'] != "image/gif") {
echo "<center><br>param name: img<br>directory file /challenge/ex";
exit;
}
$uploaddir = 'ex/';
$uploadfile = $uploaddir . basename($_FILES['img']['name']);
if (move_uploaded_file($_FILES['img']['tmp_name'], $uploadfile)) {
echo "File is valid, and was successfully uploaded.\n";
} else {
echo "File uploading failed.\n";
}
?>
<center>
<br><br>
<form action="" method="post" enctype="multipart/form-data">
<input type="file" size="20" name="img" />
<input type="submit" name="upload" value="Upload" />
</form>
但是,表单上传和按钮无法显示,当我运行此文件时,这通常发生在出现php错误时, 尝试在第二行之前添加这三行:
ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);
因此,它将是:
<?php
ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);
if($_FILES['img']['type'] != "image/gif") {
echo "<center><br>param name: img<br>directory file /challenge/ex";
exit;
}
$uploaddir = 'ex/';
$uploadfile = $uploaddir . basename($_FILES['img']['name']);
if (move_uploaded_file($_FILES['img']['tmp_name'], $uploadfile)) {
echo "File is valid, and was successfully uploaded.\n";
} else {
echo "File uploading failed.\n";
}
?>
<center>
<br><br>
<form action="" method="post" enctype="multipart/form-data">
<input type="file" size="20" name="img" />
<input type="submit" name="upload" value="Upload" />
</form>
希望会出现一些错误,以便我们可以修复它。您的意思是该按钮没有显示在表单上吗?当我运行此文件时,表单上载和按钮无法显示,还是您询问如何向用户显示一条消息,说明“上载已成功”?这是此脚本文件中的所有代码吗?请添加到文件顶部例如,在打开PHP标记之后进行测试。即使您是在配置为LIVE的服务器上开发,现在也会看到任何错误<代码>新问题通知:未定义索引:第5行的/home/buayalau/domains/co/public\u html/challenge/up2.php中的img尝试在显示错误行后使用if检查方法($\u SERVER['REQUEST\u method']='POST'){并在关闭php标记之前将其关闭($\u服务器['REQUEST\u METHOD']='POST')