我通过下面的代码将图像路径插入到数据库中，但我无法在 html 页面中显示它...图像的路径是"images/"如何显示实际图像？我很努力，但我要做的最多的是显示文件名而不是图像。
<?php
$mysqli = new mysqli("localhost", "root", "", "simple_login");
// TODO - Check that connection was successful.
$photo= "images/" . $_FILES["file"]["name"];
$stmt = $mysqli->prepare("INSERT INTO photo (photo) VALUES (?)");
// TODO check that $stmt creation succeeded
// "s" means the database expects a string
$stmt->bind_param("s", $photo);
$stmt->execute();
$stmt->close();
$mysqli->close(
?>
这是我试图显示的图像...这只显示带有路径的文件名.....
<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("simple_login", $con);
$result = mysql_query("SELECT * FROM photo");
while($row = mysql_fetch_array($result)) …我之前已经问过这个但是我似乎从来没有得到它的工作方式(我尝试了很多但完全没有成功)有人可以告诉我如何在注册时向用户的电子邮件地址发送激活链接,并且不允许用户,直到他们按照电子邮件地址中的链接激活他们的帐户？我该怎么办？我根本没有得到它...请帮帮我..
我users在数据库中的表中有什么:
1 id int(11) AUTO_INCREMENT
2 username varchar(255)
3 password char(64)
4 salt char(16)
5 email varchar(255)
register.php
// First we execute our common code to connection to the database and start the session
require("common.php");
// This if statement checks to determine whether the registration form has been submitted
// If it has, then the registration code is run, otherwise the form is displayed
if(!empty($_POST))
{
// Ensure that the user has entered a non-empty username
if(empty($_POST['username'])) …我正在尝试将文件名传递给线程函数,但它的类型在函数内转换为int
struct Data {
char file_name;
}
void *Requestor(void *args) {
struct Data *data = (struct Data*)args;
printf("%s\n", data->file_name); //says expected char* but argument is type of int
}
int file_count = 5;
struct Data files[file_count];
for (int i = 0; i < file_count; i++) {
printf("%s\n", argv[5 + i]); //this prints the file_name correctly;
files[i].file_name = argv[5 + i]; // I get: warning: assignment makes integer from pointer without a cast [-Wint-conversion when compiling
int thread = pthread_create(&(requesterThreads[i]), NULL, …<html>
<body>
<form action="upload-file.php" method="post"
enctype="multipart/form-data">
<label for="file">Filename:</label>
<input type="file" name="file" id="file"><br>
<input type="submit" name="submit" value="Submit">
</form>
</body>
</html>
这是文件上传的PHP代码..我想使用我从w3schools获得的这个PHP代码..你认为这是一个安全的文件上传代码？这是我发现的最简单的代码,它非常有用..我尝试过其他来源的几个代码,但我无法让它们工作......任何想法？
<?php
ini_set('display_errors', '0');
error_reporting(E_ALL | E_STRICT);
$allowedExts = array("jpg", "jpeg", "gif", "png");
$extension = end(explode(".", $_FILES['file']['name']));
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/png")
|| ($_FILES["file"]["type"] == "image/pjpeg"))
&& ($_FILES["file"]["size"] < 2097152)
&& in_array($extension, $allowedExts))
{
if ($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
}
else
{
echo "Upload: " . …