我设计了一个汽车销售网站。
我有一个表单,其中输入了有关汽车的某些信息,然后将其添加到数据库并显示在我的主表单上。如何将图像发送到服务器,然后使用URL作为图像的源。
干杯。
这是我的代码。
PHP:
<?php
try {
# Connect to SQLite database
$dbh = new PDO("sqlite:../Car_Sales_Network");
$make = $_POST['Make'];
$model = $_POST['Model'];
$badge = $_POST['Badge'];
$price = $_POST['Price'];
$trans = $_POST['Transmission'];
$ppl = $_POST['P_Plate_Legal'];
print_r($_POST);
$sth = $dbh->prepare('INSERT INTO Cars_On_Network
("car_make","car_model","car_badge","price","trans","P_Plate_Legal")
VALUES
(?, ?, ?, ?, ?, ?)');
$sth->execute(array($make, $model, $badge, $price, $trans, $ppl));
header("Location: ../Carsales_Network.php");
}
catch(PDOException $e) {
echo $e->getMessage();
}
?>
HTML:
<!DOCTYPE html>
<html>
<head>
<title>New Vehicle</title>
<link type="text/css" rel="stylesheet" href="New_Car_Form.css"/>
</head>
<body>
<div id="main">
<form action="Insert_Car.php" method="post" enctype="multipart/form-data">
Make:<br>
<input type="text" name="Make">
<br>
Model:<br>
<input type="text" name="Model">
<br><br>
Badge:<br>
<input type="text" name="Badge">
<br>
Price:<br>
<input type="text" name="Price">
<br>
Transmission: <br>
<input type="radio" name="Transmission" value="Manual" checked>Manual
<br>
<input type="radio" name="Transmission" value="Auto">Automatic
<br><br>
<form enctype="multipart/form-data" action="upload_file.php" method="post">
P Plate Legal: <br>
<select name="P_Plate_Legal">
<option value="Yes">Yes</option>
<option value="No">No</option>
</select>
<br>
<input type="file" name="fileToUpload" id="fileToUpload">
<input type="submit" value="Upload Image" name="submit">
</form>
<br>
<br>
<input class="submit" type="submit" value="Submit">
<br>
<a href="http://1673-itstudies/12-infotech/jsummers/Carsales_Network.php" class="myButton">Let's go back!</a>
<br>
</div>
</body>
</html>
</body>
</html>
如何将上传的图像发布到要使用的数据库中,我的数据库中有一个文件夹,上面有汽车照片。
答案 0 :(得分:0)
将文件上传到php中的服务器,使用php fileupload函数
move_uploaded_file ( string $filename , string $destination )
示例:
<?php
$uploads_dir = '/uploads';
foreach ($_FILES["fileToUpload"]["error"] as $key => $error) {
if ($error == UPLOAD_ERR_OK) {
$tmp_name = $_FILES["fileToUpload"]["tmp_name"][$key];
$name = $_FILES["fileToUpload"]["name"][$key];
move_uploaded_file($tmp_name, "$uploads_dir/$name");
}
}
?>
供参考检查move_uploaded_file()