将HTML表单提交到PHP文件
[英]Submitting HTML form to PHP file
问题描述
我在远程服务器上创建了两个文件。 一种是html形式，要求输入一些字段，另一种是php文件，它将获取所有数据并插入数据库中。
为此从单击提交按钮的html文件中，我正在调用php文件，但是我认为该文件没有得到执行，因为当我再次单击提交时，它会重新加载相同的html页面。
html：
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>MCQ Questions</title>
</head>
<body>
<form method="post" >
<p> Enter the question :</p> <input name="question" type="input"> <br><br>
<p> Enter options :</p>
Enter option 1 : <input name="opt1" type="input"> <br><br>
Enter option 2 : <input name="opt2" type="input"> <br><br>
Enter option 3 : <input name="opt3" type="input"> <br><br>
Enter option 4 : <input name="opt4" type="input"> <br><br>
<p> Enter correct answer :</p>
<input name="ans" type="input"> <br><br>
<input type="submit" value = "Submit" onClick = "uploadQuestion.php">
</form>
</body>
</html>
php文件：
<?php
$question=$_POST['question'];
$option1=$_POST['opt1'];
$option2=$_POST['opt2'];
$option3=$_POST['opt3'];
$option4=$_POST['opt4'];
$ans=$_POST['ans'];
$db_server = mysql_connect("address","username","pass");
if(!$db_server) {
die("Database connection failed: " . mysql_error());
}else{
$db_select = mysql_select_db("mlm",$db_server);
if (!$db_select) {
die("Database selection failed:: " . mysql_error());
}
}
$sql = "INSERT INTO questions (question,answer_a,answer_b,answer_c,answer_d,answer) VALUES ('$question','$option1',$option2,$option3,$option4,$ans)";
if (!mysql_query($sql)) {
die('Error: ' . mysql_error());
}
?>
我也尝试过这种方式：
<input type="submit" value = "Submit" onClick = "http://address/uploadQuestion.php">
但是没有任何效果。 怎么了 我是网络开发的初学者，有人可以帮忙吗？ 谢谢..
编辑：
$database = new Database('addredd','username','pass','handbook');
$dbConnection = $database->getDB();
$stmt = $dbConnection->prepare("insert into questions(question,answer_a,answer_b,answer_c,answer_d,answer) values(?,?,?,?,?,?)");
$stmt->execute(array($question,$option1,$option2,$option3,$option4,$ans));
我尝试使用pdo语句，但收到此错误：致命错误：在第12行的/var/www/html/uploadQuestion.php中找不到类'Database'
编辑2：
我正在尝试将一个文件上传到服务器上，并希望也将其保存在数据库中，因此，我为此创建了2个文件，一个是index.php，另一个是uploadFile.php。
正如您现在所显示的，我为此使用了pdo，但是当我再次单击上载图像时，同一页面正在加载。
index.php
<form action="index.php" method="post" enctype="multipart/form-data">
<p> Select image to upload:</p>
<input name = "file" type="file" id="fileToUpload"><br><br>
Enter chapter name :
<input name = "chapterName" type = "text"><br><br>
<input type="submit" value = "Upload Image">
</form>
<?php
if (isset($_FILES['file']['tmp_name']))
{
$ch = curl_init();
$cfile = new CURLFile($_FILES['file']['tmp_name'],$_FILES['file']['type'],$_FILES['file']['name']);
$data = array("myfile" => $cfile);
curl_setopt($ch, CURLOPT_URL, "http://host/NewProject/uploadFile.php");
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true);
curl_setopt($ch, CURLOPT_POST, true);
curl_setopt($ch, CURLOTP_POSTFIELDS, $data);
$response = curl_exec($ch);
if($response == true)
{
echo "File posted";
}
else{
echo "Error: " . curl_error($ch);
}
}
?>
uploadFile.php
<?php
ini_set('display_errors', 1);
if(isset($_FILES['myfile']['tmp_name']))
{
$path = "files/" . $_FILES['myfile']['name'];
move_uploaded_file($_FILES['myfile']['tmp_name'], $path);
$chapterName=$_POST['chapterName'];
$dbh = new PDO('mysql:host=host;dbname=database_name','username', 'password');
$stmt = $dbh->prepare("INSERT INTO chapters (title,file) VALUES (?, ?)");
$stmt->execute(array($chapterName,$path));
if ($dbh->lastInsertId())
{
echo 'File upploaded.';
}
else
{
echo 'File could not upload.';
}
}
?>
请帮助..谢谢
2 个解决方案
解决方案1
1 已采纳 2017-03-18 12:29:16
首先修复您的表单，不能将type=""命名为input可以在此处检查https://www.w3schools.com/tags/att_input_type.asp
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>MCQ Questions</title>
</head>
<body>
<form action="uploadQuestion.php" method="post" enctype="multipart/form-data">
<p> Enter the question :</p> <input name="question" type="text"> <br><br>
<p> Enter options :</p>
Enter option 1 : <input name="opt1" type="text"> <br><br>
Enter option 2 : <input name="opt2" type="text"> <br><br>
Enter option 3 : <input name="opt3" type="text"> <br><br>
Enter option 4 : <input name="opt4" type="text"> <br><br>
<p> Enter correct answer :</p>
<input name="ans" type="text"> <br><br>
<input type="submit" value="Submit">
</form>
</body>
</html>
然后你的php代码
<?php
// mysql connection
$db_server = mysql_connect("address","username","pass");
// check for mysql connection
if(!$db_server)
{
die("Database connection failed: " . mysql_error());
}
else
{
// check if database exists
$db_select = mysql_select_db("mlm",$db_server);
if (!$db_select)
{
die("Database selection failed:: " . mysql_error());
}
}
// escape post variables
$question = mysql_real_escape_string($_POST['question']);
$option1 = mysql_real_escape_string($_POST['opt1']);
$option2 = mysql_real_escape_string($_POST['opt2']);
$option3 = mysql_real_escape_string($_POST['opt3']);
$option4 = mysql_real_escape_string($_POST['opt4']);
$ans = mysql_real_escape_string($_POST['ans']);
// make query
$sql = "INSERT INTO questions (question,answer_a,answer_b,answer_c,answer_d,answer) VALUES ('$question', '$option1', '$option2', '$option3', '$option4', '$ans')";
// check if query runs
if (!mysql_query($sql)) {
die('Error: ' . mysql_error());
}
?>
或php与mysqli
<?php
// host, username, password, database name
$db_server = mysqli_connect("address", "username", "pass", "mlm");
// check for connection
if(!$db_server)
{
die("Database connection failed: " . mysqli_error($db_server));
}
// escape post variables
$question = mysqli_real_escape_string($db_server, $_POST['question']);
$option1 = mysqli_real_escape_string($db_server, $_POST['opt1']);
$option2 = mysqli_real_escape_string($db_server, $_POST['opt2']);
$option3 = mysqli_real_escape_string($db_server, $_POST['opt3']);
$option4 = mysqli_real_escape_string($db_server, $_POST['opt4']);
$ans = mysqli_real_escape_string($db_server, $_POST['ans']);
// make query
$sql = "INSERT INTO questions (question,answer_a,answer_b,answer_c,answer_d,answer) VALUES ('$question', '$option1', '$option2', '$option3', '$option4', '$ans')";
// check if query runs
if (!mysqli_query($db_server, $sql))
{
die('Error: ' . mysqli_error($db_server));
}
?>
或带有准备好的语句的php
<?php
// mysql connection
$dbh = new PDO('mysql:host=adress;dbname=database_name', 'username', 'password');
// escape post variables
$question = $_POST['question'];
$option1 = $_POST['opt1'];
$option2 = $_POST['opt2'];
$option3 = $_POST['opt3'];
$option4 = $_POST['opt4'];
$ans = $_POST['ans'];
$stmt = $dbh->prepare("INSERT INTO questions (question,answer_a,answer_b,answer_c,answer_d,answer) VALUES ( ?, ?, ?, ?, ?, ?)");
$stmt->execute(array($question, $option1, $option2, $option3, $option4, $ans));
if ($dbh->lastInsertId())
{
echo 'Sucess.';
}
else
{
echo 'Fail.';
}
?>
解决方案2
-2 2017-03-18 12:10:07
从代码更改为此
<!doctype html> <html> <head> <meta charset="utf-8"> <title>MCQ Questions</title> </head> <body> <form action="uploadQuestion.php" method="post" enctype="multipart/form-data"> <p> Enter the question :</p> <input name="question" type="input"> <br><br> <p> Enter options :</p> Enter option 1 : <input name="opt1" type="input"> <br><br> Enter option 2 : <input name="opt2" type="input"> <br><br> Enter option 3 : <input name="opt3" type="input"> <br><br> Enter option 4 : <input name="opt4" type="input"> <br><br> <p> Enter correct answer :</p> <input name="ans" type="input"> <br><br> <input type="submit" value="Submit"> </form> </body> </html>
有没有一种方法可以获取图像文件内容，而无需在html / php中提交表单？
[英]Is there a way of getting an image file contents without submitting a form in html/php?
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