в чем может быть проблема с этим кодом, некоторые данные npt отправляются в базу данных

  • Автор темы lisss2
  • 35
  • Обновлено
  • 17, May 2024
  • #1
в чем может быть проблема с этим кодом? Некоторые данные npt отправляются в базу данных.

 <?php

$host = 'localhost';

$username = 'root';

$password = '';

$datadase = 'registerfinal';

$connect = mysqli_connect($host, $username, $password) or die ('error to connect to datadase'.mysqli_error());

if ($connect) {

echo 'mysqli connect succsessfull';

}

echo '<br /><br />';

$selectdb = mysqli_select_db($connect, $datadase) or die ('unable to select datadase'.mysqli_error());

if($selectdb) {

echo 'database selected succsessfully';

}

if(isset($_POST['savedetails'])) {

$firstname = $_POST['firstname'];

$lastname = $_POST['lastname'];

$username = $_POST['username'];

$password = $_POST['password'];

$repeat_password = $_POST['repeat_password'];

$gender = $_POST['gender'];

$country = $_POST['country'];

if(isset($_POST['food'])) {

$food = $_POST['food'];

$favfood = "";

foreach($food as $meal ) {

$favfood = $meal.",";

print_r($favfood);

}

}

if(isset($_POST['imageUpload'])) {

$imageUploadname = $_FILES['imageUpload']['name'];

$imageUploadsize = $_FILES['imageUpload']['size'];

$imageUploadtmp_name = $_FILES['imageUpload']['tmp_name'];

$imageUploadtype = $_FILES['imageUpload']['type'];

$uploadFolder = "uploadFolder/";

$destinationName = rand(1000, 10000).$imageUploadname;

move_uploaded_file($imageUploadtmp_name, $uploadFolder.$destinationName);

echo "$imageUploadname";

echo "$imageUploadsize";

echo "$imageUploadtmp_name";

echo "$imageUploadtype";

echo "$destinationName";

}

var_dump($_FILES['imageUpload']) ;

$sqltwo = "INSERT INTO `registerfinaltable` (`id`, `firstname`, `lastname`, `username`, `password`, `repeat_password`,

`gender`, `food`, `country`, `imageUploadname`, `imageUploadsize`, `imageUploadtype`)

VALUES (NULL, '$firstname', '$lastname', '$username', '$password', '$repeat_password', '$gender', '$favfood', '$country',

'$destinationName', '$imageUploadsize', '$imageUploadtype')";

$results = mysqli_query($connect, $sqltwo) ;

if($results){

echo "inserted successfully";

}

echo "$sqltwo";

}

?>

<html>

<head>

<title>register</title>

</head>

<body>

<form action = "" method = "post" enctype = "multipart/form-data" >

<label>first name : <input type = "text" name = "firstname" /> </label> <br /><br />

<label>last name : <input type = "text" name = "lastname" /> </label><br /><br />

<label>username : <input type = "text" name = "username" /> </label><br /><br />

<label>password : <input type = "password" name = "password" /> </label><br /><br />

<label>repeat password : <input type = "password" name = "repeat_password" /> </label><br /><br />

<label>Male : <input type = "radio" name = "gender" value = "Male" /> </label><br /><br />

<label>Female : <input type = "radio" name = "gender" value = "Female" /> </label><br /><br />

<label>pizza : <input type = "checkbox" name = "food[]" value = "pizza"/> </label><br /><br />

<label>burger : <input type = "checkbox" name = "food[]" value = "burger"/> </label><br /><br />

<label>chips : <input type = "checkbox" name = "food[]" value = "chips"/> </label><br /><br />

<label>sausage : <input type = "checkbox" name = "food[]" value = "sausage"/> </label><br /><br />

<label>sandwich : <input type = "checkbox" name = "food[]" value = "sandwich"/> </label><br /><br />

<label>Image : <input type = "file" name = "imageUpload" /> </label><br /><br />

<select name = "country">

<?php

$sql = 'SELECT * FROM `countrie` ';

$querry = mysqli_query($connect, $sql);

while($country = mysqli_fetch_array($querry)):;

?>

<option value = "<?php echo $country['country']; ?>"><?php echo $country['country']; ?></option>

<?php endwhile;?>

</select> <br />

<input type = "submit" name = "savedetails" />

</form>

<table border = "1" bgcolor = "" width = "100%">

<tr><th>id</th><th>Firstname</th><th>Lastname</th><th>Username</th><th>Password</th><th>Password 2</th><th>Gender</th><th>Fav.

Food</th> <th>Image</th> <th>Country</th><th>imageUploadname</th><th>imageUploadsize</th><th>imageUploadtype</th></tr>

<?php

$sqldata = "SELECT * FROM registerfinaltable";

$querysqldata = mysqli_query($connect, $sqldata);

while($rows = mysqli_fetch_array($querysqldata) ):;

?>

<tr>

<td><?php echo $rows['id'];?></td>

<td><?php echo $rows['firstname'];?></td>

<td><?php echo $rows['lastname'];?></td>

<td><?php echo $rows['username'];?></td>

<td><?php echo $rows['password'];?></td>

<td><?php echo $rows['repeat_password'];?></td>

<td><?php echo $rows['lastname'];?></td>

<td><?php echo $rows['gender'];?></td>

<td><?php echo $rows['food'];?></td>

<td><?php echo $rows['country'];?></td>

<td><?php echo $rows['imageUploadname'];?></td>

<td><?php echo $rows['imageUploadsize'];?></td>

<td><?php echo $rows['imageUploadtype'];?></td>

<?php endwhile;?>

</tr>

</table>

</body>

</html>
PHP: это данные, отображаемые браузером

lisss2


Рег
06 Oct, 2010

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  • 07, Jun 2024
  • #2
строка 57, $destinationName? не $imageUploadName? в строке 55, похоже, отсутствует «destinationName» Разве переменные не должны совпадать, хотя PHP не моя сильная сторона?
 

AnnaNovikova


Рег
04 Feb, 2015

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1

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