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Próbujesz wstawić obraz wygenerowany z kamery użytkownika do bazy danych mysql za pomocą php, ale to nie działa. Używam webcam.js i wszystko działa dobrze. Kiedy użytkownik robi migawkę, obraz jest przechowywany na serwerze, jednak wkładka mysql nie wykonuje zadania wstawiania. Każdy pomysł, dlaczego nie działa?Przesyłanie obrazu wygenerowanego z kamery internetowej do bazy danych mysql z php nie działa
Zarówno kod jest pokazanie poniżej:
cam.php:
session_start();
include_once 'dbconnect.php';//connection to db
if(!isset($_SESSION['user'])){//ensures that it the true user
header("Location: index.php");
}
//display current time
//$arrival_time= date('Y-m-d H-i-s');
//echo "$arrival_time";
if (isset($_POST['send'])) {
$getname= mysql_real_escape_string($_POST['last_name']);
$idvalue= $_SESSION['myvalue'];
$update=mysql_query("UPDATE `employees`.`webcam_clockin`
SET `last_name`='$getname' WHERE image_id='$idvalue'");
if($update)
{
//run a check to verify last_name
$sql=mysql_query("SELECT users.*, employees.* FROM users
NATURAL JOIN employees
WHERE employees.last_name='$getname'");
$result=mysql_fetch_array($sql);
if($result){
$_SESSION['user'] = $result['user_id'];
header("Location: home.php");
}
else
{
?>
<script>
alert('Wrong Last Nane');
</script>
<?php
}
}
else
{
echo "Error Not done";
}
}
</style>
</head>
<body>
<div class="container">
<div align="center">
<script>
webcam.set_api_url('camsave.php');
webcam.set_quality(100); // JPEG quality (1 - 100)
webcam.set_shutter_sound(true); // play shutter click sound
</script>
<script>
document.write(webcam.get_html(640, 480));
webcam.set_hook('onComplete', 'my_callback');
function my_callback(msg)
{
document.getElementById('upload').innerHTML = msg;
}
function do_upload(){
webcam.snap();
}
function my_callback(msg) {
// extract URL out of PHP output
if (msg.match(/(https\:\/\/\S+)/)) {
var image_url = RegExp.$1;
// show JPEG image in page
document.getElementById('upload_results').innerHTML =
'<h1>Upload Successful!</h1>' +
'<h3>JPEG URL: ' + image_url + '</h3>';
}
else alert("PHP Error: " + msg);
}
</script>
</div>
<div>
<form class="form-signin" id="myForm"><br>
<h3 class="form-signin-heading">Enter Your Name. Take a Nice Picture and Submit</h3>
<input type=button class="btn btn-lg btn-primary btn-block" id="snap"
onclick="do_upload()" value="Snap">
</form>
<form action="<?php echo $_SERVER["PHP_SELF"];?>" method="post" class="form-signin" ><br>
<label for="last_name" class="sr-only">Enter Last Name</label>
<input type="text" name="last_name" id="last_name"
class="form-control" placeholder="Last Name" required autofocus>
<input type="submit" class="btn btn-lg btn-primary btn-block" name="send" id="send">
</form>
</div>
</td><td width=50> </td><td valign=top>
<div id="upload"></div>
</td></tr></table>
</div>
</body>
</html>
camsave.php:
session_start();
include_once 'dbconnect.php';
if(!isset($_SESSION['user']))
{
header("Location: index.php");
}
//get some data about this user
$res2=mysql_query("SELECT users.*, employees.* FROM users
NATURAL JOIN employees WHERE user_id=".$_SESSION['user']);
$userRow=mysql_fetch_array($res2);
if ($userRow) {
echo "correct!!";
}
//Define storage location of original images
$folder = "images/";
$filename = date('Y-m-d-H-i-s') . '.jpg';
$original = $folder.$filename;
//Get JPEG snapshot from webcam
$input = file_get_contents('php://input');
//Blank images are discarded
if(md5($input) == '7d4df9cc423720b7f1f3d672b89362be'){
exit();
}
//Retreive the snap and save to original dest.
$file= file_put_contents($original, $input);
if(!$file){
print "ERROR: Failed to write data to $filename, check permissions\n";
exit();
}
else
{
//Get the size of the image
$info = getimagesize($original);
list($width,$height) = $info;
if($info['mime'] != "image/jpeg"){//ensure we get right file extension
unlink($original);
exit();
}
//Move images to the Original folder
rename($original, "images/original/".$filename);
$emp_no = $userRow['emp_no'];//employee number
$user_id = $_SESSION['user_id'];//user ID
$image_id = NULL;
$original = "images/original/".$filename;//our image
$last_name =$_SESSION['last_name'];
$path = "images/thumbnail/".$filename;
$sql=mysql_query("INSERT INTO `webcam_clockin`
(`image_id`, `user_id`, `images`, `emp_no`, `last_name`)
VALUES ('$image_id', '$user_id','$path','$emp_no', '$last_name')");
move_uploaded_file($original, $path);
if(move_uploaded_file($filename, $original)){
echo "The file ". $original.$filename. " has been uploaded, and your information has been added to the directory";
echo "Thank You "; echo $userRow['username']; echo".\n";
}
else{
echo "Sorry, there was a problem uploading your file.";
echo "Error inserting entry data: ".mysql_error();
?>
<script>
alert('Error Inserting your details. Please, see your department manager');
</script>
<?php
}
$value=mysql_insert_id();
$_SESSION["myvalue"]=$value;
}
$url = 'https://' . $_SERVER['HTTP_HOST'] . dirname($_SERVER['REQUEST_URI']) . '/' . $filename;
print "$url\n";
I wreszcie rozwiązał problem zapisywania obrazu do bazy danych. Problemem był nieprawidłowy typ danych w phpmyadminie, który powinien być longblob.
Proszę, może ktoś pomóc to ostatnia twarz mojego projektu i jego pilne. Dzięki –
Pierwszą miarą do rozwiązania problemu jest złapanie błędów mysql i wydrukowanie ich. – Ctx
Mam włączone debugowanie php i nie wydrukowano żadnego błędu, i to samo dla mysql bez błędu wydrukowano –