私はちょうどこのエラーを以下に得ます。これはlocalhost上で動作します。何か案は? このエラーをGoogleに送信する方法についてのアドバイスは、問題の原因がわからないため、多くの面で役立ちます。要求されたURL/edit.phpはこのサーバで見つかりません
"リクエストされたURL/edit.phpがこのサーバーに見つかりませんでした。"
<?php
include ('includes/connection.php');
include ('includes/functions.php');
include ('includes/header.php');
$jobId = $_GET["id"];
$query = "SELECT * FROM Freight, WHERE id = '$jobId'";
$result = mysqli_query($connection, $query);
if (mysqli_num_rows($result) > 0) {
while ($row = mysqli_fetch_assoc($result)) {
$jobArrival = $row["Arrival"];
$jobDeparture = $row["Departure"];
$jobClient = $row["Client"];
$jobAirportOfDeparture = $row["AirportOfDeparture"];
$jobAirportOfArrival = $row["AirportOfArrival"];
$jobAdditionalInfo = $row["AdditionalInfo"];
$jobBoxType = $row["BoxType"];
$jobTemp = $row["Temp"];
// prideti is dezes table
// $pavadinimas = $row["pavadinimas"];
// $likutis = $row["likutis"];
if (isset($_POST['update'])) {
$jobArrival = validateFormData($_POST['jobArrival']);
$jobDeparture = validateFormData($_POST['jobDeparture']);
$jobClient = validateFormData($_POST['jobClient']);
$jobAirportOfDeparture = validateFormData($_POST['jobAirportOfDeparture']);
$jobAirportOfArrival = validateFormData($_POST['jobAirportOfArrival']);
$jobAdditionalInfo = validateFormData($_POST['jobAdditionalInfo']);
$jobBoxType = validateFormData($_POST['jobBoxType']);
$jobTemp = validateFormData($_POST['jobTemp']);
$query = "UPDATE Freight SET Arrival = '$jobArrival',
Departure = '$jobDeparture',
Client = '$jobClient',
AirportOfDeparture = '$jobAirportOfDeparture',
AirportOfArrival = '$jobAirportOfArrival',
AdditionalInfo = '$jobAdditionalInfo',
BoxType = '$jobBoxType',
Temp = '$jobTemp'
WHERE id = '$jobId'";
$result = mysqli_query($connection, $query);
if ($result) {
header("Location: formdisplay.php");
} else {
"Klaida" . mysqli_error($connection);
}
}
}
}
else {
echo "Nera irasu!!!!!!!!!!!!!!";
}
if (isset($_POST['istrinti'])) {
$query = "DELETE FROM Freight WHERE id ='$jobId'";
$result = mysqli_query($connection, $query);
if ($result) {
header("Location: formdisplay.php?alert=deleted");
}
else {
echo "Error" . mysqli_error($connection);
}
}
mysqli_close($connection);
?>
<h1> Iraso koregavimas</h1>
<table>
<tr>
<td>Arrival</td>
<td>Departure</td>
<td>Client</td>
<td>Airport Of Departure</td>
<td>Airport Of Arrival</td>
<td>Additional Info</td>
<td>Box Type</td>
<td>Temp</td>
</tr>
<tr>
<form method="POST" action="<?php echo htmlspecialchars($_SERVER['PHP_SELF']); ?>?id=<?php echo $jobId; ?>">
<td>
<input type = "text"
value = "<?php echo $jobArrival ?>"
name = "jobArrival">
</td>
<td>
<input type = "text"
value = "<?php echo $jobDeparture ?>"
name = "jobDeparture">
</td>
<td>
<input type = "text"
value = "<?php echo $jobClient ?>"
name = "jobClient">
</td>
<td>
<input type = "text"
value = "<?php echo $jobAirportOfDeparture ?>"
name = "jobAirportOfDeparture">
</td>
<td>
<input type = "text"
value = "<?php echo $jobAirportOfArrival ?>"
name = "jobAirportOfArrival">
</td>
<td>
<input type = "text"
value = "<?php echo $jobAdditionalInfo ?>"
name = "jobAdditionalInfo">
</td>
<td>
<input type = "text"
value = "<?php echo $jobBoxType ?>"
name = "jobBoxType">
</td>
<td>
<input type = "text"
value = "<?php echo $jobTemp ?>"
name = "jobTemp">
</td>
<td>
<input type = "submit"
name = "update"
value = "update"
href = "formdisplay.php"></td>
<td>
<input type ="submit"
name ="istrinti"
value ="istrinti"
href ="formdisplay.php">
</td>
</form>
</tr>
</table>
ありがとう!ポストははるかに良く見えます。 – DotOnI