アクションの結果を表示するフォームを送信した後にページを更新する

Question

私は自分のコードに問題があると思うが、ジョブプロファイルの「プロファイル」と2つ目のボタンを使って利用できるジョブのリストを表示するページがあるジョブを適用するには、適用されたジョブを使用可能なすべてのジョブのリストから消滅させることです。コードはlocalhost上で正常に動作しますが、サーバーにアップロードした後、適用をクリックした後にページを更新して、アクションの結果を確認する必要があります。これはAvailableJobs.phpのコードです：

<?php $sql1= "SELECT * FROM Job"; // select all the Job 
$result1= mysqli_query($conn,$sql1);     
while($row1=mysqli_fetch_array($result1)) 
{ $JobName= $row1['JobName']; 
    $JID= $row1['JobID']; 
    $OrgID= $row1['OrgID']; // get the Org ID for the available Job 

$sql10=" SELECT * FROM JobStudent WHERE JobID='$JID' AND StudentID='$SID'"; 
$resultss = mysqli_query($conn,$sql10); 
$numResults = mysqli_fetch_array($resultss); 
if ($numResults == 0) { 

$sql2="SELECT * FROM RequestedOrganization WHERE OrgID='$OrgID'"; 
$result2= mysqli_query($conn,$sql2);      
    while($row2=mysqli_fetch_array($result2)) 
{ $OrgGPA= $row2['GPA']; 
    $OrgTrack= $row2['Track']; 
    $Priority= $row2['priorityGT']; 
} 
?> 
<li><input disabled style="border:none; margin-bottom: 10px; width:100%;" name="JobName" value="<?php echo $row1['JobName'];?>"> </li> <?php 
echo "<form action='AvailableJobAction_page.php' method='POST'> 
<input type='hidden' name='tempId' value='$JID'/> 
<input style='border-radius: 12px; margin-bottom: 10px; background-color: #ccc;' type='submit' class='right' name='submit-btn' value='Apply' onclick='return func1()'/> 
</form>"; 
echo "<form action='OrgProfileSS.php' method='POST'> 
<input type='hidden' name='temppId' value='$JID'/> 
<input type='hidden' name='tempppId' value='$OrgID'/> 
<input style='border-radius: 12px; background-color: #ccc;' type='submit' name='submit-btnn' class='right' value='Profile'/> 
</form><br/>"; ?> 

この実行ページAvailableJobAction_page.phpです：

$USER=$_SESSION['login_user']; 

if (isset($_POST['submit-btn'])){ 
$Jobid=$_POST['tempId']; // J ID 


$Sql="SELECT * FROM Job WHERE JobID='$Jobid'"; 
$Result2= mysqli_query($conn,$Sql); 
while ($Row1 =mysqli_fetch_array($Result2)){ 
$JN= $Row1['JobName']; 
$orgid= $Row1['OrgID']; 

$Sql10="SELECT * FROM Employer WHERE EmployerID='$orgid'"; 
$Result10= mysqli_query($conn,$Sql10); 
while ($Row10 =mysqli_fetch_array($Result10)){ 
$EmpEmail= $Row10['EmailAccount']; 
} 

}  

$sql="SELECT * FROM Student WHERE KsuEmailStuent='$USER'"; // Get S ID 
$result2= mysqli_query($conn,$sql); 
while ($row1 =mysqli_fetch_array($result2)){ 
$StID= $row1['StudentID']; 
} 

$sql4="SELECT * FROM JobStudent WHERE StudentID='$StID' AND JobID='$Jobid'"; 
$result4= mysqli_query($conn,$sql4); 
$rows= mysqli_num_rows($result4); 
if ($rows>0){ 
echo "<script type='text/javascript'>alert('You Already Applied before!')</script>"; 
echo "<script>window.location.href='AvailableJobs.php';</script>"; 
}else{ 
$sql4="INSERT INTO JobStudent (JobID,StudentID,Status) VALUES ('$Jobid','$StID','Pending')"; 
$result4= mysqli_query($conn,$sql4); 

$sql1="SELECT * FROM Job WHERE JobID='$Jobid'"; 
$result1= mysqli_query($conn,$sql1); 
while ($row2 =mysqli_fetch_array($result1)){ 
$Nos= $row2['NoOfAppliedstudnets']; 
} 
$Nos++; 
$sql3= "UPDATE Job SET NoOfstudnets='$Nos' WHERE JobID='$Jobid'"; 
$result3= mysqli_query($conn,$sql3); 

$to = $USER; // Send Email with the new Pass 
$subject = "Applied done sucsessfully!"; 
$message = "Your request have been sent, Please wait for response from employer and note that you can view your request status in your account home"; 
$message = wordwrap($message,70); 
//$headers = "From : Admin@PTN.com"; 
if(mail($to, $subject, $message)){ 
echo "<script type='text/javascript'>alert('Applied done sucsessfully!')</script>"; 
echo "<script>window.location.href='AvailableJobs.php';</script>"; 
}else{ 
echo "<script type='text/javascript'>alert('Erorr in send Email ')</script>"; 
echo "<script>window.location.href='AvailableJobs.php';</script>"; 
} 

$to = $EmpEmail; // Send Email with the new Pass 
$subject = "New student have Applied!"; 
$message = "New student have Applied to this Job ".$JN ; 
$message = wordwrap($message,70); 
//$headers = "From : Admin@PTN.com"; 
mail($to, $subject, $message); 
} 
} 
else echo "<script type='text/javascript'>alert('ERROR!')</script>"; 

Answer 1

変更部分のコード：

$sql10=" SELECT * FROM JobStudent WHERE JobID='$JID' AND StudentID='$SID'"; 
$resultss = mysqli_query($conn,$sql10); 

変更チェックには、 mysqli_fetch_arrayの代わりに最初のmysqli_num_rows

$numResults = mysqli_num_rows($resultss); 
if ($numResults == 0) {