未定義の変数：sql in C：\ xampp \ htdocs \ training.php on line 17

Question

私はIDを取得して変数に格納しようとしていますが、 変数が定義されておらず、クエリが変数のために実行されない、私はなぜ理解していない。

include 'connect.php'; 
    if(isset($_GET['OrgName'])) 
    { 
    $file=$_GET['OrgName']; 

$sql="SELECT oraganzation.OrgName,oraganzation.City,oraganzation.OrgEmail,oraganzation.OrgPhoneNO, oraganzation.Workfield, oraganzation.Trainingrecruitment, oraganzation.WebsiteLink,Student_comments.comment 
FROM oraganzation , Student_comments 
WHERE oraganzation.OrgName=$file AND oraganzation.OrgID=Student_comments.OrgID "; 
} 
$result= mysqli_query($con,$sql) or die ("could not found; ".mysqli_error ($con)); 

while ($row=mysqli_fetch_array($result)) 
{ 

    echo "<br><strong> Name : </strong>". $row['OrgName']. 
"<br><strong> 
<br><strong> City : </strong>". $row['City']. 
"<br><strong> 
Email: </strong>" . $row['OrgEmail']. 
"<br><strong> 
PhoneNO: </strong>". $row['OrgPhoneNO']. 
"<br> <strong> 
Work field: </strong> " . $row['Workfield']. 
"<br><strong> 
Training recruitment:</strong> " . $row['Trainingrecruitment']. 
"<br> <strong> 
Website Link: </strong> " . $row['WebsiteLink']. 
"<br> <strong> 
Comments: </strong> " .$row['comment']. "<br>" ; 

    } 

そして、これは私が

$sql="SELECT OrgName, City, OrgEmail, OrgPhoneNO, Workfield, Trainingrecruitment, WebsiteLink,OrgID 
    FROM oraganzation"; 

$result= mysqli_query($con,$sql) or die ("could not found; ".mysqli_error($con)); 

while ($row=mysqli_fetch_array($result)) 
{ 

?> 

<div class="content "> 
<a href="training.php?name=<?php echo $row['OrgName'] ?> "><?php echo $row['OrgName'] ;?> </a> 

     <?php 
echo "<br><strong> City : </strong>". $row['City']. 
    "<br><strong> 
    Email: </strong>" . $row['OrgEmail']. 
    "<br><strong> 
    PhoneNO: </strong>". $row['OrgPhoneNO']. 
    "<br> <strong> 
    Work field: </strong> " . $row['Workfield']. 
"<br><strong> 
Training recruitment:</strong> " . $row['Trainingrecruitment']. 
    "<br> <strong> 
    Website Link: </strong> " . $row['WebsiteLink']. 


    "</div> ";} ?> 

Answer 1

$ sqlをのみ場合

if(isset($_GET['OrgName'])) 

あなたが閉鎖を移動する必要が定義されている名前を取得するためのhrefを入れて、前のページからの私のコードですOrgNameが設定されていなくても、$ sqlの後ろにかっこ、クエリーの実行後に$ sqlを定義する