CODEWITHSUNDEEP
phpmysqli
VoxVola
VoxVola

Reputation: 59

PHP passing data from 1 page to another page

How do I pass id from 1st page to 2nd page?

In 1st Page I have query out my table

1st table: userdetails
            userID
            Username
            Email
            Address
            NationalID
            View More

In 2nd Page I need to mix 2 table into 1

2nd table: userimage
imageID
UserID
imageupload

For view more is not from my table is from this code.

<?php echo "<a href='userdetails.php?id=".$result['userID']."'>"?> View More </td>

Below is my 2nd page code 

 $sql = "SELECT userdetails.imageupload
FROM userdetails
INNER JOIN user ON userdetails.memberpostID = user.memberpostID";

  $query = mysqli_query($db,$sql);

And this is the output on 2nd page

<?php
while($result=mysqli_fetch_array($query,MYSQLI_ASSOC))
{
DATA OUTPUT 
}

But the output always show all data from userimage

Question that I want: I wanted that if user click on "VIEW MORE" on userID 1 , it will show userID details + imageupload on my 2nd page.

Upvotes: 0

Views: 62

Answers (1)

Fawzan
Fawzan

Reputation: 4849

You need to pass the userID to the query. 

$userID = $_GET['id'];

WRONG WAY

$sql = "SELECT userdetails.imageupload
        FROM userdetails
        INNER JOIN user ON userdetails.memberpostID = user.memberpostID WHERE userdetails.userID='".$userID."'";

CORRECT WAY

$sql = "SELECT userdetails.imageupload
    FROM userdetails
    INNER JOIN user ON userdetails.memberpostID = user.memberpostID WHERE userdetails.userID=:userID";

$query = $db->prepare($sql);
$query->execute(array(':userID' => $userID));

Note: User parameterized query to pass in the userID from the $_GET request to prevent SQL Injection. Don't concatenate variables, instead bind the parameters.

Upvotes: 2

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