I am using php and mysql to display all the user information of different users and i have a button which gets the id which will be redirected to another page and the id will be displayed in the url. What i am trying to do now is display the user information with the selected id on the redirected page
JavaScript
x
$sql = "SELECT * FROM users";
if($result = mysqli_query($link, $sql)){
if(mysqli_num_rows($result) > 0){
while($row = mysqli_fetch_array($result)){
echo "<div class='users-data'>";
echo "<p>" . $row['username'] . "</p>";
echo "<p>" . $row['full_name'] . "</p>";
echo "<p>" . $row['age'] . "</p>";
echo "<p>" . $row['gender'] . "</p>";
echo "<p>" . $row['email'] . "</p>";
echo "<p>" . $row['medical_condition'] . "</p>";
echo "<img src=images/".$row['image'] ."/>";
echo '<td><a href="view-user-information.php?id='.$row['id'].'"><button>View Details</button></td>';
echo "</div>";
}
// Free result set
mysqli_free_result($result);
} else{
echo "No records matching your query were found.";
}
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
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Answer
Use this code;
JavaScript
if(isset($_GET['u'])){
$id = $_GET['u'];
$sql = "SELECT * FROM `users` WHERE `id` = '$id'";
$result = mysqli_query ($link,$sql);
$row = mysqli_fetch_assoc($result);
echo "<div class='users-data'>";
echo "<p>" . $row['username'] . "</p>";
echo "<p>" . $row['full_name'] . "</p>";
echo "<p>" . $row['age'] . "</p>";
echo "<p>" . $row['gender'] . "</p>";
echo "<p>" . $row['email'] . "</p>";
echo "<p>" . $row['medical_condition'] . "</p>";
echo "<img src=images/".$row['image'] ."/>"
}
else
{
//you can redirect it to back to the previous page if no id exist in url;
header("LOCATION:index.php"); // change the index.php to your privious page url
}