I found this issue is related to my AJAX posting method. If I post this form using the standard method (action=””) it works fine it polulates my database according to which record I select in the list to post. But when I use the AJAX Post method it will only poplulate the values with the first record on the SQL Results list. I made my input fields visible and all the posted content is unique and as expected. Below is the full code.
<?php
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
/// Check SQL connection
if ($conn->connect_error) {
echo"my error";
die();
}
// Get the the data from the database
$sql = "SELECT * FROM clients WHERE";
$result = $conn->query($sql);
// Check database connection first
if ($conn->query($sql) === FALSE) {
echo"my error";
exit();
}
else if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo'
<div class="col-md-3 bid-section">
<form action="" method="post" > <!-- Favourites -->
<input type="text" class ="name" name="name" value="'.$row["name"].'">
<input type="text" class ="surname" name="surname" value="'.$row["surname"].'">
<input type="text" class ="country" name="country" value="'.$row["country"].'">
<a class ="favourite-btn" >
<div class="'.$favicon.'"></div>
</a>
</form>
</div> <!-- Column 3 END -->';
}
}
mysqli_free_result($result);
mysqli_close($conn);
?>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.5.1/jquery.min.js"></script>
<script>
AJAX script
$(document).ready(function(){
$('.favourite-btn').click(function(event){
event.preventDefault();
var field1= $('.name').val();
var field2= $('.surname').val();
var field3= $('.country').val();
$.post('mydirectory/add_names_sql.php', {name:field1, surname:field2 , country:field3},
// Alert Success
function(data){
// Alerts the results to this Div
$('.favourite-btn').html(data);
});
});
});
</script>
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Answer
$(‘.name’).val()` is the value of the first element with that class. You need to select the element that’s in the same form as the button that was clicked.
var form = $(this).closest("form");
var field1= form.find('.name').val();
var field2= form.find('.surname').val();
var field3= form.find('.country').val();
You could also get all the form inputs with:
var formdata = $(this).closest("form").serialize();
And in the callback function, you also need to update the appropriate element.
form.find('.favourite-btn').html(data);