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Datos no puestos en el párrafo JS/AJAX/PHP

Mi script php hace eco

Connection successful{"name":"James","dateOfBirth":"2002-10-10","companyname":"Apple"}

usando

echo json_encode(array('name'=>$row['Name'], 'dateOfBirth'=>$row['dateOfBirth'], 'companyname'=>$row['companyname']));

En Javascript, los diferentes valores deberían mostrarse en una etiqueta, pero las etiquetas devuelven indefinidos

 $(function() { $.ajax({ type: "post", url: 'url', data: {usermail: 42}, success: function(response){ nameLabel.innerHTML = response.name; dateLabel.innerHTML = response.dateOfBirth; companyLabel.innerHTML = response.companyname; console.log(response); } }); });

Ese es el código php:

 <?php $host = ''; $database = ''; $user = ''; $pswd = ''; header("Access-Control-Allow-Origin: http://127.0.0.1:5500"); $conn = new mysqli($host, $user, $pswd, $database); if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } echo "Connection successful"; $user = $_POST['usermail']; $sql = $conn->prepare("SELECT Name, dateOfBirth, companyname FROM Customers WHERE user_id = ?"); $sql->bind_param("i", $user); $sql->execute(); $result = $sql->get_result(); if($result->num_rows === 0) exit('No rows'); while($row = $result->fetch_assoc()) { echo json_encode(array('name'=>$row['Name'], 'dateOfBirth'=>$row['dateOfBirth'], 'companyname'=>$row['companyname'])); } $sql->close(); $conn->close(); ?>
about 4 years ago · Juan Pablo Isaza
1 answers
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0

Elimina la línea echo "Connection successful"; de su código PHP .

about 4 years ago · Juan Pablo Isaza Report
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