I’m using mysqli before to my query and now I convert it to mysqli prepared statements. I’m trying to update a particular data with upload image and I don’t know why I get the error
mysqli_stmt_bind_result(): Number of bind variables doesn’t match number of fields in prepared statement in line 30
Also, how can I execute the query in mysqli query like mysqli_query($conn, $query)
Below is the code of my UPDATE query:
JavaScript
x
if (isset($_POST['submit'])) {
$imageName = mysqli_real_escape_string($conn, $_FILES["latest_photo"]["name"]);
$imageData = mysqli_real_escape_string($conn, file_get_contents($_FILES["latest_photo"]["tmp_name"]));
$imageType = mysqli_real_escape_string($conn, $_FILES["latest_photo"]["type"]);
if (substr($imageType, 0,5) == "image") {
$query = "UPDATE `crew_info` SET `updated_photo` = ? WHERE `id` = ?";
$stmt = mysqli_prepare($conn, $query);
mysqli_stmt_bind_param($stmt, 'ss', $imageData, $_GET['id']);
mysqli_stmt_execute($stmt);
mysqli_stmt_bind_result($stmt, $id, $updated_photo);
//HOW CAN I EXECUTE THE QUERY HERE?
echo "Image Uploaded";
}
else {
echo "Image is not uploaded!";
}
}
In the code above, there is a comment line on how to execute the query. How can I do that?
When I click the button Upload, it says that the image is uploaded but does not appear in the database. Why is that?
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Answer
For procedural way
JavaScript
$query = "UPDATE `crew_info` SET `updated_photo` = ? WHERE `id` = ?";
$stmt = mysqli_prepare($conn, $query);
// you should use i instead of s for id
mysqli_stmt_bind_param($stmt, 'si', $imageData, $_GET['id']);
mysqli_stmt_execute($stmt);
Try this out in object-oriented style
JavaScript
$query = "UPDATE `crew_info` SET `updated_photo` = ? WHERE `id` = ?";
$stmt = $conn->prepare($query);
$stmt->bind_param("si", $imageData, $_GET['id']);
$stmt->execute();