Database select query statement question.

Asked 2 years ago, Updated 2 years ago, 33 views

<?php
    $dbc = mysqli_connect('127.0.0.1', 'root', 'apmsetup', 'sh')
                or die('Error Connecting to MySQL server.');

    mysqli_query($dbc, "set names utf8;");
    $title = $_POST['item_title'];

    $query = " select * from db where title LIKE '%title%' ";

    $result = mysqli_query($dbc, $query)
                or die('Error Querying databases.');

    $json = array();
    if(mysqli_num_row($result)){
        while($row=mysqli_fetch_assoc($result)){
            $json['list'][] = $row;
        }
        mysqli_free_result($result);
    }
    echo json_encode($json);
    mysqli_close($dbc);
?>

The code is a query that allows me to get the value of title using param.put("item_title",title) from the android java file and get all the values that match the value from the title column. Please check if any of the above codes have a grammar error. I don't think there's anything wrong with the quotation marks of SELECT inquiries, but I don't know exactly where the error occurs.

php android

2022-09-22 21:56

1 Answers

Does error in the question mean something that cannot be asked?

I'm not sure if it's not worth it.

First of all, please revise the following query as below.

$result = mysqli_query($dbc, $query)
                or die('Error Querying databases.');
$stmt = $dbc->prepare("SELECT * FROM db WHERE title LIKE ?");
$stmt->bind_param("s","%{$title}%");
$stmt->execute(); // or die('Error Querying databases.');
$result = $stmt->get_result();


2022-09-22 21:56

If you have any answers or tips


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