dongtaoxue4674 2017-08-13 23:16
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Php到MySQL数据库连接

Im not sure whats going wrong with this code. I have a paragraph stored in the database under content, but for some reason I can't get it to display. Any help would be greatly appreciated, I'm a newer developer and welcome any feedback.

<?php
require_once('db_connection.php'); //connection credentials

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

$sql = "SELECT content FROM content";
$result = mysqli_query($conn, $sql);

while($row = mysqli_fetch_row($result)) { 
    echo $row;
}

mysqli_close($conn);
?>
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  • dongmopu6734 2017-08-13 23:20
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    echo $row; you're missing the column's object's array itself that you want to echo.

    What you want is echo $row[0];

    However, both column and table name are the same, so make sure that that is indeed correct.

    As per the manual:

    Example pulled from the manual:

    $query = "SELECT Name, CountryCode FROM City ORDER by ID DESC LIMIT 50,5";
    
    if ($result = mysqli_query($link, $query)) {
    
        /* fetch associative array */
        while ($row = mysqli_fetch_row($result)) {
            printf ("%s (%s)
    ", $row[0], $row[1]);
        }
    
        /* free result set */
        mysqli_free_result($result);
    }
    

    If the above failed, then that could mean that your query may have failed and you need to check for errors on the query, using mysqli_error($conn).

    Reference:

    Same thing goes for the connection.

    Reference:

    Example from the manual:

    <?php
    $link = mysqli_connect("127.0.0.1", "my_user", "my_password", "my_db");
    
    if (!$link) {
        echo "Error: Unable to connect to MySQL." . PHP_EOL;
        echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
        echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
        exit;
    }
    
    echo "Success: A proper connection to MySQL was made! The my_db database is great." . PHP_EOL;
    echo "Host information: " . mysqli_get_host_info($link) . PHP_EOL;
    
    mysqli_close($link);
    ?>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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