警告：mysqli_fetch_array（）要求参数1为mysqli_result，[复制]中给出布尔值

问题描述：

This question already has an answer here:

Reference - What does this error mean in PHP? 34 answers

I have a problem here is my page which i getting the error from

<?php
        // retrive post
        include('config.php');
        include ('function.php');
        dbConnect();

        $query = mysqli_query($GLOBALS["___mysqli_ston"], 
            'SELECT *
            FROM post
            WHERE post_id = 1');
        $row = mysqli_fetch_array($query);
    ?>

and the error is showing this line

$row = mysqli_fetch_array($query);

and the second error is in this code

<div class="comment-block">
         <?php while($comment = mysqli_fetch_array($comment_query)): ?>
            <div class="comment-item">
                <div class="comment-avatar">
                    <img src="<?php echo avatar($comment['mail']) ?>" alt="avatar">
                </div>
                <div class="comment-post">
                    <h3><?php echo $comment['name'] ?> <span>said....</span></h3>
                    <p><?php echo $comment['comment']?></p>
                </div>
            </div>
        <?php endwhile?> 
        </div>

error line is this

 <?php while($comment = mysqli_fetch_array($comment_query)): ?>

pleas help me i am very new to mysqli

</div>

答

There are a lot of possible reasons. Main reason: The query failed. You may have tried to access a table, a column, that is not existing.

Use die($query) and try it out in phpMyAdmin and see if it really executes.

Alternatively using mysqli_error($conn_object) will also contain some information about what happened.

Also, while using the loop, try to give in this format:

while (false != ($comment = mysqli_fetch_array($comment_query))

Snippet

A typical database query, I would write this way:

<?php
    // Connect to the MySQL Server
    $conn = mysqli_connect("localhost", "root", "password", "database") or die("Cannot Connect to MySQL Server. Error: " . mysqli_connect_error());

    // Once connection is established, build a query
    $query = "SELECT * FROM `users`";
    // Execute the query and store it in the results
    $results = mysqli_query($conn, $query);

    // Only continue if there are rows in your result.
    if (mysqli_num_rows($results))
        // Loop through the results, but don't assign directly inside while
        while (false != ($data = mysqli_fetch_array($results)))
            // Display data using the column name
            echo "User: " . $data["User"];
    // Handle the empty scenario
    else
        echo "No users found.";
?>

答

In mysqli_connect() you should pass only hostname to its first argument and pass port as an optional fifth parameter. Also, mysqli would have you specify the database in the mysqli_connect() call instead of having a separate function analogous to mysql_select_db().

check here for more info and answer to this.

警告：mysqli_fetch_array（）要求参数1为mysqli_result，[复制]中给出布尔值

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