2019-04-14 18:28
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I am trying to get and display two values from the database, 'cat_name', and 'cat_description' and display those in a table. A far as I can tell the code is fine. I am pretty new to PHP and have little experience with it.

I have tried going through many questions, and tried many possible answers. I cant really tell how many attempts I have tried to fix this issue.

    $query = $con->prepare("SELECT cat_id, cat_name, cat_description FROM categories");

    $result = $query->fetchAll(PDO::FETCH_ASSOC);
        //something went wrong, display the error
        echo 'Error'; 
        if($result == 0)
            echo 'No categories defined yet.';
            echo '<table border ="1">
                    <th>Last Topic</th>


            while($row = mysqli_fetch_array($result)){
            echo '<tr>';
            echo '<td class="leftpart">';
                echo '<h3><a href="category.php?id">' . $row['cat_name'] . '</a></h3>' . $row['cat_description'];
            echo '</td>';
            echo '<td class="rightpart">';
                echo '<a href="topic.php?id=">Topic subject</a> at 10-10';
            echo '</td>';
            echo '</tr>';


I am supposed to expect a table showing two columns one showing Categories (hyperlinked) with a small description of the category taking from the database. And another column for topics within that category but have not been able to get that far as of yet.

Expected outcome:

Expected outcome

When I run this code all I seem to achieve is an error.

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, array given

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1条回答 默认 最新

  • doxd96148 2019-04-14 18:42

    Once you have run

    $result = $query->fetchAll(PDO::FETCH_ASSOC);

    $result will be an array of the records from the query, so when you get to

    while($row = mysqli_fetch_array($result)){

    you already have the data so $result is not a result set, but an array of data. So instead you can replace this line with...

    foreach ( $result as $row ) {
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