Question

我使用jquery-ui标签使用php但是它无法正常工作，我无法确定问题，请帮助我。

<div id="tabs">
    <div class="row">
        <div class="col-md-4">

        <?php
        $con = mysqli_connect("localhost", "admin", "123456", "gazette");

        if (mysqli_connect_errno()) {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
        }
        $result = mysqli_query($con, "SELECT * FROM gazette_details");
        while ($row = mysqli_fetch_array($result)) {

            echo "<ul>";
            $id = '#tabs-' . $row['id'];
            echo "<li>" . "<a href='$id'>" . $row['gazette_id'] . "</a>" . "</li>";
            echo "</ul>";
        }
        mysqli_close($con);
        ?>

        </div>

        <div class="col-md-4">
            <div id="tabs-12">1</div>
            <div id="tabs-13">2</div>
            <div id="tabs-14">3</div>
        </div>
    </div>
    <div class="col-md-4">.col-md-4</div>
</div>

我使用了这个网站示例：http://jqueryui.com/tabs/

Answer 1

你可以尝试

<?php   
    $ARR_RESULT = array();
    $con = mysqli_connect("localhost", "admin", "123456", "gazette");

    if (mysqli_connect_errno()) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    $result = mysqli_query($con, "SELECT * FROM gazette_details");
    while ($row = mysqli_fetch_array($result)) {
        $ARR_RESULT[] = $row;

    }
    mysqli_close($con);

?>

<div id="tabs">
    <ul>
        <?php
            foreach($ARR_RESULT as $arr)
            {
                echo '<li><a href="#tabs-'.$arr['id'].'">'.$arr['gazette_id'].'</a></li>';
            }
        ?>
    </ul>

        <?php
            foreach($ARR_RESULT as $arr)
            {
                echo '<div id="tabs-'.$arr['id'].'">';
                echo '<p>Tab-'$arr['id'].'</p>';
                echo '</div>';
            }
        ?>


</div>

如何使用PHP创建jquery选项卡？

1 个答案: