Question

我是新手，还在学习php＆amp; mysql。搜索一整天并尝试不同的教程但没有任何反应。到目前为止，我只有这个，我从我的数据库中获取文本。如何在这里制作上一个和下一个按钮？

if($q = mysqli_query($con, 'SELECT * FROM joke WHERE `id` = ' . mysqli_real_escape_string($con,$_GET['id']))){
    if($row = mysqli_fetch_array($q)){
        echo nl2br($row['text']);
        echo '<div id="data">Date ' . $row['date'] . "</div>';
    } else {
        echo 'Not found';
    }
} else {
    echo mysqli_error($con);
}

echo "</div>";
$query = "select * from joke order by RAND() LIMIT 1";
$result = mysqli_query($con, $query) or die("Query failed: " . mysqli_errno($con));
while ($row = mysqli_fetch_array($result, MYSQL_BOTH)){
    echo '<a href="single.php?id='.$row['id'].'"class="random">Random</a>';
}

Answer 1

如果我理解正确你想要显示基于id的下一个和之前的笑话，
所以你想从笑话中选择max（id），其中id＆lt;当前的身份
和...从笑话中选择min（id），其中id＆gt;当前的身份

$currentId = mysqli_real_escape_string($con,$_GET['id']);
if($q = mysqli_query($con, 'SELECT *,
(SELECT IFNULL(max(id),-1) FROM joke WHERE `id` < '.$currentId.') as previousid,
(SELECT IFNULL(min(id),-1) FROM joke WHERE `id` > '.$currentId.') as nextid
  FROM joke WHERE `id` = ' . $currentId)){
    if($row = mysqli_fetch_array($q, MYSQL_BOTH)){
        echo nl2br($row['text']);
        echo '<div id="data">Date ' . $row['date'] . '</div>';
        if ($row['previousid'] > -1){
            echo '<a href="single.php?id='.$row['previousid'].'"class="random">Previous</a>';
        }
        if ($row['nextid'] > -1){
            echo '<a href="single.php?id='.$row['nextid'].'"class="random">Next</a>';
        }
    } else {
        echo 'Not found';
    }
} else {
    echo mysqli_error($con);
}

echo "</div>";
$query = "select * from joke order by RAND() LIMIT 1";
$result = mysqli_query($con, $query) or die("Query failed: " . mysqli_errno($con));
while ($row = mysqli_fetch_array($result, MYSQL_BOTH)){
    echo '<a href="single.php?id='.$row['id'].'"class="random">Random</a>';
}

上一个和下一个按钮

1 个答案: