类别不起作用

时间：2014-02-07 11:03:06

标签： php mysql

所以我尝试学习php并决定创建一个我添加图像的网站，将它们保存在文件夹和id，名称，类型，mysql中的路径中。然后在页面上显示。到目前为止，我已上传表单，我可以上传和保存图像。我也在页面上成功展示了它们。

现在我正在尝试制作类别 - 自然，有趣......等等。所以我在主表中添加了一个字段 - ＆gt; img_category。 我也疯了第二张桌子 - 猫咪cat_id和cat_name字段。使用它来显示页面上的类别：

         <?php
        $q = mysqli_query($con,"select * from cats");
        while ($res = mysqli_fetch_assoc($q))
        {
            echo '<a href="pic.php?cat_id='. $res['cat_id'] .'">'.$res['cat_name'].'<br/>';
        }

现在，当我点击某个类别链接以仅从此类别加载图片时，我该如何制作？ 我已经设法制作了这样的东西，但它不像预期的那样工作

<?php
        $q = mysqli_query($con,"select * from cats");
        while ($res = mysqli_fetch_assoc($q))
        {
            echo '<a href="pic.php?cat_id='. $res['cat_id'] .'">'.$res['cat_name'].'<br/>';
        }

        ?>
        <hr>
        <?php
        $cat_id = $_GET['cat_id'];
        $query = "SELECT * FROM images JOIN cats ON images.img_category = cats.cat_id WHERE cats.cat_id = '$cat_id'";
        $result = mysqli_query($con, $query) or die("Query failed: " . mysqli_errno($con));

        $line = mysqli_fetch_array($result, MYSQL_BOTH);
        if (!$line) echo '';
        $previd = -1;
        $currid = $line[0];
        if (isset($_GET['id'])) {
            do {
                $currid = $line[0];
                if ($currid == $_GET['id']) break;
                $previd = $currid;
                $line = mysqli_fetch_array($result, MYSQL_BOTH);
            } while ($line);
        }

        if ($line) {
            echo "<div id=\"picture\">";

            echo "<img style=\"width:100%;margin:0 auto;\" src=\"upload/".$line['name']."\" /></a><br />";
            echo "<div id=\"caption\">".$line['caption']."</div><br />";
        }
        else echo "There is no images!\n";

        if ($previd > -1) echo '<a href="pic.php?cat_id='.$previd.'" class="prev_pic"><span>Prev</span></a>';
        echo str_repeat('&nbsp;', 5);

        $line = mysqli_fetch_array($result, MYSQL_BOTH);

        $query = "select * from images 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="pic.php?cat_id='.$row['id'].'"class="random">Random</a>';
        }
        echo str_repeat('&nbsp;', 5);
        if ($line) echo '<a href="pic.php?cat_id='.$line[0].'" class="next_pic"><span>Next</span> </a><br /><br />';

        echo "</div>";

        ?>

结果是：

当显示类别中的图像时，如果我点击“下一步”按钮，我会得到相同的图像。 如果类别中没有图像，我会得到所有回声，例如链接白色，检查的最后一个类别的ID：没有像链接的图像，如果我点击它，我会加载最后一个类别ID。在我的情况下，我有8个类别，所以ID = 8。 任何帮助都很感激！ 感谢的

编辑： 好的这一行：

echo '<a href="pic.php?cat_id='.$line[0].'" class="next_pic"><span>Следваща</span> </a>

pic.php在哪里？cat_id = ...我认为是错的。在这里，我必须采取下一个图像ID而不是下一个类别ID但是如何改变图像呢？如果我把它作为pic.php？id = ...我得到空页。 我不明白。我知道这是凌乱的代码，但我现在能做的最好。

编辑2： 我做了这样的事。现在，您可以帮助我如何为下一个图像进行查询，因为现在没有获得下一个图像并保持不变。 

$cat_id = $_GET['cat_id'];
        $cat_id = mysqli_real_escape_string($con, $cat_id);
        $query = "SELECT * FROM images JOIN cats ON images.img_category = cats.cat_id WHERE cats.cat_id = '$cat_id'";
        $result = mysqli_query($con, $query) or die("Query failed: " . mysqli_errno($con));
$prevSQL = mysqli_query($con,"SELECT cat_id FROM cats WHERE cat_id < $cat_id ORDER BY cat_id DESC LIMIT 1") or die (mysqli_error($con));
        $nextSQL = mysqli_query($con, "SELECT cat_id FROM cats WHERE cat_id > $cat_id ORDER BY cat_id ASC LIMIT 1") or die (mysqli_error($con));
        $prevobj=mysqli_fetch_object($prevSQL);
        $nextobj=mysqli_fetch_object($nextSQL);
        $pc = mysqli_fetch_object(mysqli_query($con, "SELECT COUNT(cat_id) as pid FROM cats WHERE cat_id<$cat_id ORDER BY cat_id DESC")) or die (mysqli_error($con));
        $nc = mysqli_fetch_object(mysqli_query($con, "SELECT COUNT(cat_id) as nid FROM cats WHERE cat_id>$cat_id ORDER BY cat_id ASC")) or die (mysqli_error($con));
        $prev=$pc->pid>0 ? '<a href="pic.php?cat_id='.$cat_id.'&id='.$prevobj->cat_id.'">Prev</a> |' : '';
        $next=$nc->nid>0 ? '<a href="pic.php?cat_id='.$cat_id.'&id='.$nextobj->cat_id.'">Next</a>' : '';
        $row = mysqli_fetch_array($result);
        echo "<div id=\"picture\">";
        echo "<img src=\"upload/" . $row['name']  . "\" alt=\"\" /><br />";
        echo $row['caption'] . "<br />";
        echo "</p>";
        echo $prev;
        echo $next;

2 个答案:

答案 0 :(得分：2)

正如你所说，我猜错误在于：

echo '<a href="pic.php?cat_id='.$line[0].'" class="next_pic"><span>Следваща</span> </a>

我认为应该是：

echo '<a href="pic.php?cat_id='.$line[0].'&id='.$next_picture_id.'" class="next_pic"><span>Следваща</span> </a>

修改 您的代码应如下所示：

<?php
    $q = mysqli_query($con,"select * from cats");
    while ($res = mysqli_fetch_assoc($q))
    {
        echo '<a href="pic.php?cat_id='. $res['cat_id'] .'">'.$res['cat_name'].'<br/>';
    }

    ?>
    <hr>
    <?php
    $cat_id = $_GET['cat_id'];
    $query = "SELECT * FROM images JOIN cats ON images.img_category = cats.cat_id WHERE cats.cat_id = '$cat_id'";
    $result = mysqli_query($con, $query) or die("Query failed: " . mysqli_errno($con));

    $line = mysqli_fetch_array($result, MYSQL_BOTH);
    if (!$line) echo '';
    $previd = -1;
    $currid = $line[0];
    if (isset($_GET['id'])) {
        do {
            $currid = $line[0];
            if ($currid == $_GET['id']) break;
            $previd = $currid;
            $line = mysqli_fetch_array($result, MYSQL_BOTH);
        } while ($line);
    }

    if ($line) {
        echo "<div id=\"picture\">";

        echo "<img style=\"width:100%;margin:0 auto;\" src=\"upload/".$line['name']."\" /></a><br />";
        echo "<div id=\"caption\">".$line['caption']."</div><br />";
    }
    else echo "There is no images!\n";

    if ($previd > -1) echo '<a href="pic.php?cat_id='.$previd.'" class="prev_pic"><span>Prev</span></a>';
    echo str_repeat('&nbsp;', 5);

    $line = mysqli_fetch_array($result, MYSQL_BOTH);

    $query = "select * from images 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="pic.php?cat_id='.$row['id'].'"class="random">Random</a>';
    }
    echo str_repeat('&nbsp;', 5);
    if ($line) echo '<a href="pic.php?cat_id='.$line[0].'&id='.$line[0].'" class="next_pic"><span>Next</span> </a><br /><br />';

    echo "</div>";

    ?>

答案 1 :(得分：1)

试试这个

<?php
        if(isset($_GET['cat_id'])){
            $cat_id = $_GET['cat_id'];
            $query = "SELECT * FROM images WHERE img_category = '$cat_id'";
            $result = mysqli_query($con, $query) or die("Query failed: " . mysqli_errno($con));

            $line = mysqli_fetch_array($result, MYSQL_BOTH);
            if (!$line) echo '';
            $previd = -1;
            $currid = $line[0];
            if (isset($_GET['id'])) {
                $previous_ids = array();
                do {
                    $previous_ids[] = $line[0];
                    $currid = $line[0];
                    if ($currid == $_GET['id']) break;
                    $previd = end($previous_ids);
                    $line = mysqli_fetch_array($result, MYSQL_BOTH);
                } while ($line);
            }

            if ($line) {
                echo "<div id=\"picture\">";

                echo "<img style=\"width:100%;margin:0 auto;\" src=\"upload/".$line['name']."\" /><br />\r";
                echo "<div id=\"caption\">".$line['caption']."</div><br />";
            }
            else echo "There is no images!\n";

            if ($previd > -1)
                echo '<a href="pic.php?cat_id='.$cat_id.'&amp;id='.$previd.'" class="prev_pic"><span>Prev</span></a>';
            echo str_repeat('&nbsp;', 5);

            $line = mysqli_fetch_array($result, MYSQL_BOTH);

            $query = "select * from images WHERE img_category = '$cat_id' 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="pic.php?cat_id='.$cat_id.'&amp;id='.$row['id'].'"class="random">Random</a>';
            }
            echo str_repeat('&nbsp;', 5);
            if ($line) echo '<a href="pic.php?cat_id='.$cat_id.'&amp;id='.$line[0].'" class="next_pic"><span>Next</span> </a><br /><br />';

            echo "</div>\r";
        }
        ?>
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