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PHP 图片库

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我已经设置了一个新的图像库,其中图像存储在/images/文件夹中的服务器上,相应的细节存储在 mysql 数据库(标题,描述,imagesrc)中。

然而,我要做的是将服务器上文件夹中存储的所有图像拉回来,而不必在 html 中列出它们。我已经让 php 拉回正确的字段并正确填充图库,但是当一个新图像被添加到库或其中一个原始图像被更新时,它显然不会在我的网站上通过。

我的代码是:

<div id="galleria"><!-- Begin Galleria -->
        <div>
                <a href="<?php mysql_select_db("dbname", $con);

$result = mysql_query("SELECT * FROM images WHERE id='1'");

while($row = mysql_fetch_array($result))
  {
  echo $row['imagesrc'];
  }

mysql_close($con);
?>">
                <img src="<?php mysql_select_db("dbname", $con);

$result = mysql_query("SELECT * FROM images WHERE id='1'");

while($row = mysql_fetch_array($result))
  {
  echo $row['imagesrc'];
  }

mysql_close($con);
?>" alt='' title='' />
        </a>
        <strong><?php mysql_select_db("dbname", $con);

$result = mysql_query("SELECT * FROM images WHERE id='1'");

while($row = mysql_fetch_array($result))
  {
  echo $row['title'];
  }

mysql_close($con);
?></strong>
        <span><?php mysql_select_db("dbname", $con);

$result = mysql_query("SELECT * FROM images WHERE id='1'");

while($row = mysql_fetch_array($result))
  {
  echo $row['desc'];
  }

mysql_close($con);
?></span>
        </div>
        <div>
                <a href="<?php mysql_select_db("dbname", $con);

$result = mysql_query("SELECT * FROM images WHERE id='1'");

while($row = mysql_fetch_array($result))
  {
  echo $row['imagesrc'];
  }

mysql_close($con);
?>">
                <img src="<?php mysql_select_db("dbname", $con);

$result = mysql_query("SELECT * FROM images WHERE id='1'");

while($row = mysql_fetch_array($result))
  {
  echo $row['imagesrc'];
  }

mysql_close($con);
?>" alt='' title='' />
        </a>
        <strong><?php mysql_select_db("dbname", $con);

$result = mysql_query("SELECT * FROM images WHERE id='1'");

while($row = mysql_fetch_array($result))
  {
  echo $row['title'];
  }

mysql_close($con);
?></strong>
        <span><?php mysql_select_db("dbname", $con);

$result = mysql_query("SELECT * FROM images WHERE id='1'");

while($row = mysql_fetch_array($result))
  {
  echo $row['desc'];
  }

mysql_close($con);
?></span>
        </div>
        <div>

在循环中检索此信息的最佳方法是什么 - 我不是非常熟练的 PHP(正如您从我的代码中看到的那样!)所以非常感谢有关在此脚本中构建此类循环以填充库的任何帮助和指导在文件夹中的所有图像上。

谢谢!!

JD

3 回答

  • 0

    对于像这样的简单查询,您可以遍历整个表并在一个 while 循环中填充您的库,如下所示:

    <?php
    
    mysql_select_db("dbname", $con);
    
    $result = mysql_query("SELECT * FROM images");
    
    while($row = mysql_fetch_array($result))
    {
        echo '<a href="' . $row['imagesrc'] . '">' . $row['title'] . '</a>';
    }
    
    mysql_close($con);
    ?>
    

    哪个印刷品:

    <a href="imagesrc1">Imagetitle1</a>
    <a href="imagesrc2">Imagetitle2</a>
    <a href="imagesrc3">Imagetitle3</a>
    ...
    

    你可以在 while 循环中混合你需要的任何 div,span 和其他字段(如$row [5]),例如。

    while($row = mysql_fetch_array($result))
    {
        echo '<div>';
        echo '<a href="' . $row['imagesrc'] . '">';
        echo '<img src="' . $row['imagesrc'] . '" />';
        echo '</a>';
        echo '<strong>' . $row['title'] . '</strong>';
        echo '<span>' . $row['desc'] . '</span>';
        echo '</div>';
    }
    

    哪个会打印:

    <div>
        <a href="imagesrc1"><img src="imagesrc1" /></a>
        <strong>imagetitle1</strong>
        <span>desc1</span>
    </div>
    <div>
        <a href="imagesrc2"><img src="imagesrc2" /></a>
        <strong>imagetitle2</strong>
        <span>desc2</span>
    </div>
    <div>
        <a href="imagesrc3"><img src="imagesrc3" /></a>
        <strong>imagetitle3</strong>
        <span>desc3</span>
    </div>
    

    希望这有帮助!

  • 0

    您正在通过 mysql_close()关闭 MySQL 连接,因此这不起作用。您要执行的步骤是:

    mysql_connect
    mysql_select_db
    mysql_query (or multiple queries, if placement matters)
    mysql_close
    
  • 0

    你不需要再次执行查询只需执行一次,当在循环中获取 thr 记录时,使用 if(如果需要)根据您的要求放置 div。我正在向您展示我最近使用的代码示例。

    <?php
    $conn=mysqli_connect(DBHOST,DBUSER,"",DB);
    $query1="select dev_image,dev_name from developers";
    $result=mysqli_query($conn,$query1);
    $cnt=0;
    while($row=mysqli_fetch_array($result))
        //$dev_name=row['dev_name'];
    {               
       $dev_image=$row['dev_image'];
       $dev_name=$row['dev_name'];
        if($cnt%4==0) {
        echo "<div class=\"project_main\">";
    }
    if($cnt%4==0) { 
            echo "<div class=\"project_img_main\">";
            echo "<div class=\"project_img1\">";
            echo "<img src=\"$dev_image\" alt=\"\" title=\"Project-1\" border=\"none\" />";
            echo "</div>";
            echo "<div class=\"project_img_name\">";
            echo "<p align=\"center\" class=\"txt1\">".$dev_name."</p>";
            echo "</div>";
        echo "</div>";
    } else {
            echo "<div class=\"project_img_main1\">";
            echo "<div class=\"project_img1\">";
            echo "<img src=\"$dev_image\" alt=\"\" title=\"Project-1\" border=\"none\" />";
        echo "</div>";
        echo "<div class=\"project_img_name\">";
            echo "<p align=\"center\" class=\"txt1\">".$dev_name."</p>";
        echo "</div>";
            echo "</div>";
    }
    
    $cnt++;
    if($cnt%4==0) {
        echo "</div>";
    }
    }
    
    ?>
    

    我希望 you..just 做出相应的帮助。

    谢谢
    拉杰什

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