从数据库中显示多个图像？ （PHP，MySQL）

Question

如果需要，我试图显示多个图像。

我不是要求勺子喂食，而是要朝着正确的方向前进。

我当前的echo打印数据库列：

    <table style="margin:0 5%;border-left: 1px solid #000;" id="monsterstable" cellpadding="4" cellspacing="0" width="90%"><thead> <tr>
  <th class="tabletop" width="150">NPC Picture</th>
  <th class="tabletop">NPC Name </th>
<th class="tabletop">NPC Drops</th>
</tr></thead> 
<?php
    $query = $_GET['query']; 
    // gets value sent over search form

    $min_length = 3;
    // you can set minimum length of the query if you want

    if(strlen($query) >= $min_length){ // if query length is more or equal minimum length then

        $query = htmlspecialchars($query); 
        // changes characters used in html to their equivalents, for example: < to &gt;

        $query = mysql_real_escape_string($query);
        // makes sure nobody uses SQL injection

        $raw_results = mysql_query("SELECT * FROM npcs
            WHERE (`npc_name` LIKE '%".$query."%')") or die(mysql_error());

        // * means that it selects all fields, you can also write: `id`, `title`, `text`
        // articles is the name of our table

        // '%$query%' is what we're looking for, % means anything, for example if $query is Hello
        // it will match "hello", "Hello man", "gogohello", if you want exact match use `title`='$query'
        // or if you want to match just full word so "gogohello" is out use '% $query %' ...OR ... '$query %' ... OR ... '% $query'

        if(mysql_num_rows($raw_results) > 0){ // if one or more rows are returned do following

            while($results = mysql_fetch_array($raw_results)){
            // $results = mysql_fetch_array($raw_results) puts data from database into array, while it's valid it does the loop

             echo '<form action="search.php" method="GET">
        <input type="text" name="query" />
        <input type="submit" value="Search" />
    </form>';
             echo "<br>Showing results for <b>'".$query."'</b><br>";
                echo '<tr><td class="tablebottom"><img src="'.$results['image'].'" width="100px" height="100px""></td><td class="tablebottom">'.$results['npc_name'].'</td>
<td class="tablebottom"><img src="'.$results['location'].'" width="20px" height="20px"></td></tr>';
                // posts results gotten from database(title and text) you can also show id ($results['id'])
            }

        }
        else{ // if there is no matching rows do following
            echo 'No results<br><form action="search.php" method="GET">
        <input type="text" name="query" />
        <input type="submit" value="Search" />
    </form>';

        }

    }
    else{ // if query length is less than minimum
        echo '<form action="search.php" method="GET">
        <input type="text" name="query" />
        <input type="submit" value="Search" />
    </form><br><br>Minimum length is '.$min_length.'<br>';

    }
?>
</table>

我的SQL：

http://i.stack.imgur.com/XEquI.png

如何添加多个图像并使其回显？

Answer 1

您想要遍历MySQL行并回显每个img标记。

foreach($results as $result){
    echo "<img src='$results->image' width='20px' height='20px'>";
}

调整您已经循环浏览结果的方式，无论您是否正在使用PDO或Mysqli或诸如此类的东西。这假设您获得了一个名为$results的结果数组，其中每个元素都包含一个行对象，其中键image包含您的图片网址。

编辑：

那么，如果我正确地解释您的数据结构，那么就是这样：

echo '<form action="search.php" method="GET">
            <input type="text" name="query" />
            <input type="submit" value="Search" />
        </form>';
echo "<br>Showing results for <b>'".$query."'</b><br>";

while($results = mysql_fetch_array($raw_results)){
    echo '<tr>
        <td class="tablebottom">
            <img src="'.$results['image'].'" width="100px" height="100px">
        </td>
        <td class="tablebottom">'.$results['npc_name'].'</td>
        <td class="tablebottom">
            <img src="'.$results['location'].'" width="20px" height="20px">
        </td>';
}

echo '</tr>';

    // posts results gotten from database(title and text) you can also show id ($results['id'])
}

尽管如其他人所述，mysql已在PHP中弃用，但应使用mysqli或PDO。