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PODisplay image from server in a div (newest first) PHP
 

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  1. PODisplay image from server in a div (newest first) PHP
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    <p>I want to display images in from server. In the database you can find the description and name of the image. the images should be displayed in a div where always the newest is on top. As the number of images increase so does the number of div's. so, the oldest will be at the bottom. I have read a number of posts and forums but I was unable to get the logic and the code on how to do it. I hope you can help me with my problem. Thanks a lot for your answer.</p> <p>the only code i have is the formatting of the dive's:</p> <pre><code>&lt;div class="gallery"&gt; &lt;div class= "group"&gt; &lt;div class="images"&gt; &lt;/div&gt; &lt;div class="details"&gt; &lt;/div&gt; &lt;/div&gt; &lt;div class= "group"&gt; &lt;div class="images"&gt; &lt;/div&gt; &lt;div class="details"&gt; &lt;/div&gt; &lt;/div&gt; &lt;div class= "group"&gt; &lt;div class="images"&gt; &lt;/div&gt; &lt;div class="details"&gt; &lt;/div&gt; &lt;/div&gt; &lt;/div&gt; </code></pre> <p>it's something like this:</p> <p>newest image | some details here</p> <p>newer image | some datails here</p> <p>new image | some details here</p> <p>old image | some details here </p> <p><strong>EDIT 1</strong></p> <p>I combined Lauri Elias' and iamde_coder's answer. I come up with this code which works almost similar to what is wanted. the only problem is that it displays the item (image&amp;details in a div) 4 times. How can I eliminate the three? thanks!</p> <pre><code>$image_query = mysql_query ("SELECT filename, story FROM tbl_contest ORDER BY time DESC"); while($image_data = mysql_fetch_array($image_query)){ $imageName = stripslashes(mysql_real_escape_string($image_data['filename'])); $imageDetails = stripslashes(mysql_real_escape_string($image_data['story'])); $count = 0; foreach($image_data as $imageName) { echo '&lt;div class="group"&gt;'; echo '&lt;div class="images"&gt;&lt;img src="/Mainfolder/image_entry/'.$imageName.'"&gt;&lt;/img&gt;&lt;/div&gt;'; echo '&lt;div class="details"&gt;'.$imageDetails.'&lt;/div&gt;&lt;/div&gt;'; $count ++; } } </code></pre>
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    1. USJben Kaye
    1. USJben Kaye
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    1. COwhere is your code to fetch the images from database ?
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    2. COYou just need to use an `ORDER BY ` in your MySQL query, and order by ImageDate. Then just use a for loop in php to loop through the already ordered records. Optionally you can sort the fetched data with php, but I would do that on the DB server instead.
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      1. USHanlet Escaño
    3. CO@Hanlet thanks for you answer! I will try the logic you suggested and keep you posted to what happens. thanks
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      1. USJben Kaye
 

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