Fetching mysql records to a html table using PHP

Question

I am trying to fetch my mysql records into a table, each cell must have one image only.

The problem is that the photo duplicated 5 times according to the number I need to fill each row.

Here is my code:

<table width="%" border="1" bordercolor="#F7F7F7" align="center" cellspacing="10" cellpadding="5">

       <tr>

       <?
       for($i=0;$i<=5;$i++)
       { 

         echo "<td width=125 height=125><img width=125 height=125 src=images/".$info['photo'] ."></td>"; 
         } }?>
       </tr>
     </table>

How can I correct this code to make each photo fetched one time for each cell?

***** EDIT *****

I put the while inside the table, and the fetching is okay now, but it still fetch in the same row, I need to make something to stop fetching until I have 5 cells in the same row, then continue to fetch in a new row.

 <table width="%" border="1" bordercolor="#F7F7F7" align="center" cellspacing="10" cellpadding="5">

       <tr>


       <?

       while($info = mysql_fetch_array( $data  )) 
 {  
 ?>

        <td width=125 height=125 ><img width=125 height=125 src=images/<? echo ($info['photo']); ?>></td>


     <?   } ?>
       </tr>

duplicate of How to create php 2 column table with values from the database? — Your Common Sense
– Your Common Sense, Commented Oct 20, 2011 at 13:08

Thunraz · Accepted Answer · 2011-10-20 13:08:29Z

1

You're not changing $info['photo'] inside the loop. That's why you're echoing the same photo five times.

Depending how your code looks like you can modify your code like this:

$result = mysql_query($your_query);

while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
    echo("<td width=\"125\" height=\"125\"><img width=\"125\" height=\"125\" src=\" images/". $row["photo"] ."></td>");
}

edited Oct 20, 2011 at 13:08

answered Oct 20, 2011 at 13:02

Thunraz

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Comments

Dennis · Accepted Answer · 2011-10-20 13:06:28Z

1

This does not seem to be all the relevant code to answer this question. You are looping 6 times in your 'for' loop and are fetching the same picture '$info['photo']' all 6 times.

You want to go to the next MySQL query every time you loop. So have a

$info = mysql_fetch_row($result)

inside your loop.

answered Oct 20, 2011 at 13:06

Dennis

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Comments

Arny · Accepted Answer · 2011-10-20 13:11:01Z

1

Try add $info = mysql_fetch_row() before echo inside your for-loop. Also check $info != false to ensure that you have enough images (5 items) to fill cells.

answered Oct 20, 2011 at 13:11

Arny

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2 Comments

Bader H Al Rayyes Over a year ago

i tried that but it did not work, all images still not available

Arny Over a year ago

Maybe, $info = mysql_fetch_row($data) will work? For your edit - declare $i variable, assign 0 and inside loop add code if (++$i >= 5) {echo "</tr><tr>"; $i = 0; }

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