Displaying an image with link underneath in the table

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Displaying an image with link underneath in the table

corkalom25

    $chkuserid = mysql_query("select UserID from users where Username = '".$_SESSION['Username']."'");
    $row = mysql_fetch_array($chkuserid);
    $finall = $row['UserID'];
    $result = mysql_query("SELECT PhotoDesc, photoDateAdded, PhotoString from photo where photographerID='".$finall."'");
    if (!$result) {
    die("Database query failed: " . mysql_error());
  }

  //
  // 4. Use returned data
  //

  // while ($row = mysql_fetch_array($result)) {
  //  echo $row[0]." ".$row[1]."<br />";
  //}

  ?>

<?php

//
//Set up table headings
//
echo "<table border=5 align=center>";
echo "<tr>";
echo '<td>Photo Desc.</td>';
echo '<td>Date&Time Added</td>';
echo '<td>Photo</td>';
echo '<tr>';


//
//Display data returned by the query
//

$i=0;
while ($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo '<td>'.mysql_result($result,$i,0).'</td>';
  echo '<td>'.mysql_result($result,$i,1).'</td>';
  $image = mysql_result($result,$i,2);
  echo '<td>'.'<img src="http://'.$image.'">'.
  [COLOR="#FF0000"].'<a href="http://'.$image.'">View>'.[/COLOR] <==== this is line 185
  '</td>';
  echo '</tr>';
  $i++;
  }
echo "</table>";

Parse error: syntax error, unexpected '.' in /srv/disk7/1259829/www/blablabla/imgup.php on line 185

I want to display an image and just below it a link to display it full size in seperate tab.

When I get rid of link code I can display the images but when I add link code I get this error message.

coding123

There is a dot which is not needed on line 185. Just remove it (first dot you encounter).

outsource

Hi,
you need to change your following code
echo '<td>'.'<img src="http://'.$image.'">'.
.'<a href="http://'.$image.'">View>'.
'</td>';
to

echo '<td>'.'<img src="http://'.$image.'">'.
'<a href="http://'.$image.'">View</a>'. '</td>';

There is extra dot and ending tag of a is missing.

if you want to read tutorials on webdevelopment please also vist http://www.websourceblog.com.

arronlee

How about creating a web-based image viewer to help display an image with link underneath? I am not very sure about it. But you can google it and have a try. And I am also testing about the relevant programs these days. We can aommunicate about it later. Good luck.

Best regards,
Arron

Error404

You need the . for concatenation but when you're specifying HTML elements inside an echo statement, it's easier to use " " as it will automatically make the distinction between PHP variables and HTML elements. This will make your life easier and will (hopefully) avoid these sorts of errors. For example:

echo("<td><img src = $yourSrc /><a href = $yourHref >View</a></td>");

simar

<table cellspacing="0" cellpadding="0" border="0" width="500">
<tr><td><img src="nav1.gif"><img src="nav2.gif"><img src="nav3.gif"><img
src="nav4.gif"><img src="nav5.gif"></td></tr>
<tr><td style="background: red;">
<img src="smallred.gif" height="1" width="1"></td></tr>
<tr><td>
<p style="margin: 0.5em;">This is text in another cell of the
table. Within the textthere is an icon <img src="icon2.gif">
that indicates a link to another site. It's very worldly. Lorem
ipsum, dolor sit amet...</p></td></tr></table>