PHP SQL Display images

Question

I've written a php code to display all the images. But there's is something wrong in the code and I can't fix it. It's kind of a syntax error but I've wasted hours over it and still mixing up the "quotes(')"..here's my php code:

while($row = mysql_fetch_array($display_query)){
    print "<tr><td>".$row['itemid']."</td><td><img src="resources/wh/'.$row['itemid'].'.png"/></td><td>".$row['description']."</td><td>";
    print "&#8377;".$row['cost']."</td></tr>";
}

first check with firebug,is image is loading or not?if not try with one image and go for multiple images — Mahesh.D
– Mahesh.D, Commented Jun 9, 2012 at 4:49
while($row = mysql_fetch_array($display_query)){ echo "<tr><td>".$row['itemid']."</td><td>"; echo '<img src="resources/wh/'.$row['itemid'].'.png" title="'.$row['itemid'].'"/>'; echo "</td><td>".$row['description']."</td><td>"; echo "₹".$row['cost']."</td></tr>"; } — xan
– xan, Commented Jun 9, 2012 at 4:57

dvir · Accepted Answer · 2012-06-09 08:31:56Z

4

"<tr><td>".$row['itemid']."</td><td><img src="resources/wh/'.$row['itemid'].'.png"/></td><td>".$row['description']."</td><td>";

Should be

"<tr><td>".$row['itemid'] . '</td><td><img src="resources/wh/'.$row['itemid'].'.png"/></td><td>'.$row['description']."</td><td>";

But mix ' and " make your code a mess, you could use HEREDOC to make it more readable.

while($row = mysql_fetch_array($display_query)){
  echo <<<EOF
<tr>
    <td>{$row['itemid']}</td>
    <td><img src="resources/wh/{$row['itemid']}.png"/></td>
    <td>{$row['description']}</td>
    <td>&#8377;{$row['cost']}</td>
</tr>
EOF;
}

edited Jun 9, 2012 at 8:31

dvir

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answered Jun 9, 2012 at 4:47

xdazz

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2 Comments

Sam Peacey Over a year ago

Or, since he's using ", "<tr><td>{$row['itemid']}</td><td><img src='resources/wh/{$row['itemid']}.png'/></td><td>{$row['description']}</td><td>"; but yeah, HEREDOC is tidier. :)

dvir Over a year ago

I would point users to some explanation of HEREDOCs, as it's not something a new programmer will just run into - en.wikipedia.org/wiki/Here_document#PHP

smm · Accepted Answer · 2012-06-09 08:59:26Z

2

Your concatenation of string and quotation is not right. Try this -

while($row = mysql_fetch_array($display_query)){
    print "<tr><td>" . $row['itemid']."</td><td><img src=" . 'resources/wh/' .$row['itemid']. ".png'/></td><td>".$row['description'] . "</td><td>";
    print "&#8377;".$row['cost']."</td></tr>";
}

edited Jun 9, 2012 at 8:59

answered Jun 9, 2012 at 4:59

smm

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2 Comments

dvir Over a year ago

Your concatenation of string and quotation is not right - you added an extra single-quote before the file extension, and for some reason separated the path into it's own string

smm Over a year ago

Oh! it should be after the file extension. Thanks

Shaikh Farooque · Accepted Answer · 2012-06-09 06:52:30Z

1

The simplest solution is just make proper pencuation in your code.

The correct code is:

print "<tr><td>".$row['itemid']."</td><td><img src='resources/wh/".$row['itemid'].".png'/></td><td>".$row['description']."</td><td>";

answered Jun 9, 2012 at 6:52

Shaikh Farooque

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Comments

mousumi deb · Accepted Answer · 2014-01-26 10:14:01Z

0

from db we get column value of rating $number=$row->rating ;

if we want TO print image as rating value in a table row then then

 $number=$row->rating   ;   
 $middle="";
 $first="<td width='200' align='left'>";
  for($x=1;$x<=$number;$x++) {

   $middle=$middle.img($fullimage_properties);  
}


if (strpos($number,'.')) {
    $middle=$middle.img($halfimage_properties); 
    $x++;
}
while ($x<=5) {
    $middle=$middle.img($blankimage_properties);    ;
    $x++;
}


echo $last=$first.$middle."</td>";

answered Jan 26, 2014 at 10:14

mousumi deb

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