CODEWITHSUNDEEP
phprow
user2260431
user2260431

Reputation: 75

how to show default image if user have no photo

example if user have no photo in our database then row show default photo how can i do this please help me thanks

 $result = mysql_query($query) or die(mysql_error());
    $rows = mysql_num_rows($result);

      echo "<table class='hovertable' border='1' cellpadding='0' cellspacing='0'>";
echo "<tr><th>Photo</th><th>Teacher Detail</th></tr>";

if ($rows > 0) {
        while($row = mysql_fetch_array($result)) {

    echo "<tr><td align='center'>";
    echo "<img src=images/".$row['photo'] ." width='90' height='90'></a></td>";
    echo '<td valign="top">Teacher Name: ';
    echo $row['name'], '<br/>';
    echo 'Teacher No: ';
    echo $row['teacherno'], '<br/>';
    echo 'Father Name: ';
    echo $row['fathername'], '<br/>';
    echo '</td></tr>';

        }
} else {
        echo "<tr><td colspan=\"5\">No results found!</td></tr>";
}

echo "</table>"

Upvotes: 0

Views: 3387

Answers (6)

Dave
Dave

Reputation: 892

Just check to see if $row['photo'] is not empty or contains a string. If we get a match, then the user has an image, otherwise, we can output the default image. A nice sample follows.

The Code

if (!empty($row['photo']) AND isset($row['photo'])){
    // Image is not empty and there is data contained within the array value
    echo "<img src='images/" . $row['photo'] . "' width='90' height='90' />";
} else {
    // There is no data contained within the value, show the default image
    echo "<img src='images/default_image.png' width='90' height='90' />";
}

What's going on?

using empty() we use this to determine if a string (or an array) is empty, if it is, this will return FALSE. isset() checks to see if the variable is set, if it isn't, it will return FALSE.

-- snip -- User changed question, answer changed.

Upvotes: 0

Praveen kalal
Praveen kalal

Reputation: 2129

echo "<table class='hovertable' border='1' cellpadding='0' cellspacing='0'>";
echo "<tr><th>Photo</th><th>Teacher Detail</th></tr>";

if ($rows > 0) {
    while ($row = mysql_fetch_array($result)) {

        echo "<tr><td align='center'>";
        if (isset($row['photo']) && !empty($row['photo'])) {
            echo "<img src='images/" . $row['photo'] . "' width='90' height='90'></a></td>";
        } else {
            echo "<img src='images/noimage.jpg' width ='90' height = '90'></a></td>";
        }
        echo '<td valign="top">Teacher Name: ';
        echo $row['name'], '<br/>';
        echo 'Teacher No: ';
        echo $row['teacherno'], '<br/>';
        echo 'Father Name: ';
        echo $row['fathername'], '<br/>';
        echo '</td></tr>';
    }
} else {
    echo "<tr><td colspan = \"5\">No results found!</td></tr>";
}

echo "</table>";

Upvotes: 1

Zmart
Zmart

Reputation: 1183

This does not appear to be a complete section of the code at hand.

However, I would recommend removing the die(mysql_error()) if your query is intact, and then after the two echo statements try something like:

 } else {
    echo "<img src='images/default_image.png' width='90' height='90'>";
 }

Also of note is you have a terminating </a> without an opening <a> tag. This is not valid HTML.

Upvotes: 0

Anjoola
Anjoola

Reputation: 128

Try doing this:

echo "<img src=images/". $row['photo'] === NULL ? "default_img" : $row['photo'] .
"width='90' height='90'></a></td>";

Upvotes: 1

Itay Gal
Itay Gal

Reputation: 10834

if ($rows > 0){
     echo 'the user photo';
else{
     echo 'default photo';
}

Upvotes: 0

abhij89
abhij89

Reputation: 625

<?php 
   if(isset($row['photo'] && $row['photo'])
   { 
       echo "<tr><td align='center'>";
       echo "<img src=images/".$row['photo'] ." width='90' height='90'></a></td>";
    }
   else
   {
     echo "<tr><td align='center'>";
    echo "<img src=images/Sample.jpg" width='90' height='90'></a></td>";
   }
  ?>

Upvotes: 2

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