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[SOLVED] Quick Question: Echo An Image From MySQL, I got it but...


suttercain

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Quick Question: Echo An Image From MySQL, I got it but...

I need to add tb_ in front of the 'image.jpg' so it'll actually echo 'tb_image.jpg'

 

CODE

echo "<td><img src=$row[cover]></td>";

 

I tried:

echo "<td><img src=$rowtb_[cover]></td>";

But of course it didn't work...

 

Any suggestions? Thanks.

Hey MadTechie,

 

Thanks for the reply. I tried both but neither worked out for me:

 

echo "<td><img src=tb_$row[cover]></td>";

resulted in the following image path:

http://localhost/Superman/tb_images/520027.jpg (Should be http://localhost/Superman/images/tb_520027.jpg to work)

 

And the other code:

$X = "tb_".$row[cover];
<td><img src=$X></td>;

resulted in the following image path:

http://localhost/Superman/tb_

 

I know, I have tried different ways but so far no luck. Any other suggestions?

 

Thanks.

OK the reason for that is because $row[cover]  = "images/520027.jpg"

 

so try this

<?php

$X = split("/",$row[cover]);
echo "<td><img src=".$X[0]."tb_".$X[1]."></td>";

?>

 

 

if this fails can you also post the results of a

<?php

print_r($row[cover]);

?>

Thanks Mad Techie,

 

I tried both of those but neither seems to work. The first still lead to:

http://localhost/Superman/tb_

 

And the print_r function echoed the root directory of the image within the table.

 

 

Hi MadTechie, Should the code be included in the while loop:

 

Here is my entire code:

 

<?php
#specify the connection information
$db_server ="localhost";
$db_name = "comics";
$username = "root";
$password = NULL;        
#
#The following lines do not need to be edited
#
#make the connection.  If there is a problem, print out a helpful error message
$dbh = @mysql_connect($db_server,$username,$password) or die 
("Connection to $db_server with login '$username'/'$password' failed.");

if ($dbh = TRUE) {
echo "Connected!";
}

#select the database.  If the database is not found on the server, let us know
$db = @mysql_select_db($db_name) or die 
("Connection made. But database '$db_name' was not found.");

// Begin your table outside of the array
echo "<table width='100%' border='0' cellpadding='1' cellspacing='0'>
    <tr>
        <td width='110'><b></b></td>
    </tr>
<tr>
        <td width='110'><b></b></td>
	<td width='110'><b>WRITERS</b></td>
	<td width='110'><b>PUBLISHED</b></td>
    </tr>";

// Perform an statndard SQL query:
$res = mysql_query("SELECT cover, title, email, publish_date FROM comics ORDER BY title ASC") or die (mysql_error());


// Assuming $res holds the results from your query:
$class = 'even';
while ($row = mysql_fetch_assoc($res)) {
  $class = $class == 'even' ? 'odd' : 'even';
  echo "<tr class=\"$class\">\n";
  echo "<td><img src=$row[cover]></td>"; <---IMAGE BEING ECHOED
  echo "</tr>\n";
  echo "<tr class=\"$class\">\n";
  echo "<td>$row[title]</td>\n";
  echo "<td>$row[email]</td>\n";
  echo "<td>$row[publish_date]</td>\n";
  echo "</tr>\n";
}

?>

Ahh ha

 

try this

<?php

#specify the connection information

$db_server ="localhost";

$db_name = "comics";

$username = "root";

$password = NULL;       

#

#The following lines do not need to be edited

#

#make the connection.  If there is a problem, print out a helpful error message

$dbh = @mysql_connect($db_server,$username,$password) or die

("Connection to $db_server with login '$username'/'$password' failed.");

 

if ($dbh = TRUE) {

echo "Connected!";

}

 

#select the database.  If the database is not found on the server, let us know

$db = @mysql_select_db($db_name) or die

("Connection made. But database '$db_name' was not found.");

 

// Begin your table outside of the array

echo "<table width='100%' border='0' cellpadding='1' cellspacing='0'>

    <tr>

        <td width='110'><b></b></td>

    </tr>

<tr>

        <td width='110'><b></b></td>

<td width='110'><b>WRITERS</b></td>

<td width='110'><b>PUBLISHED</b></td>

    </tr>";

 

// Perform an statndard SQL query:

$res = mysql_query("SELECT cover, title, email, publish_date FROM comics ORDER BY title ASC") or die (mysql_error());

 

 

// Assuming $res holds the results from your query:

$class = 'even';

while ($row = mysql_fetch_assoc($res)) {

  $class = $class == 'even' ? 'odd' : 'even';

  echo "<tr class=\"$class\">\n";

  //echo "<td><img src=$row[cover]></td>"; #<---IMAGE BEING ECHOED

 

  $X = split("/",$row['cover']);                                #<----NEW!!

  echo "<td><img src=".$X[0]."tb_".$X[1]."></td>";    #<----NEW!!

 

  echo "</tr>\n";

  echo "<tr class=\"$class\">\n";

  echo "<td>$row[title]</td>\n";

  echo "<td>$row</td>\n";

  echo "<td>$row[publish_date]</td>\n";

  echo "</tr>\n";

}

 

?>

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