Help required regarding images in php

i am developing a site for used car purchasing. i want to display the images in the home page of the website.. i make 2 tables one for car info and other for images...primary key of car info is id and foriegn key of the image table is id.. i am storing the picture in a folder and saving the name of the picture in the database..
here is my code plz tell me where is the problem...


$search = "SELECT *FROM cars_info , front_image Where cars_info.id = front_image.id";
$result = mysql_query($search,$con) or die(mysql_error());
$row = mysql_num_rows($result);




for($i=0;$i<$row;$i++)
{

$value = mysql_fetch_array($result);
{
echo "<tr>";
echo "<td bgcolor='#F2F9FF'><div align='center'>";
echo '<a href="car.php?id='.$value{"id"}.'">';
echo '<img src="front/'.$value{"name"}.' width="40" height="40"/>';



//**name is the column name of the image table where i am storing the image name**/


echo "</a>";
echo "</div></td>";
echo "<td bgcolor='#F2F9FF'><div align='center'>";
echo '<a href="car.php?id='.$value{"id"}.'">';
echo $value{"Make"};
echo "</a>";
echo "</div></td>";
echo "<td bgcolor='#F2F9FF'><div align='center'>";
echo '<a href="car.php?id='.$value{"id"}.'">';
echo $value{"model"};
echo "</a>";
echo "</div></td>";
echo "<td bgcolor='#F2F9FF'><div align='center'>";
echo '<a href="car.php?id='.$value{"id"}.'">';
echo $value{"price"};
echo "</a>";
echo "</div></td>";/****************/
echo "</tr>";
}

}
?>



thanks in advance..waiting for experts opinion

 

only image column is giving problem.. i mean its not picking up the images from the foldel...its only displaying the small hyperlink

 

You need to check your syntax for arrays, see: http://php.net/manual/en/language.types.array.php

$value["price"]

Code (markup):

is the right syntax, not:

$value{"price"}

Code (markup):

 

but this syntax is working all the other information except images are displaying correctly

 

Hmm, that's weird way. I personally haven't tried it that way (because arrays are always [] to me)..
Okay. Replace:

echo '<img src="front/'.$value{"name"}.' width="40" height="40"/>'; 

Code (markup):

with this:

echo '<img src="front/'.$value{"name"}.'" width="40" height="40"/>'; 

Code (markup):

 

thanks a lot problem have been solved
one double qaute was missing so thats y it was not displaying