PHP foreach function and getting data from MySql

[MP145]

Let me explain what i am trying to accomplish here. I am trying to get image data from a table in MySql and display it on my page. There are 5 rows of images in the database table, my code is as follow;

<?php

$db="news";
$link = mysql_connect("localhost","xxx","yyyy");
if (! $link)
die("Couldn't connect to MySQL");
mysql_select_db($db , $link)
or die("Couldn't open $db: ".mysql_error());
$result = mysql_query( "SELECT image1, image2, image3, image4, image5 FROM newsitem WHERE newsID=6" )
or die("SELECT Error: ".mysql_error());
$num_rows = mysql_num_rows($result);

while ($get_info = mysql_fetch_row($result)){

foreach ($get_info as $field)

print "<img src=\"images/news/$field\"><br>";

}

mysql_close($link);
?>

It displays the image but what i want is how to stop displaying when there is no image.

 

Not all the 5 image rows has data in it, now what i get is, just say there is data (imagename) on 2 rows ie. image1 and image2 where else the rest 3 does not have data in it. The code above prints <img src> for all the 5 images with and without data, which makes the X mark appear on the display page when there is no data (imagename)

 

Should i use and if or elseif statement ? How can i do it?

[MP145]

OK, i am not sure if this is the right way but it seems like working. I just added the if statement

 

<?php

$db="news";
$link = mysql_connect("localhost","root","kannan");
if (! $link)
die("Couldn't connect to MySQL");
mysql_select_db($db , $link)
or die("Couldn't open $db: ".mysql_error());
$result = mysql_query( "SELECT image1, image2, image3, image4, image5 FROM newsitem WHERE newsID=6" )
or die("SELECT Error: ".mysql_error());
$num_rows = mysql_num_rows($result);

while ($get_info = mysql_fetch_row($result)){

foreach ($get_info as $field)

if (!empty($field)) {
print "<img src=\"images/news/$field\"><br>";
}

}

mysql_close($link);
?>

 

Is this the correct way of doing it?

[Bendude14]

if it works then stick with it... There is more than one way to skin a rabit..

[MP145]

Agreed