Don't display image if not in DB

[sean14592]

Hi, I have stored my image name in db after uploading it, so this is all well for the first 2 images as they are complusary. Though for example image 8 that is not required. How do I stop it from being shown if the database field is empty?

<?php

$connect = mysql_connect($host,$username,$password) or die("Error connecting to Database!". mysql_error());
mysql_select_db($database,$connect) or die("cannot select database!".mysql_error());


//photo8
$check8 = mysql_query("select photo8url from properties where id='$id'") or die(mysql_error());
$result8 = mysql_query($check8);
if (mysql_num_rows($check8) > 0) {
     $a_record = mysql_fetch_array($check8);
     $photo8url = $a_record['photo8url'];
 $photo8url = preg_replace('/\s+/', '%20', $photo8url);

} ?>

<?php  

			if ($result8 = ""){}
			else{
			echo ("<img src=". $id ."/images/".$photo8url." width=\"235\" border=\"1\">");
			}	?>

[sean14592]

Please someone, I really need some help. If you want yu can contact me on Xfire = sean14592

[Lumio]

SQL Injections could easily be done with your script! Check if the id is numeric otherwise DON'T execute this page.

 

To your quest:

<?php
   $res = mysql_query('YOUR QUERY');
   if (mysql_num_rows($res)) {
      //View image
   }else {
      echo 'Image not found';
   }
?>

[porta325]

Try

$check8 = mysql_query("select photo8url from properties where id='$id'" AND yourimagefiels IS NOT NULL) or die(mysql_error());

[sean14592]

Hi, I have tried the following:

 

$res = mysql_query("select photo8url from properties where id='$id' AND photo8url =' ' ") or die(mysql_error());
if (mysql_num_rows($res)) {
echo ("<img src=". $id ."/images/".$photo8url." width=\"235\" border=\"1\">");
}else {
echo 'Image not found';
}

 

 

 

Though its trying to show a image when the data base is empty.

 

Cheers

Sean Preston

[sean14592]

Please somebody, I have proberly just missed something stupidly.

 

 

Sean

[sean14592]

I have also tried....

 

$connect = mysql_connect($host,$username,$password) or die("Error connecting to Database!". mysql_error());
mysql_select_db($database,$connect) or die("cannot select database!".mysql_error());

//photo1
$check1 = mysql_query("select photo1url from properties where id='$id'") or die(mysql_error());
$result1 = mysql_query($check1);
if (mysql_num_rows($check1) > 0) {
$a_record = mysql_fetch_array($check1);
$photo1url = $a_record['photo1url'];
$photo1url = preg_replace('/\s+/', '%20', $photo1url);
}
else {
$photo1_check = "1";
}




if ($photo1_check = "1") {
echo ("no image!!!!!!");
}else {
echo ("<img src=". $id ."/images/".$photo1url." width=\"235\" border=\"1\">");
}
?>

 

 

This shows "no image" even if a image has been stored,lol

 

This is driving me crazyyyy!

 

Sean Preston

[CrustyDOD]

select photo8url from properties where id='$id' AND photo8url <> ''

 

if "photo8url" contains something the mysql_num_rows will be 1 else 0

[robrave]

maybe the image is cached on the browser?

[sean14592]

maybe the image is cached on the browser?

 

Nope, it aint that.

[sean14592]

select photo8url from properties where id='$id' AND photo8url <> ''

 

if "photo8url" contains something the mysql_num_rows will be 1 else 0

 

 

Tried, dosent work :(

Thanks anyway

Sean Preston

[sean14592]

I have figued out (so I think) that somet is wrong here

 

$check8 = mysql_query("select photo8url from properties where id='$id'") or die(mysql_error());
if (mysql_num_rows($check8) > 0) {
     $a_record = mysql_fetch_array($check8);
     $photo8url = $a_record['photo8url'];
 $photo8url = preg_replace('/\s+/', '%20', $photo8url);
 }
else {
$photo8_check = "1";
}

 

because if I....

 

echo $photo8_check;

 

I should see the number '1', as in the database field Photo8url using the id is completely empty.

 

Do you understand what I mean?

 

Cheers

Sean

[CrustyDOD]

Before mysql_query line put this:

 

echo "select photo8url from properties where id='$id'";

 

Then you will see if $id even is set to something.