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I’m making a page for a friend. He will add some text an upload some pictures to the page. All this is saved in MySql. He have the possibility to upload 5 pictures to each text. If he the only upload 2 pictures we will have some boxes with a red x.

 

Then I made this.

 

and I believe I have forgotten something.

 

 

 

 <?php if($row_Bilder['bilde1'] != "") { ?>    <div align="center"><a href="<?php echo $row_Bilder['bilde1']; ?>"><img src="<?php echo $row_Bilder['bilde1']; ?>" width="80" height="80" border="0"></a></div>  <?php } ?></td><td width="103">  <?php if($row_Bilder['bilde2'] != "") { ?>    <div align="center"><a href="<?php echo $row_Bilder['bilde2']; ?>"><img src="<?php echo $row_Bilder['bilde2']; ?>" width="80" height="80" border="0"></a></div>  <?php } ?>    </tr> 

.

 

This is almost working.

 

the only thing is that picture nr 1 ($row_Bilder['bilde1') will not show on the page. But picture 2 is working. STRANGE

 

This has been killing me. I need really some help.

 

 

Take a look.

vanvara.net

 

 

 

 

 

 

 

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https://forums.phpfreaks.com/topic/1836-images-with-null-value-in-mysql/
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  • 2 weeks later...

if there is a null in the image varialbe it will not display the image but the HTML is still there so you get a image missing x on the browser. Try a n if statement to check that the image exits first if so display it other wise display a default image or don't sent any html.

 

if(row_Bilder[1}(

img src=<?php echo $row_Bilder['bilde1];

} else{

display either a default image or no image.

on one of my sites I use this;

 

if($row[4]==""){

echo"<td class=\"producttext\"><a href=\"details.php?id=$row[0]&storeID=$storeID&currency=$currency\"><img src=\"../images/savers/image_not_available.jpg\"></a></td>";

}else{

echo "<TD valign=\"top\" class=\"producttext\"><a href=\"details.php?id=$row[0]&storeID=$storeID&currency=$currency\"> <img src=\"../images/savers/$row[4]\"></a> </TD>";

}

 

where row[4] would be my image variable

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