Programming Forums > Web Development and Design > PHP

TecBrain · Feb 19th, 2006, 11:52 AM

This is the code, not sure why its not working, I defined

$img = "<img src="images/">;

to be the path then i simply added to

(Toggle Plain Text)

print"<td width=48 bgcolor=#EEEEEE height=22 align=left  height=22 class=style5>$img$imgpath</td>";

Complete File:

(Toggle Plain Text)

<?
$query = "SELECT ID, countryname, currency, imgpath FROM countries order by ID"; 
$result = mysql_query($query);
$num_rows = mysql_num_rows( $result );
$img = "<img src="images/">;
while($row = mysql_fetch_array($result)) 
{

	$ID = $row["ID"];  
	$countryname = $row["countryname"];
   	$currency = $row["currency"];
   	$imgpath = $row["imgpath"];
	
	
	print"<tr>";
		print"<td width=48 bgcolor=#EEEEEE height=22 align=left  height=22 class=style5>$ID</td>";
		print"<td width=48 bgcolor=#EEEEEE height=22 align=left  height=22 class=style5>$countryname</td>";
		print"<td width=48 bgcolor=#EEEEEE height=22 align=left  height=22 class=style5>$currency</td>";
		print"<td width=48 bgcolor=#EEEEEE height=22 align=left  height=22 class=style5>$img$imgpath</td>";
		print"<td width=48 bgcolor=#EEEEEE align=center height=22 align=left  height=22 class=style5><div align=center >Edit</div></td>";
		print"<td width=48 bgcolor=#EEEEEE align=center height=22 align=left  height=22 class=style5><div align=center ><input type=checkbox name=cb[] value=ID></div></td>";
	print"</tr>";
	}
?>

DaWei · Feb 19th, 2006, 11:56 AM

View your source code in the browser where you think you have a nice valid image path and specification. Then investigate concatenation.

TecBrain · Feb 19th, 2006, 12:02 PM

Quote:

Originally Posted by DaWei

View your source code in the browser where you think you have a nice valid image path and specification. Then investigate concatenation.

Am running the site in my local host machine, what i was thinking is the following, having an images folder store the images in there, add the name of the image in my database, then view it by adding the path... is that right way of thinking...!!

DaWei · Feb 19th, 2006, 12:06 PM

Well, I think it would have stuck better if you had peeped at it, but consider this:
Print these adjacent to each other: $img$imgpath
you get this: <img src="images/"> up against whatever was in the DB. Probably something like this: <img src="images/">myImage.jpg.

Don't think that's what you're going to want.

TecBrain · Feb 19th, 2006, 12:11 PM

(Toggle Plain Text)

	$imgpath = "<img src=\"images/$imgpath\">";

Works now...

Feb 19th, 2006, 11:52 AM	#1
TecBrain Hobbyist Programmer Join Date: Sep 2004 Location: Cyprus Posts: 147 Rep Power: 5	Displaying images This is the code, not sure why its not working, I defined (Toggle Plain Text) $img = "<img src="images/">; $img = "<img src="images/">; to be the path then i simply added to (Toggle Plain Text) print"<td width=48 bgcolor=#EEEEEE height=22 align=left height=22 class=style5>$img$imgpath</td>"; print"<td width=48 bgcolor=#EEEEEE height=22 align=left height=22 class=style5>$img$imgpath</td>"; Complete File: (Toggle Plain Text) <? $query = "SELECT ID, countryname, currency, imgpath FROM countries order by ID"; $result = mysql_query($query); $num_rows = mysql_num_rows( $result ); $img = "<img src="images/">; while($row = mysql_fetch_array($result)) { $ID = $row["ID"]; $countryname = $row["countryname"]; $currency = $row["currency"]; $imgpath = $row["imgpath"]; print"<tr>"; print"<td width=48 bgcolor=#EEEEEE height=22 align=left height=22 class=style5>$ID</td>"; print"<td width=48 bgcolor=#EEEEEE height=22 align=left height=22 class=style5>$countryname</td>"; print"<td width=48 bgcolor=#EEEEEE height=22 align=left height=22 class=style5>$currency</td>"; print"<td width=48 bgcolor=#EEEEEE height=22 align=left height=22 class=style5>$img$imgpath</td>"; print"<td width=48 bgcolor=#EEEEEE align=center height=22 align=left height=22 class=style5><div align=center >Edit</div></td>"; print"<td width=48 bgcolor=#EEEEEE align=center height=22 align=left height=22 class=style5><div align=center ><input type=checkbox name=cb[] value=ID></div></td>"; print"</tr>"; } ?> <? $query = "SELECT ID, countryname, currency, imgpath FROM countries order by ID"; $result = mysql_query($query); $num_rows = mysql_num_rows( $result ); $img = "<img src="images/">; while($row = mysql_fetch_array($result)) { $ID = $row["ID"]; $countryname = $row["countryname"]; $currency = $row["currency"]; $imgpath = $row["imgpath"]; print"<tr>"; print"<td width=48 bgcolor=#EEEEEE height=22 align=left height=22 class=style5>$ID</td>"; print"<td width=48 bgcolor=#EEEEEE height=22 align=left height=22 class=style5>$countryname</td>"; print"<td width=48 bgcolor=#EEEEEE height=22 align=left height=22 class=style5>$currency</td>"; print"<td width=48 bgcolor=#EEEEEE height=22 align=left height=22 class=style5>$img$imgpath</td>"; print"<td width=48 bgcolor=#EEEEEE align=center height=22 align=left height=22 class=style5><div align=center >Edit</div></td>"; print"<td width=48 bgcolor=#EEEEEE align=center height=22 align=left height=22 class=style5><div align=center ><input type=checkbox name=cb[] value=ID></div></td>"; print"</tr>"; } ?> __________________ Personal Portfolio TecBrain Support Forum Linux VS Windows ... Dont Even Think of it .. Distribution: Slackware if (OS==Linux) return success There are 10 kinds of people, those who can read binary numbers and those who can't.

Feb 19th, 2006, 11:56 AM	#2
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	View your source code in the browser where you think you have a nice valid image path and specification. Then investigate concatenation. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Feb 19th, 2006, 12:06 PM	#4
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	Well, I think it would have stuck better if you had peeped at it, but consider this: Print these adjacent to each other: $img$imgpath you get this: <img src="images/"> up against whatever was in the DB. Probably something like this: <img src="images/">myImage.jpg. Don't think that's what you're going to want. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Feb 19th, 2006, 12:11 PM	#5
TecBrain Hobbyist Programmer Join Date: Sep 2004 Location: Cyprus Posts: 147 Rep Power: 5	(Toggle Plain Text) $imgpath = "<img src=\"images/$imgpath\">"; $imgpath = "<img src=\"images/$imgpath\">"; Works now... __________________ Personal Portfolio TecBrain Support Forum Linux VS Windows ... Dont Even Think of it .. Distribution: Slackware if (OS==Linux) return success There are 10 kinds of people, those who can read binary numbers and those who can't.

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Programming Forums > Web Development and Design > PHP

Displaying images