[FalconFX]

Hello there.

I need to put in a Defaul picture defined per MYSQL.
So i set it up that the Default is the path to the picture in my MYSQL Table.

Now i try to get it displayed with the following command :

Code:

<br><img scr="'.$info[categorypic].'"><br>

But it doesnt Display the picture. Anything you see what i did wrong ?
Thanks

BTW i am PHP Ultra newbie.

[alabaster_lynch]

Originally Posted by FalconFX

Hello there.

I need to put in a Defaul picture defined per MYSQL.
So i set it up that the Default is the path to the picture in my MYSQL Table.

Now i try to get it displayed with the following command :

Code:

<br><img scr="'.$info[categorypic].'"><br>

But it doesnt Display the picture. Anything you see what i did wrong ?
Thanks

BTW i am PHP Ultra newbie.

You didn't really give us anything to work with...so the only thing I can think of is did you echo that statement like this

Code:

echo '<br><img scr="'.$info[categorypic].'"><br>';

if you did, try to give some more information like your query or something else. Its hard to help otherwise.

Hope this helps.....Jose

[FalconFX]

The picture "nanews.gif" is defined as default picture under "categorypic" in the MYSQl Database. So if the user doesnt give a link to a category picture this picture should be displayed.

the code for the coulmn is : category VCHAR (50) NOT NULL DEFAULT "nanews.gif",

Now i try to get the picture Path out of the MYSQL database and then the picture displayed. Hard to explain sorry. I give you a link whats happeining right now :

http://zas.za.funpic.de/zwnews/newsview2.php

Thanks for any help

Heres a Code Segment :

Code:

	while($info = mysql_fetch_assoc($select)){         //holt info von Database
   		echo'<table width="73%" border="1">
  <tr>
    <td width="50%" align="left">'.$info[newstitle].'</td>
    <td width="21%">'.$info[newscategory].'</td>
  </tr>
  <tr>
    <td align="left" valign="top">'.$info[newscontent].'
	</td>
    <td height="100%" align="center" valign="top">	
	<br><img scr="'.$info[categorypic].'"><br>
	<br>ID : '.$info[newsid].'<br>
	<br>USER :'.$info[user].'<br>
	</td>
  </tr>
  <tr>
    <td align="left">&nbsp;</td>

[alabaster_lynch]

Originally Posted by FalconFX

The picture "nanews.gif" is defined as default picture under "categorypic" in the MYSQl Database. So if the user doesnt give a link to a category picture this picture should be displayed.

the code for the coulmn is : category VCHAR (50) NOT NULL DEFAULT "nanews.gif",

Now i try to get the picture Path out of the MYSQL database and then the picture displayed. Hard to explain sorry. I give you a link whats happeining right now :

http://zas.za.funpic.de/zwnews/newsview2.php

Thanks for any help

Heres a Code Segment :

Code:

	while($info = mysql_fetch_assoc($select)){         //holt info von Database
   		echo'<table width="73%" border="1">
  <tr>
    <td width="50%" align="left">'.$info[newstitle].'</td>
    <td width="21%">'.$info[newscategory].'</td>
  </tr>
  <tr>
    <td align="left" valign="top">'.$info[newscontent].'
	</td>
    <td height="100%" align="center" valign="top">	
	<br><img scr="'.$info[categorypic].'"><br>
	<br>ID : '.$info[newsid].'<br>
	<br>USER :'.$info[user].'<br>
	</td>
  </tr>
  <tr>
    <td align="left">&nbsp;</td>

Two things one, you have this:

Code:

<br><img scr="'.$info[categorypic].'"><br>

but at the top of this post, you say the column is name just category, not categorypic.

Also, where is the pic located, is it in the same directory as this script or in a subdirectory.

Hope this helps....Jose O. Estrada

[FalconFX]

Nah sorry for that.
The pic is in the same path as the script is and categorypic is the right name

So any further ideas ?

[Jaffizzle]

ur going to laugh cuz its just a simple typo

scr = src

[FalconFX]

hmmm that didnt work either....
How would you do it ?

Lets say someone stores a path to a image in your database and you want your webpage to display this image. What would be your code for displaying it ?

[alabaster_lynch]

Originally Posted by FalconFX

Nah sorry for that.
The pic is in the same path as the script is and categorypic is the right name

So any further ideas ?

Are you sure that your db connection and query is correct, I just don't see anything else that would be the problem.

Jose

[alabaster_lynch]

you could check your query by simply adding this after completing it

[code]
If(!$query)
echo 'query wasn't sucessful';

that will tell you if the query is ok.

Jose

[Jaffizzle]

i mean you have <img sCR=...>

when it should be <img sRC=...>

your PHP code is totally fine, its ur HTML