Photos Not View

[sweetangel]

I try now... ;) GooD MoRNINg BuDDy :D

[inet]

Try This:

$photo = mysql_fetch_array(mysql_query("SELECT imageurl FROM ibwf_gallery WHERE uid='".$chat."' ORDER BY RAND() "));

echo "<img src=\"$photo[0]\" width=\"50\" height=\"60\" alt=\"$photo[0]\"/>";
echo "[$ds]<a href=\"chat.php?action=say2&sid=$sid&who=$chat[0]&rid=$rid&rpw=$rpw\">$optlink</a> - ";
echo "".strtolower($tosay)."
";


this is ok!

[sweetangel]

with this code I view the photo...but are only my photos at all nicknames...:o

[inet]

with this code I view the photo...but are only my photos at all nicknames...:o[/b]

yes this works... right?

<div class='codetop'>CODE</div><div class='codemain' style='height:200px;white-space:pre;overflow:auto'>$photo = mysql_fetch_array(mysql_query("SELECT imageurl FROM ibwf_gallery WHERE uid=&#39;".$uid."&#39; ORDER BY RAND() "));
</div>


or use this one................................... this will help you...............

for all the pictures in your ibwf gallery this works fine for me on my fourth domain!

<div class='codetop'>CODE</div><div class='codemain' style='height:200px;white-space:pre;overflow:auto'>$photo = mysql_fetch_array(mysql_query("SELECT imageurl FROM ibwf_gallery ORDER BY RAND() "));
</div>

<div class='codetop'>CODE</div><div class='codemain' style='height:200px;white-space:pre;overflow:auto'>$photo = mysql_fetch_array(mysql_query("SELECT imageurl FROM ibwf_gallery ORDER by RAND() LIMIT 1"));
</div>

[Guest]

this is my code:

  $photo = mysql_fetch_array(mysql_query("SELECT imageurl FROM ibwf_gallery WHERE uid='".$who."' ORDER BY RAND() "));

              echo "<img src=\"$photo\" width=\"50\" height=\"60\" alt=\"photo[0]\"/>";
              echo "[$ds]<a href=\"chat.php?action=say2&amp;sid=$sid&amp;who=$chat[0]&amp;rid=$rid&amp;rpw=$rpw\">$optlink</a> - ";
              echo "".strtolower($tosay)."
";

but photos not open....any1 can help me to fix this problem?? [/b]

post full code

[inet]

<div class='codetop'>CODE</div><div class='codemain' style='height:200px;white-space:pre;overflow:auto'>$photo = mysql_fetch_array(mysql_query("SELECT imageurl FROM ibwf_gallery ORDER BY RAND() limit 1"));

echo "<img src=\"$photo[0]\" width=\"50\" height=\"60\" alt=\"$photo[0]\"/>";</div>

problem solved..

[sweetangel]

but I wont add the gallery pics random before nickname at chat page...but this pics are from user write msg gallery...

[sweetangel]

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

[inet]

but I wont add the gallery pics random before nickname at chat page...but this pics are from user write msg gallery...[/b]

use this final one hope this finaly resolves your problem

do not random the image but rather it displays image of the chatter



<div class='codetop'>CODE</div><div class='codemain' style='height:200px;white-space:pre;overflow:auto'>$photo = mysql_fetch_array(mysql_query("SELECT avatarImageofChatter FROM ibwf_users WHERE uid=&#39;".$who.&#39;" "));</div>

<div class='codetop'>CODE</div><div class='codemain' style='height:200px;white-space:pre;overflow:auto'>$photo = mysql_fetch_array(mysql_query("SELECT avatar FROM ibwf_users WHERE uid=&#39;".$who.&#39;""));</div>

<div class='codetop'>CODE</div><div class='codemain' style='height:200px;white-space:pre;overflow:auto'>$photo = mysql_fetch_array(mysql_query("SELECT mypicture FROM ibwf_users WHERE uid=&#39;".$uid.&#39;""));</div>

[sweetangel]

I have add this:

$photo = mysql_fetch_array(mysql_query("SELECT avatar FROM ibwf_users WHERE uid=&#39;".$who.&#39;"));

but:

Parse error: syntax error, unexpected &#39;"&#39;

[inet]

I have add this:

$photo = mysql_fetch_array(mysql_query("SELECT avatar FROM ibwf_users WHERE uid=&#39;".$who.&#39;"));

but:

Parse error: syntax error, unexpected &#39;"&#39;

[/b]

try
this one!!!
$photo = mysql_fetch_array(mysql_query("SELECT avatar FROM ibwf_users WHERE uid=&#39;".$who.&#39;"&#39;));

[sweetangel]

at this line:

$photo = mysql_fetch_array(mysql_query("SELECT avatar FROM ibwf_users WHERE uid=&#39;".$who.&#39;"&#39;));

I have:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

[inet]

at this line:

$photo = mysql_fetch_array(mysql_query("SELECT avatar FROM ibwf_users WHERE uid=&#39;".$who.&#39;"&#39;));

I have:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource[/b]

because you dont have this avatar in your sql... you must add it ....


what script you use today?

[Guest]

Ok, now try this:

$photo = mysql_fetch_array(mysql_query("SELECT imageurl FROM ibwf_gallery WHERE uid=&#39;".$chat[0]."&#39; ORDER BY RAND() "));

echo "<img src=\"$photo[0]\" width=\"50\" height=\"60\" alt=\"$photo[0]\"/>";
echo "[$ds]<a href=\"chat.php?action=say2&amp;sid=$sid&amp;who=$ chat[0]&amp;rid=$rid&amp;rpw=$rpw\">$optlink</a> - ";
echo "".strtolower($tosay)."
";

[sweetangel]

Ok, now try this:

$photo = mysql_fetch_array(mysql_query("SELECT imageurl FROM ibwf_gallery WHERE uid=&#39;".$chat[0]."&#39; ORDER BY RAND() "));

echo "<img src=\"$photo[0]\" width=\"50\" height=\"60\" alt=\"$photo[0]\"/>";
echo "[$ds]<a href=\"chat.php?action=say2&amp;sid=$sid&amp;who=$ chat[0]&amp;rid=$rid&amp;rpw=$rpw\">$optlink</a> - ";
echo "".strtolower($tosay)."
";[/b]

this work how I can add a image by default if 1 users not have photo ????