HI again,
OK, now I have a page pictures.php
and I am using following code :
[PHP]printf("<a href=\"pictures .php?name=%s\"> ",urlencode($va l[0]));
print ("<img src=\"$val[0]\" width=100%)/></a>");[/PHP]
the problem is no value is stored in $name in pictures.php
I don't get anything when i print $name.
Need a help.
Thanks
11 1225 Markus 6,050
Recognized Expert Expert
There's no reference to '$name'
=/
There's no reference to '$name'
=/
I have included following line
[PHP]print ("$name");[/PHP]
in pictures.php
just after above code.
thanks
Markus 6,050
Recognized Expert Expert
$name is empty?
I'm really sorry, it's just i can't know what your code is.. you have to show me.
OK,
here is the code I'm using
[PHP]<?php
$loginid=$_SESS ION['loginid'];
mysql_connect(' localhost','use r','user');
mysql_select_db ('test');
$query="select name from pictures where loginid='$login id'";
$result=mysql_q uery($query);
print("<table width=95% align=center cellspacing=4> <tr>");
$count=0;
while($val=mysq l_fetch_array($ result))
{
if ($count++<5)
print("<td width=20%>");
else { print("</tr><tr><td width=20%>"); $count=0; }
printf("<a href=\"pictures .php?name=%s\"> ",urlencode($va l[0]));
print ("<img src=\"$val[0]\" width=100%)/></a>");
print("</td>");
}
if ($count<5)
while ($count++<5) print ("<td width=20%>  ;</td>");
print("</tr></table>");
print ("\n<br/><center><img src=\" $name \"><br/>");
?>[/PHP]
When I see the source code in my browser it doesn't show anything in place of $name in the last line
Thanks
OK,
here is the code I'm using
[PHP]<?php
$loginid=$_SESS ION['loginid'];
mysql_connect(' localhost','use r','user');
mysql_select_db ('test');
$query="select name from pictures where loginid='$login id'";
$result=mysql_q uery($query);
print("<table width=95% align=center cellspacing=4> <tr>");
$count=0;
while($val=mysq l_fetch_array($ result))
{
if ($count++<5)
print("<td width=20%>");
else { print("</tr><tr><td width=20%>"); $count=0; }
printf("<a href=\"pictures .php?name=%s\"> ",urlencode($va l[0]));
print ("<img src=\"$val[0]\" width=100%)/></a>");
print("</td>");
}
if ($count<5)
while ($count++<5) print ("<td width=20%>  ;</td>");
print("</tr></table>");
print ("\n<br/><center><img src=\" $name \"><br/>");
?>[/PHP]
When I see the source code in my browser it doesn't show anything in place of $name in the last line
Thanks
Hi,
Well i didnt find any place in your code where you are assigning any value to '$name' so it will be empty..... what do you exactly want '$name' to print??
if i'm not understanding your code then let me know
Regards,
RP
Hi,
Well i didnt find any place in your code where you are assigning any value to '$name' so it will be empty..... what do you exactly want '$name' to print??
if i'm not understanding your code then let me know
Regards,
RP
Well, I think I should explain my code.
I am trying to make a page for pictures in which user will see thumnail-size pictures first. And when he clicks the picture, he will see the actual size of image later on the same page.
Ok, now I have included this code in line no. 14
[PHP]#
printf("<a href=\"pictures .php?name=%s\"> ",urlencode($va l[0]));[/PHP]
Isn't this suppose to assign value of urlencode($val[0]) to name ??
and when clicked in the link, isn't this code supposed to pass value of $name to pictures.php (which is the same page in my case).
So, I need the value of $name so as to know which image has the user clicked.
I think, that clears my problem and your doubt.
But please do help me
Thanks
Markus 6,050
Recognized Expert Expert
Have you tried printing just $val[0]?
Also, don't say $name.. it makes people look for a variable called $name in your script when it isnt there. You're just reffering to the name= in the url.
Hi,
thanks for your suggestions.
I have tried printing the value of $val[0] as well and it's working fine.
I can see in the page source as well as the url like pictures.php?na me=something but it seems name is not taking that value.
later when I include
[PHP]print ("\n<br/><center><img src=\" $name \"><br/>");[/PHP]
name doesn't have any value.
Thanks
Markus 6,050
Recognized Expert Expert
Are you using $_GET['name'];
To retrieve the name?
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