[SOLVED] view image

[squiblo]

i cannot complete echoing $imagelocation without getting an error i have tried everything, i would also like the image to be 100px be 100px

<font face="Arial">
<?php

//get data
$button = $_GET['submit'];
$search = $_GET['search'];

if (!$button)
header("location:unititled.php");
else
{
if (strlen($search)<3)
	header("location:unititled.php");
else
{
	echo "You searched for <b>$search</b><hr size='1'>";

	//connect to our database
	mysql_connect("localhost","","");
	mysql_select_db("");


		//explode our search term
		$search_exploded = explode(" ",$search);

		foreach($search_exploded as $search_each)
		{

		//construct query
		$x++;
		if ($x==1)
		    $construct .= "username LIKE '%$search_each%'";
		else
		    $construct .= " OR username LIKE '%$search_each%'";			
	    
		}


	//echo out construct

	$construct = "SELECT * FROM members WHERE $construct";
	$run = mysql_query($construct);

	$foundnum = mysql_num_rows($run);

	if ($foundnum==0)
	   echo "No results found.";
	else
	{
	   echo "$foundnum result(s) found!<p>";
	   
	   while ($runrows = mysql_fetch_assoc($run))
	   {
	 //get data
                $state = ucwords($runrows['state']);
                $url = $runrows['url'];
                $username = ucwords($runrows['username']);
   		$imagelocation = $runrows['imagelocation'];

                echo "
                <b>$username</b><br>	
                $state<br>
                <a href='$url'>View Profile</a><br><br>

                ";
		echo "<hr width='300' align='left'>";
	   
             }


           }   



        }	



}	



?>
</font>

[priti]

what is the error you are receiving and please point where in you are trying to echo $imagelocation in code?

[squiblo]

i do not get an error in the code because i have taken the errors out, i am trying to echo $imagelocation here...

//get data
                $state = ucwords($runrows['state']);
                $url = $runrows['url'];
                $username = ucwords($runrows['username']);
         $imagelocation = $runrows['imagelocation'];

                echo "
                <b>$username</b><br>   
                $state<br>
                <a href='$url'>View Profile</a><br><br>
                                                                       // <------------echo $imagelocation here
                ";

[Philip]

Well, right off the bad, you have something outputting (<font face="Arial">) to the browser before your header function.

[squiblo]

ok i have just got rid of the font tags, how can i echo $imagelocation properly?

[priti]

what is stored in database in imagelocation? can you post example?

[squiblo]

in imagelocation, there are image paths, like "./profilepictures/nameofpicture.format"

[priti]

make sure

 

> query is correctly generated

> query is returning correct output

 

echo "

                <b>$username</b><br> 

                $state<br>

                <a href='$url'>View Profile</a><br><br>

                                                                      // <------------echo $imagelocation here

                ";

 

in echo you are echoing image location ?????

 

echo "$username-$state-$imagelocation"

 

what does above line prints. Please make sure that your query returning correct data.

[squiblo]

i tried this but resulted in another error...

 //get data
                $state = ucwords($runrows['state']);
                $url = $runrows['url'];
                $username = ucwords($runrows['username']);
                $imagelocation = ['imagelocation'];

                echo "
                <b>$username</b><br>	
                $state<br>
                <a href='$url'>View Profile</a><br><br>
                <img src =$imagelocation width='100' height='100' border='0'>
                ";
                 echo "<hr width='300' align='left'>";

[priti]

<img src ="$imagelocation" width='100' height='100' border='0'>

 

you are missing with " " around $imagelocation

 

what is the error!!!

[squiblo]

sorry i do not know the error, i just get a blank page how do i find out?

[priti]

before displaying image do

 

header("Content-type:image/jpeg"); please take care of image format.

 

and see if it displays

[squiblo]

like this?

 //get data
                $state = ucwords($runrows['state']);
                $url = $runrows['url'];
                $username = ucwords($runrows['username']);
                $imagelocation = ['imagelocation'];


                echo "
                <b>$username</b><br>	
                $state<br>
                <a href='$url'>View Profile</a><br><br>
	header("Content-type:image/jpeg");
	<img src =$imagelocation width='100' height='100' border='0'>
                ";
                 echo "<hr width='300' align='left'>";

 

i still get an error

[priti]

try this

 

  echo "<b>$username</b><br>    $state<br> <a href='$url'>View Profile</a><br><br>";

  header("Content-type:image/jpeg");

  <img src ="$imagelocation" width='100' height='100' border='0'/>

 

Do let me know error if any error pops up

[squiblo]

i get an error with this...

echo "
	<b>$username</b><br>    
	$state<br> 
	<a href='$url'>View Profile</a><br><br>
	";
  		header("Content-type:image/jpeg");
  		<img src ="$imagelocation" width='100' height='100' border='0'/>

 

how can i find out what line the error is on?

[RichardRotterdam]

i do not get an error in the code because i have taken the errors out

What do you mean with taken the errors out? Did you turn off error reporting or did you fix the errors you received?

 

i tried this but resulted in another error...

sorry i do not know the error, i just get a blank page how do i find out?

Here you say you get an error but you don't know the error. How do you know it's an error to begin with then? Do you concider a blank screen as an error