How to insert image in php

[vikamjain]

Hii all
It might be a silly doubt but I am puzzled a lot by it.
Basically my application need to display uploaded file from user.
The name of image file is variable. Lets say the path to uploaded file is stored in
Following four are various combination of $s

$S. Now when i used following commands
$s="http://localhost/myfolder/imagefolder/image.jpeg"       //with quotations
$s="c://xampp/htdocs/myfolder/imagefolder/image.jpeg"   
$s=http://localhost/myfolder/imagefolder/image.jpeg         //without quotations
$s=c://xampp/htdocs/myfolder/imagefolder/image.jpeg

Following four commands are various syntax used to display image corresponding to each of the above $s

echo "img src="$S" width="60" height="80" ";
echo "img src='$S' width='60' height='80' ";
echo "img src=$S width='60' height='80' ";
echo "img src=".$S. "width='60'". "height='80' ";

[zorgi]

echo "<img src='$S' width='60' height='80' />";
 

[vikamjain]

@ Zorgi Sorry, In my question i forgot to put <> but in actual i have tried this command
echo "<image src='$S' width='60' height='80' />";
but it did not work for me

[johny10151981]

give the actual code, following the rule. your provided code is disturbing.

[zorgi]

If you did this:

echo "<img src='$S' width='60' height='80' />";
 

and image is not displayed than your path $S must be wrong.

[vikamjain]

helllo jorgi I used follwing four kind of pathnames in $s

$s="http://localhost/myfolder/imagefolder/image.jpeg" //with quotations
$s="c://xampp/htdocs/myfolder/imagefolder/image.jpeg"
$s=http://localhost/myfolder/imagefolder/image.jpeg //without quotations
$s=c://xampp/htdocs/myfolder/imagefolder/image.jpeg

but none of them worked

[zorgi]

Well its difficult for me to tell where are your image files but it looks to me as if you should do some reading on absolute and relative paths. There is loads of articles on internet on the subject. Just google for it. Here is one i found. I have't read it but I hope it helps.

http://support.internetconnection.ne...le_Paths.shtml

Googd luck

[kovik]

<image> and <img> are not the same.

[vikamjain]

<h2> Here is your Quiz</h2>
<?php
$i=0;
 
echo "<table border=\"1\" align=\"top\">";
echo "<tr><th>Question no</th>";
echo "<th>Question </th>";
echo "<th>Hint</th>";
echo "<th>Answer</th>";
echo "<th>Picture</th></tr>";
echo "<tr><td>";
while($i<$num)
{
  $qnn= mysql_result($result,$i, "Question_no");
  $qn= mysql_result($result,$i, "Question");
  $hint=mysql_result($result,$i, "Hint");
  $ans=mysql_result($result,$i, "Answer");
  $pic=mysql_result($result,$i, "Picture");
  $pic1="http://localhost/test2/uploaded_files/1264644980-11_17250697.jpg";
  echo $pic; 
  // $pic on echo command gives http://localhost/test2/uploaded_file...1_17250697.jpg
   echo $pic1;
  // $pic1 on echo command gives http://localhost/test2/uploaded_file...1_17250697.jpg 
 
        echo "<tr><td>";
    echo $qnn;
    echo "</td><td>";
    echo $qn;
    echo "</td><td>";
        echo $hint;
        echo "</td><td>";
        echo $ans;
        echo "</td><td>";
        echo "<img src='$pic1' />";
        echo "</td></tr>";      
 
$i+=1;
 
}
 
?>

//Here Value of Picture is passed using following command in some other file
// $s='http://' . $_SERVER['HTTP_HOST'] . $directory_self.'/uploaded_files/'.$now.$_FILES[$fieldname]['name'];

[vikamjain]

here all the connection to database are done in some other file. This file was included in that file so connection won't be a problem. $pic1 is showing the picture while $pic is not showing the picture when used in the command
echo "<img src='$pic or $pic1' />";

[kovik]

Have you tried verifying that the image URL is correct by directly visiting it?

[vikamjain]

As I said before I have given the actual url value in $pic1. which is working fine and when I used $pic which has the same value as $ pic1(which is evident from this fact that both are giving same thing when echo command is used)

[zorgi]

Hm, when things are this confusing I consult firebug. http://getfirebug.com/

[kovik]

You said that neither is working. You'll need to be much clearer if you want to get this solved.

If one works and the other doesn't, then they are NOT equal, regardless of what you think. Double-check them by copying the URL that they echo from the HTML in the <img> tag. Also, compare the two values using the "===" operator.