view Upload image

Question

niths -1 Posting Whiz in Training

13 Years Ago

hi all,
i am having a page where i can upload images.so now my problem is i hav to view the image that is uploaded and that should display on the web page.

My code is uploading an image in the images folder. so now i need to display that image below browse button.
thank u..

<html>
<body>
  <form name="newad" method="post" enctype="multipart/form-data"  action="test.php">
 <table>
     <tr><td><input type="file" name="image"></td></tr>
     <tr><td><input name="Submit" type="submit" value="View image"></td></tr>
 </table>    
 </form>
</body>
</html>               
<?php
 define ("MAX_SIZE","1000"); 
 function getExtension($str) 
 {
         $i = strrpos($str,".");
         if (!$i) 
         { 
             return ""; 
         }
         $l = strlen($str) - $i;
         $ext = substr($str,$i+1,$l);
         return $ext;
 }
 $errors=0;
 if(isset($_POST['Submit'])) 
 {
     $image=$_FILES['image']['name'];
     if ($image) 
     {
         $filename = stripslashes($_FILES['image']['name']);
         $extension = getExtension($filename);
         $extension = strtolower($extension);
         if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) 
         {
             echo '<h1>Unknown extension!</h1>';
             $errors=1;
         }
         else
         {
            $size=filesize($_FILES['image']['tmp_name']);
            if ($size > MAX_SIZE*1024)
            {
                echo '<h1>You have exceeded the size limit!</h1>';
                $errors=1;
            }
            $image_name=time().'.'.$extension;
            $newname="images/".$image_name;
            $copied = copy($_FILES['image']['tmp_name'], $newname);
            if (!$copied) 
            {
                echo '<h1>Copy unsuccessfull!</h1>';
                $errors=1;
            }
         }
     }
 }
 if(isset($_POST['Submit']) && !$errors) 
 {
     echo "<h1>File Uploaded Successfully! Try again!</h1>";
 }

 ?>

php

2 Contributors
11 Replies
148 Views
12 Hours Discussion Span
Latest Post 13 Years Ago Latest Post by niths

All 11 Replies

vibhaJ 126 Master Poster

13 Years Ago

For displaying it back you should use mysql db.

So once image is uploaded on server i.e. @ echo "<h1>File Uploaded Successfully! Try again!</h1>";
$newname will be saved in table.

After that using select query that $newname will be fetched and shown in image src tag.

Edited 13 Years Ago by vibhaJ because: n/a

vibhaJ 126 Master Poster

13 Years Ago

where did you place this code? rather post complete page.
What error you are getting??

vibhaJ 126 Master Poster

13 Years Ago

At line no 61:
echo $sql="INSERT INTO images(name,size,path,comments) VALUES ('$filename','$s','$newname','data is not present')"; exit;
Check this echo query in sql window. Does it works successfully?

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niths -1 Posting Whiz in Training · Answer 1 · 2010-07-14T14:01:00+00:00

include 'connection.php';
     $sql="INSERT into images(name,size,path,comments) VALUES ('$filename','$s','$newname','data is not present')";
     if(!mysql_query($sql,$con))
     {
        die('Error:' . mysql_error());
     }

i am getting the output as 'Error' and the data is not inserted in to the database. so can anyone please

niths -1 Posting Whiz in Training · Answer 2 · 2010-07-14T15:38:20+00:00

<html>
<body>
  <form name="newad" action="test.php" method="POST" enctype="multipart/form-data">
    <p>Upload Image: <input type="file" name="image"><br>
    <font size="1">Click browse to upload a local file</font><br>
    <br>
    <input type="submit" name="Submit" value="View Image">
    </form>     
</body>
</html>               
<?php
include 'connection.php';
 define ("MAX_SIZE","1000"); 
 function getExtension($str) 
 {
         $i = strrpos($str,".");
         if (!$i) 
         { 
             return ""; 
         }
         $l = strlen($str) - $i;
         $ext = substr($str,$i+1,$l);
         return $ext;
 }
 $errors=0;
 if(isset($_POST['Submit'])) 
 {
     $image=$_FILES['image']['name'];
     if ($image) 
     {
         $filename = stripslashes($_FILES['image']['name']);
         $extension = getExtension($filename);
         $extension = strtolower($extension);
         if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) 
         {
             echo '<h1>Unknown extension!</h1>';
             $errors=1;
         }
         else
         {
            $size=filesize($_FILES['image']['tmp_name']);
            $s=round($size/1024);
            if ($size > MAX_SIZE*1024)
            {
                echo '<h1>You have exceeded the size limit!</h1>';
                $errors=1;
            }
            $image_name=$filename;
            $newname="images/".$image_name;
            $copied = copy($_FILES['image']['tmp_name'], $newname);
            if (!$copied) 
            {
                echo '<h1>Copy unsuccessfull!</h1>';
                $errors=1;
            }
         }
     }
    // echo $filename;
    //  echo $s;  
    //  echo $newname;   
     $sql="INSERT INTO images(name,size,path,comments) VALUES ('$filename','$s','$newname','data is not present')";
     $data=mysql_query($sql,$con);
     if(!$data)
     {
        die('Could not enter data:' . mysql_error());
     }
 }
 if(isset($_POST['Submit']) && !$errors) 
 {
     echo "<h1>File Uploaded Successfully! Try again!</h1>";
 }
  
 ?>

i am getting error as
"Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\Photoediting\test.php on line 62".

niths -1 Posting Whiz in Training · Answer 3 · 2010-07-14T16:01:03+00:00

i had done that i am not getting any error the query is correct.even i had done like this also

echo $sql;

and in the browser when i uploaded any image i am getting that name, size and path correctly.but the only thing is they are not inserting into data base.

vibhaJ 126 Master Poster · Answer 4 · 2010-07-14T16:06:55+00:00

then there is missing with your $con. Execute any simple insert query just for testing connection and check that insert works or not?

niths -1 Posting Whiz in Training · Answer 5 · 2010-07-14T16:38:53+00:00

i removed "$con" from

$data=mysql_query($sql,$con);

line 62. and now i got it.

niths -1 Posting Whiz in Training · Answer 6 · 2010-07-14T17:02:33+00:00

now how can i display the uploaded image below the browser button..

vibhaJ 126 Master Poster · Answer 7 · 2010-07-14T17:11:13+00:00

lets this page is index.php.

<?
 include 'connection.php';
 define ("MAX_SIZE","1000"); 
 function getExtension($str) 
 {
         $i = strrpos($str,".");
         if (!$i) 
         { 
             return ""; 
         }
         $l = strlen($str) - $i;
         $ext = substr($str,$i+1,$l);
         return $ext;
 }
 $errors=0;
 if(isset($_POST['Submit'])) 
 {
     $image=$_FILES['image']['name'];
     if ($image) 
     {
         $filename = stripslashes($_FILES['image']['name']);
         $extension = getExtension($filename);
         $extension = strtolower($extension);
         if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) 
         {
             echo '<h1>Unknown extension!</h1>';
             $errors=1;
         }
         else
         {
            $size=filesize($_FILES['image']['tmp_name']);
            $s=round($size/1024);
            if ($size > MAX_SIZE*1024)
            {
                echo '<h1>You have exceeded the size limit!</h1>';
                $errors=1;
            }
            $image_name=$filename;
            $newname="images/".$image_name;
            $copied = copy($_FILES['image']['tmp_name'], $newname);
            if (!$copied) 
            {
                echo '<h1>Copy unsuccessfull!</h1>';
                $errors=1;
            }
         }
     }    
     if(!$data)
     {
        die('Could not enter data:' . mysql_error());
     }
 }
 if(isset($_POST['Submit']) && !$errors) 
 {
     echo "<h1>File Uploaded Successfully! Try again!</h1>";
	 $sql="INSERT INTO images(id,name,size,path,comments) VALUES (null,'$filename','$s','$newname','data is not present')";
     $data=mysql_query($sql);
	 $LastId = mysql_insert_id();
	 header("Location:index.php?id=".$LastId);
	 exit;
 }
 if(isset($_GET['id']))
 {
 	$q= " select * from images where id=".$_GET['id'];
	$rs = mysql_query($q);
	$sar = mysql_fetch_array($rs);
	$imgPath = $sar['path'];
 }
  
 ?>
 <html>
<body>
  <form name="newad" action="test.php" method="POST" enctype="multipart/form-data">
    <p>Upload Image: <input type="file" name="image"><br>
    <font size="1">Click browse to upload a local file</font><br>
    <br>
    <input type="submit" name="Submit" value="View Image">	
	<? if($imgPath!=''){?><img src="<?=$imgPath;?>"><? } ?>
    </form>     
</body>
</html>

niths -1 Posting Whiz in Training · Answer 8 · 2010-07-14T17:24:34+00:00

i should get all the images present in the database. so can we get by that code..?

view Upload image

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