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Problem could not show the image

Discussion in 'PHP' started by t6werserv, Jan 24, 2010.

0
  1. #1
    Problem could not show the image from a database
    

$sql = "SELECT content FROM upload WHERE id=$id";
if(!($result=mysql_query($sql))) showerror;
$data=mysql_fetch_array($result);
echo'<img border=0 src='.$data['content'].' width=248 height=166>';
    Code (markup):

     
    t6werserv, Jan 24, 2010 IP
  2. hasanbasri

    hasanbasri Peon

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    #2 
    $sql = "SELECT content FROM upload WHERE id=$id";
$result=mysql_query($sql);
$data=mysql_fetch_array($result);
if($data) 
echo'<img border=0 src='.$data['content'].' width=248 height=166>';
else
showerror();
    Code (markup):
     
    hasanbasri, Jan 24, 2010 IP
  3. RDJLabs

    RDJLabs Peon

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    #3
    You have a single quote problem in syntax use following code


    
$sql = "SELECT content FROM upload WHERE id=$id";
if(!($result=mysql_query($sql))) showerror;
$data=mysql_fetch_array($result);
echo "<img border=0 src=".$data['content']." width=248 height=166>";
    PHP:
     
    RDJLabs, Jan 24, 2010 IP
  4. t6werserv

    t6werserv Peon

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    #4
    result:


    Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in D:\AppServ\www\myscripts\uploads\view.php on line 15

    Fatal error: Call to undefined function showerror() in D:\AppServ\www\myscripts\uploads\view.php on line 19
     
    t6werserv, Jan 24, 2010 IP
  5. t6werserv

    t6werserv Peon

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    #5 
    


$sql = "SELECT content FROM upload WHERE id=$id";
if(!($result=mysql_query($sql))) showerror;
$data=mysql_query($result);
echo "<img border=0 src=".$data['content']." width=248 height=166>";
    Code (markup):
    not error but no view picture!
     
    t6werserv, Jan 24, 2010 IP
  6. ibg

    ibg Member

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    #6 
    
$sql = "SELECT content FROM upload WHERE id=$id";
if(!($result=mysql_query($sql))) showerror;
$data=mysql_fetch_assoc($result);
echo '<img border="0" src="'.$data['content'].'" width="248" height="166">';
    PHP:
     
    ibg, Jan 24, 2010 IP
  7. t6werserv

    t6werserv Peon

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    #7
    print data no image

    «¾ئئئزززتتت–¢¯\m}ٌٍِؤؤؤ„™©¨ش‎PMJٌîéذذذ“’’êىîx½‏¯¯¯¢££ججج Z…؛e~•o¬î‎ُْSSS™±عi§îنمâûّْ‚‚‚£»م¢،ں¤¤¤»»»KJJ*¸هزذد×ظعFNXكغضصصصOe{861ôٌٍëëيçههGEB؟هے?²ےجےكٍےïùےمممًًًىىىهههںےàààO¹ےدىےيييêêê¥ےâââëëëççç™ےüüüèèè÷÷÷نننééé/¬ے¸¸¸_؟ےِِِوووٌٌٌoإے‏‏‏لللںظےڈزے‎‎‎¯كےةةةُِ÷ذشطًًٌ‎ü‎ًٌٌü‎ün¯ّe“؟]کهييپ_‹¾Vi‚dddmjhUٹ½z´ٌ7¯¬D€°ôc <د جS‡çہ@Ù¾ ¦5™‰پ L0"W#BØ·V Hھôا€™Â¥:ض±ˆ%پˆ0²مشڈ?‰ٌ2يً©‹KPTI+’Ú¯ّمہق{y00<#D‰ے9؛ôرÙ¾Ú˜ظ†ôڈشw¨كے†ْ*Ù†$Û’Ú¯
     
    t6werserv, Jan 24, 2010 IP
  8. ibg

    ibg Member

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    #8
    Then your content is wrong or you're storing the whole image in the database.
     
    ibg, Jan 24, 2010 IP
  9. astkboy2008

    astkboy2008 Peon

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    #9
    you must print your content in other header with image format header
    and put this new page in scr
     
    astkboy2008, Jan 24, 2010 IP
  10. astkboy2008

    astkboy2008 Peon

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    #10
    you must print your content in other header with image format header
    and put this new page in scr
     
    astkboy2008, Jan 24, 2010 IP