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 Topic: PHP help |
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peacemaker
Joined: 23 May 2008
Posts: 80
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 Posted: Mon 26 Jan '09 2:28 Post subject: PHP help |
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Hi all
the problem i am facing is different, i am trying to display images form database,i m giving code that i wrote
| Code: |
$sql = "select * from upload ";
$result = mysqli_query($link,$sql) or die ("Could not access DB: " . mysqli_error());
echo "<table border=1>";
while ($row = mysqli_fetch_array($result))
{
//header("content-type: image/jpeg");
echo $row['name'] . "<br /> <br /><br />";
echo $row['content'];
//echo "<img src=\"images/". $row['content'] . "\" alt=\"\" /> <br /></td></tr>";
//echo "</p>";
//echo "</div>";
}
echo"</table>";
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when i use the above code then the result come with whole bineary charactred,but the number of images shown are correct all 5 images are shown.
but if i remove the comment header statement then nothing happens
but just message is dispaly as below
"http://localhost/.../gallery.php"
I am confused wats happening
plz help me wat can be the problem. as the header statement is very important but if i uncommnet it above message come and if i comment it then every thing comes ok but it shows whole binary data insted of image.
so plz correct my code if u can
or any one has simple code as i am using BLOB datatype and images are getting uploaded properly in database.
thanks
in advance
the help will be really appriciated as i am unable to find solution since long time
[/code] |
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James Blond
Moderator
Joined: 19 Jan 2006
Posts: 7371
Location: Germany, Next to Hamburg
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| Posted: Mon 26 Jan '09 11:02 Post subject: |
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if you use a header you can't put out any code before that! The table is not correct before that. Also you can't give out text/html and binary data in the file.
correct would be something like this:
| Code: |
<html>
....
<img src="your.php?picture_name=picure1" />
...
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your.php
| Code: |
<?php
$name=mysql_real_escap_string($_GET['picture_name']);
$sql = "select * from upload WHERE `name`='$name';";
$result = mysqli_query($link,$sql) or die ("Could not access DB: " . mysqli_error());
header("content-type: image/jpeg");
echo $row['content'];
?>
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(Code not tested) |
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peacemaker
Joined: 23 May 2008
Posts: 80
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| Posted: Tue 27 Jan '09 1:24 Post subject: Still the same problem |
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hi James
thanks for the replay and correction of code, i made 2 file 1 gallery.html where i put the 1st code , but i didnt understood the significance of variable
picture_name=picture1 does u means to say i should give the database field or something or it just a variable to pass.
I made the php file withe gallery.php
and put the remaining code in it but i modified as i need array, so i put while loop and put arrry. I didnt understood from where it wil accept or get picture name, in my web site when user click on the Gallery tab it should display the pic, thats wat i want ,i cannt pass the picture_name variable any name of the pic.
so can u explain to me plz. i am nt getting it or i am making something wrong init again
i m giving the code i modified for myself, i request you to plz modify it i am really in need of this code to get sorted.
Thanks in advance
| Code: |
if(!$link)
{
die('Could not connect: ' . mysqli_error());
}
else
{
$name=mysqli_escape_string($_GET['picture_name']);
//
$sql = "select * from upload WHERE `name`='$name'";
$result = mysqli_query($link,$sql) or die ("Could not access DB: " . mysqli_error());
while ($row=mysqli_fetch_array($result))
{
header("content-type: image/jpeg");
echo $row['content'];
}
}
?>
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Another thing is in gallery.html file the Html part is properly executed but it doesnt transfer control to gallery.php.
I tried to extract from the array the other fields like size,name id. It also dont work.
thanks for the help in advance |
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James Blond
Moderator
Joined: 19 Jan 2006
Posts: 7371
Location: Germany, Next to Hamburg
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| Posted: Wed 04 Mar '09 23:29 Post subject: |
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it would be much easier if you only store the path in the db and the images themself in the file system.
But back to your code
snip
| Code: |
while ($row = mysqli_fetch_array($result)) {
echo '<img src="your.php?picture_name='. $row['picture_name'] .'" />';
}
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snip
| Code: |
<?php
$name=mysql_real_escap_string($_GET['picture_name']);
$sql = "select * from upload WHERE `name`='$name';";
$result = mysqli_query($link,$sql) or die ("Could not access DB: " . mysqli_error());
header("content-type: image/jpeg");
echo $row['content'];
?>
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peacemaker
Joined: 23 May 2008
Posts: 80
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| Posted: Mon 29 Jun '09 12:36 Post subject: Hi |
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| Hi James sorry for not replying to this reply of yours, actually i did the same thing u told me to do. I stored the paths of the images in the table and images into a folder. thanks for that. |
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