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display image file from mysql databse using php
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Join Date: Jun 2008
Posts: 2
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I'm having a problem with displaying my profile images from a database .the upload works fine,but no image is displayed:Could you please help!
I have the following code which cant work:
1. main profile page has the following script:
<?php echo '<img src="XXXXX_XXX.php?id=$Current_Person_Id">';
---------------------------------------------------------------------------------
2.the image page has the following script:
$result = mysqli_query($link,"select * from person where Person_Id='1
$result = mysqli_query($link,"select * from XXX where Person_Id='$Current_Person_Id'");
////starting the inner validation
//else if($result){
$row_count=0;
while($row = mysqli_fetch_array($result))
{
//$row_color = ($row_count % 2) ? $color1 : $color2;
$content=$row['Photo_Content'];
$name=$row['Photo_Name'];
$type=$row['Photo_Type'];
$size=$row['Photo_Size'];
$row_count++;
}
header("Content-length
size");
header("Content-type
type");
header("Content-Disposition: inline; filename=$name");
//header('filename=$name');
//Content-Disposition: attachment;
echo $content;
Note:
The variables
size,$type,$name are stored in database too.
I'm able to print all of them except the content.
Could it be my php configuration orsomething to do with images?
I've attached the image icon displayed..here....
I have the following code which cant work:
1. main profile page has the following script:
<?php echo '<img src="XXXXX_XXX.php?id=$Current_Person_Id">';
---------------------------------------------------------------------------------
2.the image page has the following script:
$result = mysqli_query($link,"select * from person where Person_Id='1
$result = mysqli_query($link,"select * from XXX where Person_Id='$Current_Person_Id'");
////starting the inner validation
//else if($result){
$row_count=0;
while($row = mysqli_fetch_array($result))
{
//$row_color = ($row_count % 2) ? $color1 : $color2;
$content=$row['Photo_Content'];
$name=$row['Photo_Name'];
$type=$row['Photo_Type'];
$size=$row['Photo_Size'];
$row_count++;
}
header("Content-length
size");header("Content-type
type");header("Content-Disposition: inline; filename=$name");
//header('filename=$name');
//Content-Disposition: attachment;
echo $content;
Note:
The variables
size,$type,$name are stored in database too.I'm able to print all of them except the content.
Could it be my php configuration orsomething to do with images?
I've attached the image icon displayed..here....
![]() |
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