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angel55

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I have problem with viewing my image.When i load the image page it give me loads of symbols.I can not get my photo field to be displayed, it gives all different sysmbol.

 

$db1 = new estate();

$db1->openDB();

$sql="SELECT * from properties";

$result=$db1->getResult($sql);

 

echo "<table border='1' >";

 

echo "<tr><th>Properties_ID</th><th>Type</th><th>Price</th><th>Address

</th><th>Postcode</th><th>Photo</th></tr>";

 

 

 

 

 

while($row = mysql_fetch_assoc($result))

{

echo "<tr>";

echo "<td>{$row['Properties_ID ']}</td><td>{$row['Type ']}</td>";

echo"<td>{$row['Price']}</td><td>{$row['Address']}</td>";

echo "<td>{$row['Postcode']}</td><td>{$row['Photo']}</td>";

echo "</tr>";

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<?php echo"<img src='{$row['photo']}' />";

xhtml way.

 

<?php echo"<img src='{$row['photo']}' >";

html way.

 

i hope the $row['photo'] from the database,points to the correct folder and file from the while loop.

 

in other words have you enter the path to the current files or is photo the file names?

 

<?php echo"<img src='{$row['photo']}' >" << no directory?

[code]<?php echo"<img src='pic/{$row['photo']}' >" << with directory

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what as binary

 

$query = "select image from se_keys where id = $id";

$result = mysql_query($query);

$row = mysql_fetch_array($result);

 

echo"$row[image]";

 

 

then

 

$query1 = "select * from se_keys";

$result1 = mysql_query($query);

$row1 = mysql_fetch_array($result1);

<img src='imget.php?id=$row[id]'> 

 

sorry very busy.

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this is my table.i did photo as blob.

CREATE TABLE IF NOT EXISTS `properties` (

  `Properties_ID` int(4) NOT NULL AUTO_INCREMENT,

  `Type` varchar(20) NOT NULL,

  `Price` float NOT NULL,

  `Address` varchar(40) NOT NULL,

  `Postcode` varchar(7) NOT NULL,

  `Photo` blob NOT NULL,

  `IsAvailable` varchar(4) NOT NULL,

  `Agent_ID` int(4) NOT NULL,

  PRIMARY KEY (`Properties_ID`),

  UNIQUE KEY `Agent_ID` (`Agent_ID`)

) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ;

There is no picture displaying just red cross.

 

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afaik you'll need a second page which will function as the image and does something like

 

<?php
header('Content-Type: image/jpeg');
//DB stuff here
echo $row['Photo'];
?>

 

Of course you could make that second page work off an $_GET['value'] and then use <img src='/secondPageScript.php?imageId' alt='' />

Or you could just not store the image in the database but save the path to it, which is a lot more resource friendly and you'll be less likely to get a mail from your host stating that you need to lower the resource use of your page by the end of the month or switch to a more expensive package.

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