problem with this code

[alexdog1805]

<?php

$host = "host";
$user = "user";
$pass = "password";


$con = mysql_connect($host , $user , $pass);

if(!$con)
{
die("Failed to connect to the server" . mysql_error());
}

$image = "images/advert.jpg";

$result = mysql_query("SELECT * FROM table WHERE image = $image");
while($row = mysql_fetch_array($result))
{
echo $row['image'];
?>


I created in database a table ''image" and i created in this table a decimal camp called 'image'. I put the image advert.jpg in image folder but it doesnt works.

It appears to me this message:
Parse error: syntax error, unexpected $end in /home/hosting/masterfifa/index.php on line 25

[zeromadpeter]

<?php
$host = "host";
$user = "user";
$pass = "password";
$link = mysql_connect($host , $user , $pass);
if($link)
{
$db_selected = mysql_select_db("databasename",$link);
if ($db_selected)
{
$image = "images/advert.jpg";
$result = mysql_query("SELECT * FROM table WHERE image='".$image."'");
while($I = mysql_fetch_array($result))
{
echo "- entry: ".$I['image']."<br/>";
}
}
}
?>

expected
- entry: images/advert.jpg <br/>
repeated the number of times in the database

changes:
added database select
changed some names ^_^
updated the SQL to enclose the WHERE statment value in '
finished teh code

[alexdog1805]

<?php
$host = "host";
$user = "user";
$pass = "password";
$link = mysql_connect($host , $user , $pass);
if($link)
{
$db_selected = mysql_select_db("databasename",$link);
if ($db_selected)
{
$image = "images/advert.jpg";
$result = mysql_query("SELECT * FROM table WHERE image='".$image."'");
while($I = mysql_fetch_array($result))
{
echo "- entry: ".$I['image']."<br/>";
}
}
}
?>

expected
- entry: images/advert.jpg <br/>
repeated the number of times in the database

changes:
added database select
changed some names ^_^
updated the SQL to enclose the WHERE statment value in '
finished teh code
the code you gave me is ok, it doesn' appears to me any error but it doesn't apears the photo, which is the thing that i want for this code to do. Please help me with this code to appears me the photo. Thanks

[zeromadpeter]

if you want it to output the html for the image then why are u using a database and not just

<img src="./images/advert.jpg" alt="image woot"/>

please can you tell me what you want to do with the database.

[alexdog1805]

if you want it to output the html for the image then why are u using a database and not just

<img src="./images/advert.jpg" alt="image woot"/>

please can you tell me what you want to do with the database.

because I am a beginner in php and I want to learn to do in php things that I can make in html, so can you tell me in php how I do that the image appears?

[boogyman]

while($I = mysql_fetch_array($result))
{
$img = sprintf("<img src=\"%s\" width=\"\" height=\"\" alt=\"\">", $I['image']);
echo $img;
}

[alexdog1805]

now I have this code:

<?php
$host = "host";
$user = "user";
$pass = "password";
$link = mysql_connect($host , $user , $pass);
if($link)
{
$db_selected = mysql_select_db("image",$link);
if ($db_selected)
{
$image = "images/advert.jpg";
$result = mysql_query("SELECT * FROM table WHERE image='".$image."'");
while($I = mysql_fetch_array($result))
{
echo "- entry: ".$I['image']."<br/>";
}
}
}
while($I = mysql_fetch_array($result))
{
$image = sprintf("<img src="images/advert.jpg" width="20" height="100" alt="20">", $I['image']);
echo $img;
}
?>


and this don't works too, It appears to me this message:

Parse error: syntax error, unexpected T_STRING in /home/hosting/masterfifa/index.php on line 21

[punstc]

here try this.


<?php
$host = "host";
$user = "user";
$pass = "password";
$link = mysql_connect($host , $user , $pass);

if($link) {
$db_selected = mysql_select_db("image",$link);
if ($db_selected) {
$image = "images/advert.jpg";
$result = mysql_query("SELECT * FROM table WHERE image='$image' ");
while($I = mysql_fetch_array($result)) {
echo "- entry: ".$I['image']."<br/>";
}
}
}

while($I = mysql_fetch_array($result)) {
$img = sprintf("<img src=\"images/advert.jpg\" width=\"20\" height=\"100\" alt=\"20\">", $I['image']);
echo $img;
}

?>


If you look at this specific line



$img = sprintf("<img src=\"images/advert.jpg\" width=\"20\" height=\"100\" alt=\"20\">", $I['image']);


The backslashes in font of the quotes are needed to tell php to ignore this quote as the end of the string I'm entering. I believe the proper name for it is escaping.

Just remember that any time you need to use a quote for a normal html like src="", alt="", etc. inside your php echo string or whatever else your entering you need to add the backslash in front of it so php will ignore the quote.

[alexdog1805]

here try this.


<?php
$host = "host";
$user = "user";
$pass = "password";
$link = mysql_connect($host , $user , $pass);

if($link) {
$db_selected = mysql_select_db("image",$link);
if ($db_selected) {
$image = "images/advert.jpg";
$result = mysql_query("SELECT * FROM table WHERE image='$image' ");
while($I = mysql_fetch_array($result)) {
echo "- entry: ".$I['image']."<br/>";
}
}
}

while($I = mysql_fetch_array($result)) {
$img = sprintf("<img src=\"images/advert.jpg\" width=\"20\" height=\"100\" alt=\"20\">", $I['image']);
echo $img;
}

?>


If you look at this specific line



$img = sprintf("<img src=\"images/advert.jpg\" width=\"20\" height=\"100\" alt=\"20\">", $I['image']);


The backslashes in font of the quotes are needed to tell php to ignore this quote as the end of the string I'm entering. I believe the proper name for it is escaping.

Just remember that any time you need to use a quote for a normal html like src="", alt="", etc. inside your php echo string or whatever else your entering you need to add the backslash in front of it so php will ignore the quote.

I tried this code and it doesn't appears me any message error, but it doesn't apeears image too. Look here http://masterfifa.3x.ro, here should appears the advert.jpg photo.

[alexdog1805]

I still can't see the photo.:(