Image display

[pdn]

Forgive me as I am absolutely a Newbie...

I am trying to query mysql and display an image I used php software to create the scripts and it runs fine in them... however I would like to customize the look somewhat.

Here is the script I have written so far:

<?php
$username="root";
$password="mypassword";
$hostname="localhost";

// Connect to Database
$dbh=mysql_connect($hostname, $username, $password)
or die("Unable to connect to Database");
$selected=mysql_select_db("rsod",$dbh)
or die("Unable to select rsod");

$result = mysql_query("SELECT id, name, address, conviction, ('<img src="../php_test/images/",' border=0>') pic FROM main_tbl");
while ($row = mysql_fetch_array($result,MYSQL_ASSOC))
{
?>
<table border="1" cellspacing="5" width="600">
<tr><th align="left" width="100"><font face="Verdana" size="2" color="#0000FF">Name:</font></th>
<td>
<?php print $row{'name'}."<br>"; ?>
</td><td rowspan="3" width="340"><?php print $row{'pic'}."<br>"; ?></td></tr>
<tr><th align="left" width="100"><font face="Verdana" size="2" color="#0000FF">Address:</font></th>
<td><?php print $row{'address'}."<br>"; ?>
</tr>
<tr><th align="left" width="100"><font face="Verdana" size="2" color="#0000FF">Conviction:</font></th>
<td>
<?php print $row{'conviction'}."<br>"; ?>
</td></tr>
</table>
<?php
}

// more to come
mysql_close($dbh);

?>

When open page this is the error I get:

Parse error: syntax error, unexpected '.' in D:\csweb\PHP_View\test.php on line 12

Any help greatly appreciated

[paulnaj]

Hi pdn,

The problem line is ...

$result = mysql_query("SELECT id, name, address, conviction, ('<img src="../php_test/images/",' border=0>') pic FROM main_tbl");

What you were tring to do with the bit in bold. Something like ...

$result = mysql_query("SELECT id, name, address, conviction, CONCAT('<img src=\"../php_test/images/\"', image_field, ' border=0>') AS pic FROM main_tbl");

... where I've had to guess the name of the field (image_field) in main_tbl that contains the name of the image. You'll have to fix that.

Notice ...
1) CONCAT() function sticks strings together.
2) The whole string of SQL is enclosed in double-quotes. So, you have to escape ALL double-quotes inside it.
3) Without a reference (image_field) to an actual image, your img src would have been just a folder name.
4) Remember the keyword 'AS'.

Come back if you don't understand anything.

Paul 🙂

[pdn]

Thank you for the reply...

I made the changes you suggested but for some reason my pic still displays as a red x. (pic is the name of my image field)

I've enclosed the current code.

Thanks again

<?php
$username="root";
$password="pass";
$hostname="localhost";

// Connect to Database
$dbh=mysql_connect($hostname, $username, $password)
or die("Unable to connect to Database");
$selected=mysql_select_db("rsod",$dbh)
or die("Unable to select rsod");

$result = mysql_query("SELECT id, name, address, conviction, CONCAT('<img

src=\"../php_test/images/\"', 'pic', ' border=0>') AS pic FROM main_tbl");
while ($row = mysql_fetch_array($result,MYSQL_ASSOC))
{
?>
<table border="1" cellspacing="5" width="600">
<tr><th align="left" width="100"><font face="Verdana" size="2"

color="#0000FF">Name:</font></th>
<td>
<?php print $row{'name'}."<br>"; ?>
</td><td rowspan="3" width="340"><?php print $row{'pic'}."<br>"; ?></td></tr>
<tr><th align="left" width="100"><font face="Verdana" size="2"

color="#0000FF">Address:</font></th>
<td><?php print $row{'address'}."<br>"; ?>
</tr>
<tr><th align="left" width="100"><font face="Verdana" size="2"

color="#0000FF">Conviction:</font></th>
<td>
<?php print $row{'conviction'}."<br>"; ?>
</td></tr>
</table>
<?php
}

// more to come
mysql_close($dbh);

?>

[paulnaj]

Ok nearly there, it's the same line again ... should be ...

<?php
$result = mysql_query("
	SELECT 	id, name, address, conviction
	, CONCAT('<img src=\"../php_test/images/', pic, '\" border=\"0\">') AS pic 
	FROM main_tbl
"); 
?>

... assuming pic is a field in main_tbl.

The problem was that you had pic single-quoted.

Sometimes, a good trick for debugging is to echo out your query to the browser to see if it (at least) 'looks' right.

P.

Oops noticed somethings else ... fixed now

[pdn]

Sorry to keep bugging...

I made the change like this:

$result = mysql_query("SELECT id, name, address, conviction, CONCAT('<img

src=\"../php_test/images/',pic, '\" border=\"0\">') AS pic FROM main_tbl");
while ($row = mysql_fetch_array($result,MYSQL_ASSOC))

but still get red x.

😕

Thanks

[paulnaj]

No problem to keep coming back, but did you try the debugging trick I suggested?

P.

[pdn]

I'm not sure how to do that... can you give me further instruction

[paulnaj]

OK, try this ...

<?php 
$username="root"; 
$password="pass"; 
$hostname="localhost"; 

// Connect to Database 
$dbh = mysql_connect($hostname, $username, $password) 
	or die("Unable to connect to Database"); 
$selected = mysql_select_db("rsod",$dbh) 
	or die("Unable to select rsod"); 

$sql = "
SELECT 	id
	, 	name
	, 	address
	, 	conviction
	, 	CONCAT('<img src=\"../php_test/images/', pic, '\" border=\"0\">') AS pic FROM main_tbl
";
echo '<pre>'.$sql'</pre>';
exit;
$result = mysql_query($sql); 
// ... rest of script
?> 

Paul 🙂

PS Please use the

 tags to enclose your code ... makes life so much easier.

[pdn]

Ok did all that...

this is the response i am getting in the browser. I'm not clear on how to interpret it.

SELECT id
, name
, address
, conviction
, CONCAT('') AS pic FROM main_tbl

there is a red x between ' ' after CONCAT(

Thanks again

[paulnaj]

Ok pdn, the place to look is in the source.

The browser is trying unsuccessfully to parse the image still, so shows you the 'result'.

In Internet Explorer: [View][Source]
In Firefox/Mozilla: [View][Page Source]

Paul 🙂

[pdn]

I know definitely that my image location is set correctly in the script as well as the image field is named pic in mysql... Any thoughts why it still isn't working.

still getting red x

[paulnaj]

Post the source up to the forum ... put

 tags around it so it displays well.

P.

[pdn]

Ok here is the source...

btw... you can view the results at: http://www.csweb.net/php_test/test.php

<?php
$username="root";
$password="pass";
$hostname="localhost";

// Connect to Database
$dbh=mysql_connect($hostname, $username, $password)
or die("Unable to connect to Database");
$selected=mysql_select_db("rsod",$dbh)
or die("Unable to select rsod");

// $result = mysql_query("SELECT id, name, address, conviction, CONCAT('<img src=\"../php_test/images/',pic, '\" border=\"0\">') AS pic FROM main_tbl");

$sql = "
SELECT id
, Name
, Address
, Conviction
, CONCAT('<img src=\"./images/',main_tbl.Pic, '\" border=\"0\">') AS 'Picture' FROM

main_tbl
";
echo '<pre>'.$sql,'</pre>';
exit;
$result = mysql_query($sql);

while ($row = mysql_fetch_array($result,MYSQL_ASSOC))
{
?>
<table border="1" cellspacing="5" width="600">
<tr><th align="left" width="100"><font face="Verdana" size="2"

color="#0000FF">Name:</font></th>
<td>
<?php print $row{'name'}."<br>"; ?>
</td><td rowspan="3" width="340"><?php print $row{'pic'}."<br>"; ?></td></tr>
<tr><th align="left" width="100"><font face="Verdana" size="2"

color="#0000FF">Address:</font></th>
<td><?php print $row{'address'}."<br>"; ?>
</tr>
<tr><th align="left" width="100"><font face="Verdana" size="2"

color="#0000FF">Conviction:</font></th>
<td>
<?php print $row{'conviction'}."<br>"; ?>
</td></tr>
</table>
<?php
}

// more to come
mysql_close($dbh);

?>

Thanks mucho

[paulnaj]

Err ... the code ...

$sql = "
SELECT id
, Name
, Address
, Conviction
, CONCAT('<img src=\"./images/',main_tbl.Pic, '\" border=\"0\">') AS 'Picture' FROM

main_tbl
";

... and the source of http://www.csweb.net/php_test/test.php are different.

Check that you are using/displaying the correct file.

P.