what is wrong with this code

**Velhari** · Nov 16 '06, 08:23 PM

Originally posted by nwebhosting

I click on a image on textimage20.php
It takes me to text4.php but it all displays is:

SELECT gameandtoyname, pricevalue, bigdescription, dollsimage.imag e_id, dollsimage.imag e_date, doll.pricetype, prices.pricetyp e, prices.priceval ue FROM dollsimage, doll, prices WHERE doll.pricetype = prices.pricetyp e and dollsimage.imag e_id = '123' and doll.gameandtoy name = '123' and doll.bigdescrip tion = '123'

Then I see the border, an X in there for the image
And then $0.00

Thank you for any help.

This is my code:

Textimage20.php

[HTML]
[PHP]
echo '<TD> <a href="http://localhost/text4.php?image _id=123"> <img border="1" height="90" width="100"
src="imagedolls .php?act=view&i id=' . $row['image_id'] . '"> </a> ' '</TD>' ;
[/PHP]
[/HTML]

text4.php :

[HTML]
[PHP]
$id = $_GET['image_id'];

$sql = ("SELECT gameandtoyname, pricevalue, bigdescription, dollsimage.imag e_id, dollsimage.imag e_date, doll.pricetype, prices.pricetyp e, prices.priceval ue
FROM dollsimage, doll, prices
WHERE doll.pricetype = prices.pricetyp e and dollsimage.imag e_id = '$id' and doll.gameandtoy name = '$id' and doll.bigdescrip tion = '$id'
");
echo "$sql ";

$row = mysql_fetch_arr ay($result);

echo '<TD> <img border="1" height="90" width="100"
src="imagedolls .php?act=view&i id=' . $row['image_id'] . '"> ';

echo ' ' .$row['gameandtoyname '] ;
echo ' ' .$row['bigdescription '] ;

$row['pricevalue'] = number_format($ row['pricevalue'],2, '.', '');
echo ' ';
echo '$' .$row['pricevalue'].

[/PHP]
[/HTML]

Hi,
I think the mistake made by you is you didn't include the following statement in your code, if this is included means it might been successfull to you

[PHP]$result = mysql_query($sq l);[/PHP]

With Regards,
Velmurugan.H

**nwebhosting** · Nov 16 '06, 09:59 PM

I do have the statement : $result = mysql_query ($sql, $conn);

Here is the code again

textimage20.php :

[HTML]
[PHP]
echo '<TD> <a href="http://localhost/text4.php?image _id=123"> <img border="1" height="90" width="100"
src="imagedolls .php?act=view&i id=' . $row['image_id'] . '"> </a> ';

echo ' ' .$row['gameandtoyname '] ;

$row['pricevalue'] = number_format($ row['pricevalue'],2, '.', '');
echo ' ';
echo '$' .$row['pricevalue']. '</TD>';

[/HTML]
[/PHP]

text4.php :

[[HTML]
[PHP]
<?php

$id = $_GET['image_id'];

$sql = ("SELECT gameandtoyname, pricevalue, bigdescription, dollsimage.imag e_id, dollsimage.imag e_date, doll.pricetype, prices.pricetyp e, prices.priceval ue
FROM dollsimage, doll, prices
WHERE doll.pricetype = prices.pricetyp e and dollsimage.imag e_id = '$id' and doll.gameandtoy name = '$id' and doll.bigdescrip tion = '$id'
");
echo "$sql ";

$result = mysql_query ($sql, $conn);

if (!$result) {
die('Invalid query: ' . mysql_error());
}

echo "<TABLE BORDER=\"1\">\n ";

$row = mysql_fetch_arr ay($result);

echo '<TD> <img border="1" height="90" width="100"
src="imagedolls .php?act=view&i id=' . $row['image_id'] . '"> ';

echo ' ' .$row['gameandtoyname '] ;
echo ' ' .$row['bigdescription '] ;

$row['pricevalue'] = number_format($ row['pricevalue'],2, '.', '');
echo ' ';
echo '$' .$row['pricevalue']. '</TD>';

[/PHP]
[/HTML]

?>

Originally posted by Velhari

Hi,
I think the mistake made by you is you didn't include the following statement in your code, if this is included means it might been successfull to you

[PHP]$result = mysql_query($sq l);[/PHP]

With Regards,
Velmurugan.H

**khalidbaloch** · Nov 17 '06, 03:59 PM

try by removing this line from text4.php

echo "$sql ";

mysql select statment is being printed out in html

good luck and take care

what is wrong with this code

what is wrong with this code

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