i am doing a site for appliance center..
i need to display all the products that the company offers, but my problem is that i cant display ALL the images in my database.. the first entry on the database is the only one that displays..
i am using mysql as database
here's the code.. tell me what's my error and pls. kindly edit it..
**imgdata = it is where i store the image.. longblob is the data type -
<?php
-
$errmsg = "";
-
if (! @mysql_connect("localhost","root",""))
-
{
-
$errmsg = "Cannot connect to database";
-
}
-
@mysql_select_db("upload");
-
-
$gotten = @mysql_query("select imgdata from pix");
-
header("Content-type: image/jpeg");
-
while ($row = mysql_fetch_array($gotten))
-
{
-
print $row['imgdata'];
-
-
}
-
mysql_free_result($gotten);
-
?>
Feb 6 '08
48 41438
Hi again...
Something like this..... for the page which displays all the images -
<?php
-
$errmsg = "";
-
if (! @mysql_connect("localhost","root",""))
-
{
-
$errmsg = "Cannot connect to database";
-
}
-
@mysql_select_db("upload");
-
-
$strSQL = "select * from pix";
-
$rsPix = mysql_query($strSQL);
-
$numRows = mysql_numrows($rsPix);
-
$i = 0;
-
-
while($i < $numRows){
-
?>
-
<h1><?php echo mysql_result($rsPix,$i,"pixTitle"); ?></h1>
-
<img src="pix.php?pixID=<?php echo mysql_result($rsPix,$i,"pixID"); ?>"/>
-
<?php
-
$i++;
-
}
-
?>
hello!! thanks for replying again..
you're great.. it worked!!
mmm.. can i ask you again?
how can i display all the images in a table of 3 columns while the rows will depend on how many images is in my database
ex:
total # of images: 25
so, there will be 3 columns and 9 rows (the last row has only 1 image)
thanks again for the reply!!
Hi,
I appreciate your gratitude.
I don't get that last question fully.
What is the table structure for this new table (you can give me the create query if you like). I can try work it out or give you a proper solution, but i need more info this time...
:)
Hi,
I appreciate your gratitude.
I don't get that last question fully.
What is the table structure for this new table (you can give me the create query if you like). I can try work it out or give you a proper solution, but i need more info this time...
:)
ah ok, i'll explain it again..
previously, you've given me the code for displaying all the images in my database.. when i tried it, it looks like this ( # for example is the image)
# # # # # # # #
i tried putting the code inside a table.
here's the code: -
<table>
-
<tbody>
-
<?
-
if($colCtr % $cols == 0)
-
{
-
?>
-
<tr>
-
<td><img src="pix.php?pid=<?php echo mysql_result($rsPix,$i,"pid"); ?>"/></td>
-
</tr>
-
<? $colCtr++;
-
}
-
?>
-
</tbody>
-
</table>
-
-
but the output looks like this:
-
#
-
#
-
#
-
#
-
#
-
#
-
#
-
#
-
what i want is to be displayed it in 3 columns while the rows will depend on the number of images in my database
ex: total number of images is 7
ex: total number of images is 11 -
display:
-
# # #
-
# # #
-
# # #
-
# #
-
ex: total number of images is 6
please help me again..
and thanks for the reply..
thank you very much..
Hi, again!
Try using the div solution...
(the output code will be alot cleaner).
If you definately need tables i have another solution. but i'm not proud of that code... (very messy)
Div solution.... - <?php
-
$errmsg = "";
-
if (! @mysql_connect("localhost","root",""))
-
{
-
$errmsg = "Cannot connect to database";
-
}
-
@mysql_select_db("upload");
-
-
$strSQL = "select * from pix";
-
$rsPix = mysql_query($strSQL);
-
$numRows = mysql_numrows($rsPix);
-
$i = 0;
-
?>
-
<div>
-
<?php
-
while($i < $numRows){
-
?>
-
<div style="width:33%: float:left">
-
<img src="pix.php?pixID=<?php echo mysql_result($rsPix,$i,"pixID"); ?>"/>
-
</div>
-
<?php
-
$i++;
-
}
-
?>
-
</div>
Table solution..... - <?php
-
$errmsg = "";
-
if (! @mysql_connect("localhost","root",""))
-
{
-
$errmsg = "Cannot connect to database";
-
}
-
-
@mysql_select_db("upload");
-
-
$strSQL = "select * from pix";
-
$rsPix = mysql_query($strSQL);
-
$numRows = mysql_numrows($rsPix);
-
$i = 0;
-
?>
-
<table>
-
<tr>
-
<?php
-
while($i < $numRows){
-
?>
-
<td>
-
<img src="pix.php?pixID=<?php echo mysql_result($rsPix,$i,"pixID"); ?>"/>
-
</td>
-
<?php
-
if ($i%3 == 0){
-
?>
-
</tr><tr>
-
<?php
-
}
-
?>
-
<?php
-
$i++;
-
}
-
?>
-
<?php
-
if ($numRows%3 > 0){
-
?>
-
<td colspan="<?php echo $numRows%3; ?>"> </td>
-
<?php
-
}
-
?>
-
</tr>
-
</table>
Again i must say the code may need a spot of debugging.
Hi, again!
Try using the div solution...
(the output code will be alot cleaner).
If you definately need tables i have another solution. but i'm not proud of that code... (very messy)
Div solution.... - <?php
-
$errmsg = "";
-
if (! @mysql_connect("localhost","root",""))
-
{
-
$errmsg = "Cannot connect to database";
-
}
-
@mysql_select_db("upload");
-
-
$strSQL = "select * from pix";
-
$rsPix = mysql_query($strSQL);
-
$numRows = mysql_numrows($rsPix);
-
$i = 0;
-
?>
-
<div>
-
<?php
-
while($i < $numRows){
-
?>
-
<div style="width:33%: float:left">
-
<img src="pix.php?pixID=<?php echo mysql_result($rsPix,$i,"pixID"); ?>"/>
-
</div>
-
<?php
-
$i++;
-
}
-
?>
-
</div>
Table solution..... - <?php
-
$errmsg = "";
-
if (! @mysql_connect("localhost","root",""))
-
{
-
$errmsg = "Cannot connect to database";
-
}
-
-
@mysql_select_db("upload");
-
-
$strSQL = "select * from pix";
-
$rsPix = mysql_query($strSQL);
-
$numRows = mysql_numrows($rsPix);
-
$i = 0;
-
?>
-
<table>
-
<tr>
-
<?php
-
while($i < $numRows){
-
?>
-
<td>
-
<img src="pix.php?pixID=<?php echo mysql_result($rsPix,$i,"pixID"); ?>"/>
-
</td>
-
<?php
-
if ($i%3 == 0){
-
?>
-
</tr><tr>
-
<?php
-
}
-
?>
-
<?php
-
$i++;
-
}
-
?>
-
<?php
-
if ($numRows%3 > 0){
-
?>
-
<td colspan="<?php echo $numRows%3; ?>"> </td>
-
<?php
-
}
-
?>
-
</tr>
-
</table>
Again i must say the code may need a spot of debugging.
hi.. i've tried the codes you've given me.. the first one still displays the images like this:
the 2nd one is the one that i need but there's something wrong, the first row has only 1 image but the succeeding rows are fine.. it displays like this:
where's the problem with the code?
and where can i put this code
<td><center><?p hp echo mysql_result($r sPix,$i,"title" ); ?></center></td>
so that the image has a title under it?
thanks again for the reply..
Hi again mate,
First we'll sort out the table then we'll stick in the title....
i think i spoted the error with my code....
try this... - <?php
-
$errmsg = "";
-
if (! @mysql_connect("localhost","root",""))
-
{
-
$errmsg = "Cannot connect to database";
-
}
-
-
@mysql_select_db("upload");
-
-
$strSQL = "select * from pix";
-
$rsPix = mysql_query($strSQL);
-
$numRows = mysql_numrows($rsPix);
-
$i = 0;
-
?>
-
<table>
-
<tr>
-
<?php
-
while($i < $numRows){
-
?>
-
<td>
-
<img src="pix.php?pixID=<?php echo mysql_result($rsPix,$i,"pixID"); ?>"/>
-
</td>
-
<?php
-
$i++;
-
if ($i%3 == 0){
-
?>
-
</tr><tr>
-
<?php
-
}
-
?>
-
<?php
-
-
}
-
?>
-
<?php
-
if ($numRows%3 > 0){
-
?>
-
<td colspan="<?php echo $numRows%3; ?>"> </td>
-
<?php
-
}
-
?>
-
</tr>
-
</table>
i've moved the increment before the if statement.
Hi again mate,
First we'll sort out the table then we'll stick in the title....
i think i spoted the error with my code....
try this... - <?php
-
$errmsg = "";
-
if (! @mysql_connect("localhost","root",""))
-
{
-
$errmsg = "Cannot connect to database";
-
}
-
-
@mysql_select_db("upload");
-
-
$strSQL = "select * from pix";
-
$rsPix = mysql_query($strSQL);
-
$numRows = mysql_numrows($rsPix);
-
$i = 0;
-
?>
-
<table>
-
<tr>
-
<?php
-
while($i < $numRows){
-
?>
-
<td>
-
<img src="pix.php?pixID=<?php echo mysql_result($rsPix,$i,"pixID"); ?>"/>
-
</td>
-
<?php
-
$i++;
-
if ($i%3 == 0){
-
?>
-
</tr><tr>
-
<?php
-
}
-
?>
-
<?php
-
-
}
-
?>
-
<?php
-
if ($numRows%3 > 0){
-
?>
-
<td colspan="<?php echo $numRows%3; ?>"> </td>
-
<?php
-
}
-
?>
-
</tr>
-
</table>
i've moved the increment before the if statement.
hello again..
i've tried the code, and it works!!
yehey!!.. thanks again!!
so, where will i have to put the code for the title of the image?
i want it to be under the image
ex:
# # #
image1 image2 image3
thanks again for the reply!!
thanks..
Hi again,
I'm getting excited its almost done!!!
can you not line break and place it under the photo? like this... - <td>
-
<img src="pix.php?pixID=<?php echo mysql_result($rsPix,$i,"pixID"); ?>"/><br/>
-
<?php echo mysql_result($rsPix,$i,"pixTitle"); ?>"/>
-
</td>
If this will do excellent, and if not i do have another solution but thats gonna take a while to code out.
Hi again,
I'm getting excited its almost done!!!
can you not line break and place it under the photo? like this... - <td>
-
<img src="pix.php?pixID=<?php echo mysql_result($rsPix,$i,"pixID"); ?>"/><br/>
-
<?php echo mysql_result($rsPix,$i,"pixTitle"); ?>"/>
-
</td>
If this will do excellent, and if not i do have another solution but thats gonna take a while to code out.
hello again, hey thanks for the help..
really, really thanks..
but im still far from being done.. hehe..
i still have to make the images serve as a link to their features..
thank you very BIG!!
No problems mate, i like helping! (keeps me sane)
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