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jamrop
Forum Commoner
Posts: 80 Joined: Fri May 16, 2003 5:38 pm
Post
by jamrop » Wed Jul 02, 2003 5:09 am
Hey i am trying to show a picture that is the name of a id key for a car. The thing i am trying to work out, is that is there is not a picture, how what i put a message saying 'awaiting picture'
code
Code: Select all
<?php
<?php $sql = 'SELECT * FROM cars ' ;
$rs=mysql_query($sql,$db);
while ($r = mysql_fetch_array($rs))
{
$carid=$r["carid"];
$make=$r["make"];
$engine=$r["engine"];
$year=$r["year"];
$des=$r["des"];
$mot=$r["mot"];
$price=$r["engine"];
?>
<tr>
<td bgcolor="#99CCFF"> <strong><font size="-1" face="Verdana, Arial, Helvetica, sans-serif">
<?php echo $make ?></strong></td>
</tr>
<tr>
<td><img src = "uploaded/<?php echo $carid?>.jpg"></td>
?>
Many thanks
[]InTeR[]
Forum Regular
Posts: 416 Joined: Thu Apr 24, 2003 6:51 am
Location: The Netherlands
Post
by []InTeR[] » Wed Jul 02, 2003 5:43 am
Code: Select all
if(!file_exists("uploaded/".$carid.".jpg")){
echo "awaiting picture";
}
jamrop
Forum Commoner
Posts: 80 Joined: Fri May 16, 2003 5:38 pm
Post
by jamrop » Wed Jul 02, 2003 6:07 am
hey thanks
i have been trying this so the x picture does not show, but it does not seem to be working
Code: Select all
<?php
if(!file_exists("uploaded/".$carid.".jpg")){?>
<img src = "uploaded/<?php echo $carid; ?>.jpg">
<?php } else {
echo "awaiting picture";
}
?>
any ideas?
many thanks
[]InTeR[]
Forum Regular
Posts: 416 Joined: Thu Apr 24, 2003 6:51 am
Location: The Netherlands
Post
by []InTeR[] » Wed Jul 02, 2003 6:09 am
Code: Select all
<?php
if(!file_exists("uploaded/".$carid.".jpg")){
echo "awaiting picture";
} else {
?>
<img src = "uploaded/<?php echo $carid; ?>.jpg">
<?
}
?>
The ! before the file_exist meens, not..
So the statement is 'if file not exists do say `awaiting picture` else show image'
jamrop
Forum Commoner
Posts: 80 Joined: Fri May 16, 2003 5:38 pm
Post
by jamrop » Wed Jul 02, 2003 6:14 am
ah i c
thanks