show image when a field value = 1

I am trying to get a script to show a particular image when a field
value = 1, otherwise it will show something else, and this is what I
came up with:

$i=0;
while ($i < $num) {

$name=mysql_result($result,$i,"name");
$date=mysql_result($result,$i,"date");
$status=mysql_result($result,$i,"status");

echo "<b>$name</b>
<br><b>$date</b><br> ";

if (status == 1) {
echo <img src='../images/yes.jpg'>
} else {
echo <img src='../images/no.jpg'>
}

;

$i++;
}


and nothing shows up, if I remove the if clause then I get the text
showing name and date. What am I doing wrong, aside from not really
knowing how to program php?

Thanks

On May 9, 10:28*am, "canaj...@gmail.com" <canaj...@gmail.com> wrote:
> I am trying to get a script to show a particular image when a field
> value = 1, otherwise it will show something else, and this is what I
> came up with:
>
> $i=0;
> while ($i < $num) {
>
> $name=mysql_result($result,$i,"name");
> $date=mysql_result($result,$i,"date");
> $status=mysql_result($result,$i,"status");
>
> echo "<b>$name</b>
> <br><b>$date</b><br> ";
>
> if (status == 1) {
> * * * * * * * * * echo <img src='../images/yes.jpg'>
> * * * * * * * * *} else {
> * * * * * * * * * echo <img src='../images/no.jpg'>
> * * * * * * * * *}
>
> ;
>
> $i++;
>
> }
>
> and nothing shows up, if I remove the if clause then I get the text
> showing name and date. *What am I doing wrong, aside from not really
> knowing how to program php?
>
> Thanks

if ( status == 1 )

is not the same as

if ( $status == 1 )

- missing $ on status
- missing double quotes on echo lines and semicolon at the end
- you have an extra semicolon there as well

you can download a manual here
http://www.php.net/download-docs.php
and give it a try

On 9 May, 14:52, noname <n...@email.com> wrote:
> - missing $ on status
> - missing double quotes on echo lines and semicolon at the end
> - you have an extra semicolon there as well
>
> you can download a manual herehttp://www.php.net/download-docs.php
> and give it a try

And
echo <img src='../images/yes.jpg'>
is not the same as
echo "<img src='../images/yes.jpg'>"

Thanks guys, the missing dollar sign was just an error on my part, it
slipped through. As for the semi colon thing I thought it was enough
to place it at the end of the if statement, but now I know better, it
has to go at the end of each echo within the if statement

if ($status == 1) {
echo "<img src=\"../images/yes.jpg\">";
} else {
echo "<img src=\"../images/no.jpg\">";
}

and now it works!!!

tommybiegs@gmail.com wrote:
> On May 9, 10:28 am, "canaj...@gmail.com" <canaj...@gmail.com> wrote:
>> I am trying to get a script to show a particular image when a field
>> value = 1, otherwise it will show something else, and this is what I
>> came up with:
>>
>> $i=0;
>> while ($i < $num) {
>>
>> $name=mysql_result($result,$i,"name");
>> $date=mysql_result($result,$i,"date");
>> $status=mysql_result($result,$i,"status");
>>
>> echo "<b>$name</b>
>> <br><b>$date</b><br> ";
>>
>> if (status == 1) {
>> echo <img src='../images/yes.jpg'>
>> } else {
>> echo <img src='../images/no.jpg'>
>> }
>>
>> ;
>>
>> $i++;
>>
>> }
>>
>> and nothing shows up, if I remove the if clause then I get the text
>> showing name and date. What am I doing wrong, aside from not really
>> knowing how to program php?
>>
>> Thanks
>
> if ( status == 1 )
>
> is not the same as
>
> if ( $status == 1 )

Indeed. At the OP: set error_reporting(E_ALL | E_STRICT) while
developing, and enable display_errors. It would probably have told you
there is no constant 'status' defined.

--
Rik Wasmus
[SPAM]
Now looking for some smaller projects to work on to fund a bigger one
with delayed pay. If interested, mail rik at rwasmus.nl
[/SPAM]