How to display image if field value= 'value'

If I understand you question correctly, you just need a PHP If statement inside the cell. Assuming you have a recordset from your database (if not just change the variable as necessary) it would look something like this:

<table>

  <tr>

    <td>Invoice Paid:</td>

    <td>

<?php If ($yourRecodset['payment']==0) { ?>

    <img src="notpaid.jpg"/>

<?php } else { ?>

    <img src="paid.jpg"/>

<?php }?>

    </td>

  </tr>

</table>

(this assumes that the value can only be 0 or 1. If that is not the case, the Else should be replaced with another If)

Hope this helps

Scott

You have nested your 'if ($clientinfo['paymentstatus']==4) {' statement inside your 'if ($clientinfo['paymentstatus']==0) {' statement. This means that the condition 4 code will never be seen as that block is only run when the value is 0.

It should read:

<?php echo $row_clientinfo['paymentstatus']; ?>

<?php if ($clientinfo['paymentstatus']==0) { ?>

<img src="../../images/notpaid.png" width="100" height="50" />

<?php }?>

<?php if ($clientinfo['paymentstatus']==4) { ?>

<img src="../../images/paid.png" width="100" height="50" />

<?php }?>

Notice the first if statement is closed before the second is reached.

Scott

Is there something im missing?

 

Yes - you have improperly (or inadvertently) nested your conditional tests -

<?php echo $row_clientinfo['paymentstatus']; ?>

<?php if ($clientinfo['paymentstatus']==0) { ?>

<img src="../../images/notpaid.png" width="100" height="50" />

<?php } ?>

<?php if ($clientinfo['paymentstatus']==4) { ?>

<img src="../../images/paid.png" width="100" height="50" />

<?php }?>

The way you had it written, the only time you would test to see if the value is 4 would be if the value==0 test SUCCEEDED.

Max Resnikoff wrote:

 

It is still not working even with your code :/

 

The first set for the not paid code is working. If i change the 0 to 1, it dissapears.

The second set doesnt want to work even if i change the equal to:1

 

Try:

<?php echo $row_clientinfo['paymentstatus']; ?>

<?php if ($clientinfo['paymentstatus']==0) { ?>

<img src="../../images/notpaid.png" width="100" height="50" />

<?php } ?>

<?php if ($clientinfo['paymentstatus']==4) { ?>

<img src="../../images/paid.png" width="100" height="50" />

<?php } ?>

Should work if you database is set to 0 and 4 respectively/

As the code in your last post stands, the only way the paid image will display is if the value is set to 4. If you want it to work with 0 and 1, where 1 is paid, you need to change

<?php if ($clientinfo['paymentstatus']==4) { ?>

to

<?php if ($clientinfo['paymentstatus']==1) { ?>

Os, I don't think we are getting through...

Infact you probably only need to test for NOT paid and just print paid if it doesnt meet the condition:

<?php echo $row_clientinfo['paymentstatus']; ?>

<?php if ($clientinfo['paymentstatus']==0) { ?>

<img src="../../images/notpaid.png" width="100" height="50" />

<?php } 

else { ?>

<img src="../../images/paid.png" width="100" height="50" />

<?php } ?>

MurraySummers wrote:

 

Os, I don't think we are getting through...

I hear ya Murray, how you doing? Not seen you around in a while - been busy?

Can you guys (MurraySummers and osgood_) even see my replies? You seem to be repeating my answers. Your last is essentially the same as my original answer to the OP and the others are verbatim code to my second reply (apart from whitespace). If wires are crossing during posting then fine but you guys just seem to not even see/acknowledge my responses?

Sorry I omitted you. Indeed your initial answer is correct.

Yes - comes in waves, you know?

The variable is not the issue. The top one contains:

<?php if ($clientinfo['paymentstatus']==4) { ?>

whereas the bottom one contains:

<?php if ($clientinfo['paymentstatus']==1) { ?>

The reason it doesn't work is because the top version is looking for a value of  4 which does not exist!

Max Resnikoff wrote:

 

I found the issue:

 

The variable was the problem.

 

Instead of this:

<?php if ($clientinfo['paymentstatus']==0) { ?>

<img src="../../images/notpaid.png" width="100" height="50" />

<?php } ?>

<?php if ($clientinfo['paymentstatus']==4) { ?>

<img src="../../images/paid.png" width="100" height="50" />

<?php } ?>

 

I put this:

 

<?php $paymentstatus = $row_clientinfo['paymentstatus']; ?>

 

<?php if ($paymentstatus ==0) { ?>

<img src="../../images/notpaid.png" width="100" height="50" />

<?php } ?>

<?php if ($paymentstatus ==1) { ?>

<img src="../../images/paid.png" width="100" height="50" />

<?php } ?>

Do you see the problem?

Max Resnikoff wrote:

 

I found the issue:

 

The variable was the problem.

 

Instead of this:

<?php if ($clientinfo['paymentstatus']==0) { ?>

<img src="../../images/notpaid.png" width="100" height="50" />

<?php } ?>

<?php if ($clientinfo['paymentstatus']==4) { ?>

<img src="../../images/paid.png" width="100" height="50" />

<?php } ?>

 

I put this:

 

<?php $paymentstatus = $row_clientinfo['paymentstatus']; ?>

 

<?php if ($paymentstatus ==0) { ?>

<img src="../../images/notpaid.png" width="100" height="50" />

<?php } ?>

<?php if ($paymentstatus ==1) { ?>

<img src="../../images/paid.png" width="100" height="50" />

<?php } ?>

The reason it doesn't work is because the top version is looking for a value of  4 which does not exist!

Sorry TwoSuits - just winding you up - no offense meant - wasn't ignoring your replies, just thought maybe the OP wasn't understanding so just really reinforcing what you wrote.

MurraySummers wrote:

 

Yes - comes in waves, you know?

Yeah, same here. All or nothing.

You do realise that the extra variable ($paymentstatus) is not required and that the below code will work exactly the same?

<?php if ($row_clientinfo['paymentstatus'] ==0) { ?>

<img src="../../images/notpaid.png" width="100" height="50" />

<?php } ?>

<?php if ($row_clientinfo['paymentstatus'] ==1) { ?>

<img src="../../images/paid.png" width="100" height="50" />

<?php } ?>

Oh, and by the way - YOU'RE WELCOME!