Image Button

[candice]

I have 2 image button which allow users to change the status in or out:

 

 

Meaning that when user click on the red icon, the status would automatically change to in and when click on the green icon, it will change to out.

 

I did some codings, but the value doesn't seem to be passing in.

<input type="image" src="images/icons/in.jpg" value="In" alt="submit"/>

 

can anyone help?

[ram4nd]

first you have to split up the image, but i don't get it what do you exactly want to do. I mean where do you want to use it and where do you want to remember the status.

[candice]

Thanks for replying. The button is actually split up,meaning its two different pictures. What i want is something like, when the user clicks on one button, the value of the button would then be captured and then the form would then be submitted, updating the status in the database.

 

it works something like creating a radio button, selecting one of the radio buttons and then pressing the update button to change the status, just that in this case, i do not want the update button, just clicking the image itself and then updating the status

[Lamez]

here is an idea.

 

create a new php page, set the images to links, and make the links like so: page.php?stat=in or out. Then create a if statement, state weather the get variable is in or out, after you stated it has been set, if the variable is in, create a query to update the database, same with out, redirect back to the previous page. Now create another if statement, ask weather in the database the user is in, echo out green in image. Same for out, I can code this for you, but it will have to wait till tomorrow.

[candice]

Thanks.. we will also try it.. but could u help us to code that part to make sure that we are on the right track?? Thanks a milion.

 

[Lamez]

I sure can, will only take about ten minuets, but nothing bad can come out of practice !

[PFMaBiSmAd]

To get the value from an image being used as a submit button, read this - http://us.php.net/manual/en/faq.html.php#faq.html.form-image

[Lamez]

<?php
ob_start();
$img_path = "img";

if(empty($_GET['s'])){

  //$q = mysql_query("SELECT * FROM `table_name`");
  //$r = mysql_fetch_array($q);
  $o_stat = $r['stat'];
   
  if($o_stat == (1)){
   echo '<a href="?s=in"><img src="'.$img_path.'/in_green.bmp"></a> <a href="?s=out"><img src="'.$img_path.'/out_red.bmp"></a>';
  }
  if($o_stat == (2)){
   echo '<a href="?s=in"><img src="'.$img_path.'/in_red.bmp"></a> <a href="?s=out"><img src="'.$img_path.'/out_green.bmp"></a>';
  }
  echo $o_stat;
}else{
  if($_GET['s'] == ("in")){
    //mysql_query("UPDATE `table_name` WHERE `stat` = '1'");
header("Location: ?");
  }
  if($_GET['s'] == ("out")){
    //mysql_query("UPDATE `table_name` WHERE `stat` = '2'");
header("Location: ?");
  }
}
?>

[candice]

Hi,

 

Thanks for the help!! I'll go and give it a try.