如何使用php和javascript显示特定结果的正确图像？

Question

i am doing one website project using php.my result are displayed like this

Detected    Result
1.  CLEAN MX       0       clean site
2.  MalwarePatrol  0       unrated site
3.  ZDB Zeus       0       suspicious site
4.  K7AntiVirus    0       clean site

i am using this php code to get this result that is

  $none = 0; $i = 0; foreach($result->scans as $key => $val) { if($i==0) { echo '<th></th>'; echo '<th></th>'; echo '<th>Detected</th>'; echo '<th>Result</th>'; } echo '<tr>'; echo '<td>'.intval($i+1).'.</td>'; echo '<td>'.$key.'</td>'; if(empty($val->detected)): echo '<td>'. $none .'</td>'; else: echo '<td>'. $val->detected .'</td>'; endif; echo '<td>'.$val->result.'</td>'; echo '</tr>'; $i++; } 

but i need to add some graphics into the result page. firstly check the result is clean site or unrated site,suspicious site.then if that site is a clean site display green light image, if that site is unrated site display yellow light,if it is suspicious site display red light image. like a www.onlinelinkscan.com 's result.

finaly if that site gets most number of green light images display overall result is good,if that site gets most number of red light images display overall result is danger else display overall result is neutral.like this

  Detected Result 1. CLEAN MX 0 clean site 2. MalwarePatrol 0 clean site 3. ZDB Zeus 0 clean site overall result:good overall result:danger overall result:neutral 

please help me friends i do not have such kind of knowledge in php and javascript.

Answer 1

No fancy php or javascript needed. For styling something it is css you need. Definitly not add img as they are content, and an icon to anacify your table would not be considered content.

I would just add a class to each row to indicate the result. Also I would keep a score to indicate the global result:

outside your foreach, prepare the score var

$score = 0;

inside the foreach:

// determine which class to add
switch $val->result {
  case 'clean site': 
    $class = "clean";
    $score++;
    break;
  case "unrated site":
    $class = 'neutral';
    break;
  case "suspicious site":
    $class = 'dirty';
    $score--;
    break;
  default:
    // perhaps you should throw an esception here
    $class = '';
    break;
} 
// add it to your row
$out .= '<tr class="' . $class .'">';

Also note that I do not echo anything yet, I store it in a variable. You would have to do that with each echo in your code. An d make sure you do not add the opening <table> tag yet, as we are going to add the 'global score class' here.

After the foreach loop has finished, you would have a global score. You could add this as a class to your table, and prepend it to your prepared output like so:

if ($score < 0) {
   $tableClass = 'dirty';
}
if ($score > 0) {
   $tableClass = 'clean';
}
if ($score == 0) {
   $tableClass = 'neutral';
}
$out = '<table class="' . $tableClass .'">' . $out;

All you need to do now is echo the $out variable.

To apply the colors or icons or whatever you could add some simple css like this:

table.neutral {
   border-color: grey;
}
table.clean {
   border-color: green;
}
...
tr.clean td:first-child{
   background-image: url(icon-clean.png) no-repeat left center;
   padding-left: 20px;
}
...

Answer 2

This is simple... First check your value, and assign an image to variable...

 if($val->result=="clean site"){$image = "green-light.png";}
 elseif($val->result=="unrated site"){$image = "orange-light.png";}
 elseif($val->result=="suspicious site"){$image = "red-light.png";}

Then echo that image inside the table.

 echo "<td><img src='images/".$image."'></td>";

This means you have to create 3 images with the names as above.

And put them in your images folder

As for the other part of your question....it will involve counting the elements either with PHP, or better with SQL....but Im not gonna do that part for you.

Heres fuller code...

if(empty($val->detected)): 
     echo '<td>'. $none .'</td>';
  else: 
     echo '<td>'. $val->detected .'</td>';
  endif; 

  if($val->result=="clean site"){$image = "green-light.png";}
 elseif($val->result=="unrated site"){$image = "orange-light.png";}
 elseif($val->result=="suspicious site"){$image = "red-light.png";}

   echo "<td><img src='images/".$image."'></td>";