php - Show different image depending on value in variable -

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just want make sure i'm going in right direction this. have image want replaced/changed if value of variable either 0/1. here code guy doing server side stuff.

<?php //requires mysql_connect create connection  $link_state = 0;  //if wish don't have check connection, may idea leave in. if ($mysql_connection['connected'] == true) {     $result = mysql_query("select * link");     //the bit looking should first row, , should 1 row     $count  = mysql_num_rows($result);     if ($count <= 0) {         //interesting...         $mysql_error['error']       = true;         $mysql_error['description'] = "error: no rows returned table 'link'";     } else {         //we should ok continue         if ($count > 1) {             $mysql_error['error']       = true;             $mysql_error['description'] = "warning: found more 1 row in 'link' table!";         }         $row        = mysql_fetch_array($result);         $link_state = intval($row['state']);     } } else {     $mysql_error['error']       = true;     $mysql_error['description'] = "error: no mysql connection!"; }  /* after completion of page, $link_state 1 of 2 things:  * 0 = offline * 1 = online  throws $mysql_error:  1 warning 2 errors  */ ?>

okay, i'm assuming little bit of code have value of either 0 or 1 in $link_state.

so can simple inline script relevant image? 

<img src="img/<?=($link_state=="0"?"off.jpg":($link_state=="1"?"on.jpg":))?>" />

any insight great :)

thanks in advance.

try this

<?php $img = ($link_state == "0") ? "off.jpg" : "on.jpg"; ?>  <img src="./img/<?php echo $img; ?>" />

also use mysqli_* since mysql_* depreciated.


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