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POhow can i get my script/jquery-ajax to continue responding to clicks?
 

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  1. POhow can i get my script/jquery-ajax to continue responding to clicks?
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    <p>Still a programming and markup newbie...</p> <p>I have several little in/out button that reads and writes back from a mysql table to track whether different users are available or unavailable. But I can only click the button once to change the status. After that, it stops responding. How can I get it to just keep working? Thanks!!!!</p> <p>list.php</p> <pre><code>&lt;html&gt; &lt;head&gt; &lt;script src="http://code.jquery.com/jquery-latest.js"&gt;&lt;/script&gt; &lt;script&gt; $(function(){ $(".rating").on("click", function(){ var status = $(this).attr("id").substr(0,1); var id = $(this).attr("id").substr(1); var data = "id="+id+"&amp;status="+status; $.ajax({ type: "POST", url: "rate.php", data: data, success: function(e){ $("#r"+id).html(e); } }) }); }); &lt;/script&gt; &lt;style&gt;.rating { cursor: pointer; }&lt;/style&gt; &lt;/head&gt; &lt;body&gt; &lt;?php include ("headers.php"); $qq = mysql_query("SELECT * FROM $io"); while($rr=mysql_fetch_array($qq)){ $id = $rr["id"]; $content = $rr["status"]; include("buttons.php"); $list .= '&lt;div style="border-bottom: 1px #32baed solid"&gt;'.$q[0].' &lt;div id="r'.$id.'"&gt;&lt;img class="rating" id="'.$q[0].$id.'" src="'.$color.'"&gt; '.$status.'&lt;/div&gt;&lt;/div&gt;&lt;br&gt;&lt;br&gt;'; } echo $list; ?&gt; &lt;/body&gt; &lt;/html&gt; </code></pre> <p>rate.php</p> <pre><code>&lt;?php include ("headers.php"); $id = $_POST["id"]; $status = $_POST["status"]; if($status == 0){ mysql_query("UPDATE $io SET status = 1 WHERE id='$id'"); } else { mysql_query("UPDATE $io SET status = 0 WHERE id='$id'"); } include("buttons.php"); $list = '&lt;img class="rating" id="'.$q[0].$id.'" src="'.$color.'"&gt; '.$status; echo $list; ?&gt; </code></pre> <p>buttons.php</p> <pre><code>&lt;?php $q = mysql_query("SELECT status FROM $io WHERE id='$id'"); $q = mysql_fetch_array($q); if($q[0]){ $color = "green.png"; } else{ $color = "red.png"; } ?&gt; </code></pre> <p>headers.php</p> <pre><code>&lt;?php $c = mysql_connect("localhost", "username", "password"); $db = mysql_select_db("likes", $c); $io = 'io'; ?&gt; </code></pre>
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    1. COIs this all the code you have? There is nothing changing the ID on the button so everytime you click the button it is using a static ID and so will not result in an alternating value. You need to update the button ID each time it is clicked so that the data will alternate.
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      1. USMike Oram
    2. COwow, i feel silly. thanks for the fast and accurate responses!
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