jQuery accessing form array <p>I have a php form and I want to display or hide certain fields dependant on what I select in a drop down. I have code below which works:</p> <pre><code>$(document).ready(function() { $('div.pupname').hide(); $('div.pupnid').hide(); if ($("#perpetrator").val == "1") { $('div.pupname').hide(); $('div.pupid').show(); else { $('div.pupid').hide(); $('div.pupname').show(); } $("#perpetrator").change(function() { if ($("#perpetrator").val() == "1") { $('div.pupname').hide(); $('div.pupid').show(); else { $('div.pupid').hide(); $('div.pupname').show(); } }); }); </code></pre> <p>I now want to have a multi row block which I am generating using php so I can generate the divs as <em>pupid1</em>, <em>pupid2</em> etc. and the field <code>$m_perpetrator</code> as an array. I've tried a few ways of accessing the array but can't get anything to work. I thought he following might be along the correct lines for accessing the elements of the array but it does nothing. </p> <pre><code>if ($("#m_perpetrator[1]").val == "1") { $('div.pupname1').show(); $('div.pupid1').hide(); } else { $('div.pupid1').show(); $('div.pupname1').hide(); } </code></pre> <p>Added</p> <p>Here is a snippet of the HTML</p> <pre><code>&lt;td&gt;&lt;select name="m_perpetrator[1]" id="m_perpetrator[1]"&gt; &lt;option value="1"&gt;Current&lt;/option&gt; &lt;option value="2" selected&gt;Former&lt;/option&gt; &lt;option value="3"&gt;Parent/Carer&lt;/option&gt; &lt;/select&gt; &lt;/td&gt; &lt;td&gt;Name&lt;/td&gt; &lt;td&gt; &lt;div class="pupname1"&gt; &lt;input type="text" size=60 name="m_pupil_name[1]" value="JOHN SMITH"&gt; &lt;/div&gt; &lt;div class="pupid1"&gt; &lt;select name="m_pupil_id[1]" onchange="getXtras(this)"&gt; &lt;option value=""&gt;Select .....&lt;/option&gt; </code></pre> <p>So basically this is a multi row form and on each row if Current is selected then I want to display pupil_id if anything else is selected then I want to display pupil name. For processing the input I run round the array m_perpetrator[] using php Jim</p>