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    copied!<p>Noob warning.</p> <p>This is an interesting question for me, because I'll most-likely need to do it over the next few months, so I thought I take an hour and give it a shot.</p> <p>Here is the html (boring):</p> <pre><code>&lt;table class="centerme"&gt; &lt;tr&gt; &lt;td&gt;First Checkbox:&lt;input type="checkbox" id="checkboxtest"&gt;&lt;/td&gt; &lt;/tr&gt; &lt;/table&gt; </code></pre> <p>Here is the javascript:</p> <pre><code>&lt;script&gt; $(document).ready(function(e) { $('#checkboxtest').change(function(){ if( $('#checkboxtest').prop('checked') ) {checkboxstatus = "YES";} else {checkboxstatus = "NO";} $.ajax({ type: "POST", url: "checkboxtestbackend.php", data: {checkboxstatus: checkboxstatus}, }) .done(function(data, textStatus, jqXHR){alert(textStatus);}) .fail(function(jqXHR, textStatus, errorThrown){alert(jqXHR+"--"+textStatus+"--"+errorThrown);}); });//end change });//end ready &lt;/script&gt; </code></pre> <p>Here is the backend php:</p> <pre><code>&lt;?php $checkboxstatus = $_POST['checkboxstatus']; $host = "xxxxxxxxxxxxxxxxx"; $user = "yyyyyyyyyyyyyyyyy"; $password = "zzzzzzzzzzzzzzzzzzzzzz"; $dbname = "qqqqqqqqqqqqqqqqqqqqqqqqqq"; $cxn = mysqli_connect($host,$user,$password,$dbname); if (mysqli_connect_errno()) {echo "No connection" . mysqli_connect_error();} $query = " UPDATE testtable SET lastname = '$checkboxstatus' WHERE localid = '158' "; $result = mysqli_query($cxn, $query) or die ("could not query database 1"); ?&gt; </code></pre> <p>When I uploaded it to my site, it seems to work (get an error message that I haven't figured out yet, but the db changes (checked it with MySQL Workbench).</p> <p>You could probably get away with just a $().POST, but I'm used to $ajax.</p> <p>It would be interesting to get some comments from the experts.</p> <p>PS - found the error - I should not have included the dataType in the ajax query.</p> <p>Works great now!</p> <p>Woohoo!</p>
 

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