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POChanging check box automatically changes the database record
 

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  1. POChanging check box automatically changes the database record
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    <p>in my database I have the filed status which is boolean type, </p> <p>thank I have a connection to the database </p> <pre><code> if (!$result = mysqli_query($con,"SELECT * FROM cursos")) { die("Error: " . mysqli_error($con)); } </code></pre> <p>and displaying the status of this field with checkboxes,</p> <pre><code> &lt;?php if($row['status'] == 1) { echo "&lt;td&gt;"."&lt;input type = 'checkbox' checked='checked' name ='complete' value= '1'/&gt;"."&lt;/td&gt;"; }else { echo "&lt;td&gt;"."&lt;input type = 'checkbox' name ='incomplete' value= '0'/&gt;"."&lt;/td&gt;"; } ?&gt; </code></pre> <p>So what I need to do is, once I check or uncheck any of thoose check boxes, the database field to be canged automatically withoud the need of submit button. </p> <p><strong>Okay I did edited my code to what you suggested and it s working now, here is the new code:</strong></p> <p>Ajax Code</p> <pre><code>&lt;script&gt; $(document).ready(function(e) { $('.checkboxtest').change(function(){ if( $('.checkboxtest').prop('checked') ) {checkboxstatus = '1';} else {checkboxstatus = '0';} $.ajax({ type: "POST", url: "checkboxtestbackend.php", data: {checkboxstatus: checkboxstatus}, }) .done(function(data, textStatus, jqXHR){alert(textStatus);}) .fail(function(jqXHR, textStatus, errorThrown){alert(jqXHR+"--"+textStatus+"--"+errorThrown);}); });//end change });//end ready &lt;/script&gt; </code></pre> <p>the html code is: </p> <pre><code>&lt;?php if($row['status'] == 1) { echo "&lt;td&gt;"."&lt;input type = 'checkbox' class='checkboxtest' checked='checked' name ='complete' value= '1'/&gt;"."&lt;/td&gt;"; }else { echo "&lt;td&gt;"."&lt;input type = 'checkbox' class='checkboxtest' name ='incomplete' value= '0'/&gt;"."&lt;/td&gt;"; } ?&gt; </code></pre> <p>and the php backend code is: </p> <pre><code>&lt;?php $checkboxstatus = $_POST['checkboxstatus']; $host = "localhost"; $user = "username"; $password = "password"; $dbname = "name"; $cxn = mysqli_connect($host,$user,$password,$dbname); if (mysqli_connect_errno()) {echo "No connection" . mysqli_connect_error();} $query = " UPDATE cursos SET status = '$checkboxstatus' WHERE id = '12'"; $result = mysqli_query($cxn, $query) or die ("could not query database 1"); ?&gt; </code></pre>
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    1. USJohn Siniger
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    1. COIf you don't want the page to refresh the only way to do this is with AJAX, if you don't mind the page refreshing then you can use javascript to auto-submit the form when the checkbox state is changed.
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      1. POChanging check box automatically changes the database record
      1. USBad Wolf
 

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