bigmark Posted March 14, 2008 Share Posted March 14, 2008 I have a script (below) that consists of a form using javascript that shows an array in a table below it. What i would like to know is can i turn this into a form with the same data but include a checkbox next to each field, so i can $POST the selections to another table in the DB. //////FORM <div class="TabbedPanelsContent"> <form> Select a Round: <select name="round" onchange="showRound(this.value)"> <option value="0">0</option> <option value="1">1</option> <option value="2">2</option> <option value="3">3</option> <option value="4">4</option> </select> </form> <p> <div id="txtHint"></div> </p> </div> /////javascript var xmlHttp function showRound(str) { xmlHttp=GetXmlHttpObject() if (xmlHttp==null) { alert ("Browser does not support HTTP Request") return } var url="select_round/getround.php" url=url+"?q="+str url=url+"&sid="+Math.random() xmlHttp.onreadystatechange=stateChanged xmlHttp.open("GET",url,true) xmlHttp.send(null) } function stateChanged() { if (xmlHttp.readyState==4 || xmlHttp.readyState=="complete") { document.getElementById("txtHint").innerHTML=xmlHttp.responseText } } function GetXmlHttpObject() { var xmlHttp=null; try { // Firefox, Opera 8.0+, Safari xmlHttp=new XMLHttpRequest(); } catch (e) { //Internet Explorer try { xmlHttp=new ActiveXObject("Msxml2.XMLHTTP"); } catch (e) { xmlHttp=new ActiveXObject("Microsoft.XMLHTTP"); } } return xmlHttp; } //// PHP <?php $q=$_GET["q"]; $con = mysql_connect('localhost', 'username', 'pass'); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("database", $con); $sql="SELECT * FROM fixtures WHERE round = '".$q."'"; $result = mysql_query($sql); echo "<table border='0'> <tr> <th>Game</th> <th align='center'>Home</th> <th align='center'>Away</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<form>"; echo "<td width='25' align='center'>" . $row['game'] . "</td>"; echo "<td width='150'>" . $row['home'] . "</td>"; echo "<td width='150'>" . $row['away'] . "</td>"; echo "</form>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> Link to comment https://forums.phpfreaks.com/topic/96089-how-to-turn-this-array-into-a-form/ Share on other sites More sharing options...
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