surajkandukuri Posted October 23, 2009 Share Posted October 23, 2009 Hi All, What I am trying to do is 2 dependent drop down boxes and when user selects submit button the values are to be passed to the database to run a insert query. Right now, I am stuck with the second drop down box because it does not show up the values based on the selection in the first drop down box. Important point here is " My second drop down box displays columns from a single row(returned from second query)" Attached is my code please do suggest me the required changes to make and also I pasted the code here, <?php include("../include/dbcommon.php"); if(isset($_GET["country"]) && is_numeric($_GET["country"])) { $country = $_GET["country"]; $emp_number=$country; echo($emp_number); } if(isset($_GET["state"]) && is_numeric($_GET["state"])) { $state = $_GET["state"]; $emp_positionheld=$state; echo ($emp_positionheld); } ?> <script language="JavaScript"> function autoSubmit() { var formObject = document.forms['theForm']; formObject.submit(); } </script> <form name="theForm" method="get"> <select name="country" onChange="autoSubmit();"> <option value="null"></option> <?php //POPULATE DROP DOWN MENU WITH Employee Names $sql = "SELECT EmployeeNumber,LastName FROM employees"; $countries = mysql_query($sql,$conn); while($row = mysql_fetch_array($countries)) { echo ("<option value=\"$row[EmployeeNumber]\" " . ($country == $row["EmployeeNumber"] ? " selected" : "") . ">$row[LastName]</option>"); } ?> </select> <br><br> <?php if($country != null && is_numeric($country) ) { ?> <select name="state" onChange="autoSubmit();"> <option value="null"></option> <?php //POPULATE DROP DOWN MENU WITH Job Position helds For a Given Employee $sql = "SELECT * FROM employee_positionheld WHERE EmployeeNumber = $country "; $states = mysql_query($sql); $row = mysql_fetch_array($states); for($k=1;$k<=6;$k++) { echo("inside for"); $temp[$k]='Position held' . ' '.$k; $queryvar=$temp[$k]; //echo($queryvar); echo ("<option value=\"$row[$queryvar]\" " . ($state == $row[$queryvar] ? " selected" : "") . ">$row[$queryvar]</option>"); } ?> </select> <?php } ?> </form> MY EMPLOYEE_POSITION HELD TABLE looks like this EmployeeNumber Employee Name Position held 1 Position held 2 Position held 3 Position held 4 Position held 5 Position held 6 34550 Suraj Entryleveltech1 Seniortech1 programmer1 0 0 0 0 [attachment deleted by admin] Link to comment https://forums.phpfreaks.com/topic/178749-unable-to-display-contents-in-second-drop-down-box/ Share on other sites More sharing options...
Viola Posted October 26, 2009 Share Posted October 26, 2009 Does the second drop-down box not show at all, or does it not show the options correctly? What does it show? Link to comment https://forums.phpfreaks.com/topic/178749-unable-to-display-contents-in-second-drop-down-box/#findComment-945004 Share on other sites More sharing options...
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