Multiple Java Script In One Page Error May 29, 2023 Post a Comment I have this code that is working originally: Please Select Solution 1: As suggested in Comments, you made mistakes in your code; 1.Binding change function with wrong selector id $("#Name2").change(function(){ Copy and the HTML is <select name="Name" required class="form-control" id="Name"> Copy Should be $("#Name").change(function(){ it will fix the problem Undefined selname at line 5 2.Name Ajax Method success function success: function(data) { $("#Schedule2").html(dataname); } Copy Should be $("#Schedule2").html(data); And in PHP; credit goes to @j08691 change $selDate $selDate = $_POST['selname']; Copy to $selname $selname = $_POST['selname']; $sql="SELECT * FROM clinic.appoint WHERE name='$selname'"; Copy As you said in question, your first Ajax Call is working fine and you are facing problem in 2nd Ajax call method, after fixing above mistakes HTML <th> <select name="Name" required class="form-control" id="Name"> <option value="">Please Select Name</option> <?php $sql3="SELECT * FROM clinic.appoint GROUP BY name"; $result3 = mysqli_query($con, $sql3) or die($sql3."<br/><br/>".mysql_error()); while($rows3=mysqli_fetch_array($result3)){?> <option value="<?php echo $rows3['name'] ?>"><?php echo $rows3['name'] ?></option> <?php } ?> </select> </th> Copy AJAX $(document).ready(function(){ $("#Name").change(function(){ var selname =$(this).val(); display_name(selname); }); // This is the function... function display_name(selname) { $("#scheduleName").html(selname); var dataString = 'selname='+ selname; $.ajax({ type: "POST", url: "getdatabyname.php", data: dataString, cache: false, success: function(data) { $("#Schedule").html(data); } }); } }); Copy PHP <?php require_once ('../include/global.php'); if($_POST['selname']) { $selname = $_POST['selname']; $sql="SELECT * FROM clinic.appoint WHERE name='$selname'"; $result = mysqli_query($con, $sql) or die($sql."<br/><br/>".mysql_error()); while($rows=mysqli_fetch_array($result)){ ?> <tr> <td scope="row"><?php echo $rows['time'] ?></td> <td scope="row"><?php echo $rows['name'] ?></td> <td scope="row"><?php echo $rows['date'] ?></td> <td scope="row"><form action='/clinic form/appoint/delete.php'=<?php echo $rows['id']; ?>' method="post"> <input type="hidden" name="id" value="<?php echo $rows['id']; ?>"> <input type="submit" name="submit1" value="Done"> </form> </td> </tr> <?php } } ?> Copy (OP reached me via email and explained what he is trying to do) Now, you are fetching result by 2 different <select> element using Ajax and trying to show the result based on <select> element and in both Ajax calls success: function you are targeting 2 different id's to show the data for each <select> Ajax call e.g //For Date Result $("#Schedule").html(data); //For Name Result $("#Schedule2").html(data); Copy I would suggest to use the same id selector in both Ajax calls success: function to show the data $("#Schedule").html(data); Copy By doing so, when you switch between select element, to show the data, it will replace the first fetched data. 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