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davidd

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Posts posted by davidd

    Populate Dropdown on the basis of another dropdown value in PHP using JQuery Ajax

    in PHP Coding Help

    Posted

    10 minutes ago, ginerjm said:

    So what does your script get back from the ajax call? Anything?   If not, add some output to the ajax script and display it in the main script when it returns so you can debug the ajax script.

    If i Inspect element(first dropdown) in Google chrome, and go to network tab, i can see that when i choose some item in first dropdown it fires ajaxfile.php. But there is also an error: 

    Warning: sqlsrv_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp64\www\site\ajaxfile.php on line 16

    Which is this line: 

    Line 16 -->while( $row = sqlsrv_fetch_array($result) ){
                     $machinery_id = $row['ID'];
                     $machinery = $row['MachineID'];

                     $machinery_arr[] = array("ID" => $machinery_id, "MachineID" => $machinery);
                   }

    Populate Dropdown on the basis of another dropdown value in PHP using JQuery Ajax

    in PHP Coding Help

    Posted · Edited by davidd

    I have a form with dropdown list that is populated with values from Sql Server table. Now i would like to use this selected item in SQL query. The results of this query should be shown in label or text field. So when a user selects item from dropdown menu, results from SQL query are shown at the same time.

    I have two dropdown list at the moment in my form. First one gets all values from a column in table in SQL Server. And the second one should get a value from the same table based on a selection in first dropdown list.

    When i load ajax.php i get 2 error mesages: 
    Notice: Undefined index: machinery in C:\wamp64\www\gecko\ajaxfile.php on line  *8*
    Warning: sqlsrv_fetch_array() expects parameter 1 to be resource, boolean given in C:\wamp64\www\site\ajaxfile.php on line  *16*

    This is my code so far. I have tried to do it with this Ajax script. But i can only get first dropdown to work. The second dropdown(sub_machinery) does not show values, when first dropdown item is selected. The second dropdown should show values from databse table with this query( *$machineryID*  is first dropdown selected item): SELECT MachineID FROM T013 WHERE Machinery=".$machineryID.

    Index.php 

        <!doctype html>
      <?PHP
    $server = "server";
    $options = array(  "UID" => "user",  "PWD" => "pass",  "Database" => 
    "database");
    $conn2 = sqlsrv_connect($server, $options);
    if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
    echo " ";
    ?>
    <html>
    <head>
    <meta charset="utf-8">
    <title>Untitled Document</title>
    </head>
    
    <body>
                <section id="formaT2" class="formaT2 formContent">
                <div class="row">
                    <div class="col-md-2 col-3 row-color remove-mob"></div>
                    <div class="col-md-5 col-9 bg-img" style="padding-left: 0; 
    padding-right: 0;">
                        <h1>Form</h1>
                        <div class="rest-text">
                           <div class="contactFrm">
                              <p class="statusMsg <?php echo 
    !empty($msgClass)?$msgClass:''; ?>"><?php echo $statusMsg; ?></p>
                               
                                <form action="connection.php" method="post">
                                    <div>machinery</div>     
    <select id="machinery">
       <option value="0">--Please Select Machinery--</option>
       <?php 
       // Fetch Department
       $sql = "SELECT Machinery FROM T013";
       $machanery_data = sqlsrv_query($conn2,$sql);
       while($row = sqlsrv_fetch_array($machanery_data) ){
          $id = $row['Id'];
          $machinery = $row['Machinery'];
          
          // Option
          echo "<option value='".$id."' >".$machinery."</option>";
       }
       ?>
    </select>
    <div class="clear"></div>
                                    <div>Sub Machinery</div>
    <select id="sub_machinery">
       <option value="0">- Select -</option>
    </select>
                                    
                                    <input type="submit" name="submit" 
    id="submit" class="strelka-send" value="Insert">
                                    <div class="clear"> </div>
                                </form>
                             </div> 
                         </div>
                    </div>
                </div> 
            </section>
             </script>
          <script type="text/javascript">
          $(document).ready(function(){

        $("#machinery").change(function(){
            var machinery_id = $(this).val();
    
            $.ajax({
                url:'ajaxfile.php',
                type: 'post',
                data: {machinery:machinery_id},
                dataType: 'json',
                success:function(response){
    
                    var len = response.length;
     
                    $("#sub_machinery").empty();
                    for( var i = 0; i<len; i++){
                        var machinery_id = response[i]['machinery_id'];
                        var machinery = response[i]['machinery'];
                        
                        $("#sub_machinery").append("<option 
    value='"+machinery_id+"'>"+machinery+"</option>");
    
                    }
                }
            });
        });
    
    });
          </script>
    </body>
    </html>

    Ajaxfile.php 

    <?php
    $server = "server";
    $options = array(  "UID" => "user",  "PWD" => "pass",  
    "Database" => "database");
    $conn2 = sqlsrv_connect($server, $options);
    if ($conn2 === false) die("<pre>".print_r(sqlsrv_errors(), true));
    echo " ";

    $machineryID = $_POST['machinery'];   // department id

    $sql = "SELECT MachineID FROM T013 WHERE Machinery=".$machineryID;
    
    $result = sqlsrv_query($conn2,$sql);
    
    $machinery_arr = array();
    
    while( $row = sqlsrv_fetch_array($result) ){
        $machinery_id = $row['ID'];
        $machinery = $row['MachineID'];
    
        $machinery_arr[] = array("ID" => $machinery_id, "MachineID" => 
    $machinery);
    }
    
    // encoding array to json format
    echo json_encode($machinery_arr);
    ?>
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