创建一个下拉列表链，该列表基于前一个输入，形成php中的数据库

Question

i am using database to populate the dorpdown lis. 我正在使用数据库来填充dorpdown lis。 now when i select the option from the first dynamic populated drom down list i want to use it to populate another dropdown list and based on that selection third dromdown list. 现在当我从第一个动态填充的drom下拉列表中选择选项时，我想用它来填充另一个下拉列表并基于该选择第三个dromdown列表。

Here's the code i am working on, 这是我正在处理的代码，

<html>
<?php
    include './config.php';
?>
<head>
    <meta charset="UTF-8">
    <title></title>
    <SCRIPT>
        $(".dropdown").hide();
        $("#testcasedata").on("change", function() {
            $(".dropdown").hide();
            var value = $("#testcasedata").val();
            $("#dropdown" + value).show();
        });
    </script>
</head>
<body>
    <form method="POST" action="createTestCase.php">
         // This is the first dropdown list 
         // It will return testsuite_id in dropdownlist
        <?php
            $sqll="SELECT testsuite_id FROM assigned_testsuite_tester Where Tester_name = 'Pritesh'";
            echo "<select class='form-control' name='testsuite'>";
            echo "<option value=''>Select One</option>";  
            foreach ($conn->query($sqll) as $row){
                echo '<option value="'.$row['testsuite_id'].'">'.$row['testsuite_id'].'</option>';
            }
            echo "</select>";
        ?>
        <input type="submit" name="submit" />
    </form>
    <?php 
    if ($_SERVER["REQUEST_METHOD"] == "POST")  
    {
        //testsuite_id will be the input variable for the second dropdownlist
        // the second dropdown list will populate with testcase_id. 
        $testsuite = filter_input(INPUT_POST, 'testsuite');
        echo $testsuite;
        if($testsuite != ''){
            $sqll="SELECT Testcase_id FROM assigned_testsuite_testcase Where testsuite_id = '$testsuite'";
            echo "<select  name='testcasedata' id='testcasedata' >";
            echo "<option value=''>Select One</option>";  
            foreach ($conn->query($sqll) as $row){
                echo '<option value="'.$row['Testcase_id'].'">'.$row['Testcase_id'].'</option>';
            }
            echo "</select>";
        }

        // with use of test case id i am retrieving the Product_id
        $testcase = filter_input(INPUT_POST, 'testcase');
            $sqll="SELECT Product_id FROM Testcase_master Where Testcase_id = '$testcase'";
            $productid='';
            foreach ($conn->query($sqll) as $row){
                $productid = $row['Product_id'];
            } 


            //now product id must be the input for the third dropdownlist.
            if($productid != ''){
                $sqll="SELECT circle.circle_id,circle.Circle_name FROM circle INNER JOIN assigned_circle_product ON assigned_circle_product.`Product_id` = '$productid'";
                echo "<select class='dropdown' name='circledata' id='circledata'>";
                echo "<option value=''>Select One</option>";  
                foreach ($conn->query($sqll) as $row){
                    echo '<option value="'.$row['Circle_id'].'">'.$row['Circle_name'].'</option>';
                }
                echo "</select>";
            }
    }
    ?>
</body>

i was able to populate the first drop-down list and based on that populate the second drop-down list.but having problem for populating third one. 我能够填充第一个下拉列表，并基于填充第二个下拉列表。但是填充第三个下载列表有问题。 I was browsing through similar questions and find out that i need to use jquery/JavaScript to do that.I don't know that much about jquery but have inserted code for that too but still having problem to populating the third drop down list. 我正在浏览类似的问题，并发现我需要使用jquery / JavaScript来做到这一点。我对jquery知之甚少但是也为此插入了代码，但仍然有问题填充第三个下拉列表。 I am stuck at this point.Please give me some guidance. 我被困在这一点上。请给我一些指导。

Answer 1

I have been learning Javascript/Jquery and this is what i have done. 我一直在学习Javascript / Jquery，这就是我所做的。 I don't know if it's a correct way or not. 我不知道这是不是正确的方式。 please look and tell me if i can do more to improve code. 请看，告诉我是否可以做更多改进代码。

First i have devided whole program into three seprate PHP files. 首先，我将整个程序分成三个单独的PHP文件。

1.Main PHP File 1.主PHP文件

<!DOCTYPE html>
<html>
<?php
    include 'header.php';
    include 'footer.php';
    include './config.php';
    ?>
    <head>
    <meta charset="UTF-8">
    <title></title>
    <script type="text/javascript">
    function showCircle(str) {
        if (str=="") {
            document.getElementById("txtHint").innerHTML="";
            return;
        }
        if (window.XMLHttpRequest) {
        // code for IE7+, Firefox, Chrome, Opera, Safari
            xmlhttp=new XMLHttpRequest();
        } else { // code for IE6, IE5
        xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
        }
        xmlhttp.onreadystatechange=function() {
            if (xmlhttp.readyState==4 && xmlhttp.status==200) {
                document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
            }
        }
        xmlhttp.open("GET","GetProduct.php?q="+str,true);
        xmlhttp.send();
    }
    function product1(str) {
        str1 = document.getElementById("project");
        if (str=="") {
            document.getElementById("txtHint").innerHTML="";
            return;
        }
        if (window.XMLHttpRequest) {
        // code for IE7+, Firefox, Chrome, Opera, Safari
            xmlhttp=new XMLHttpRequest();
        } else { // code for IE6, IE5
        xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
        }
        xmlhttp.onreadystatechange=function() {
            if (xmlhttp.readyState==4 && xmlhttp.status==200) {
                document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
            }
        }
        xmlhttp.open("GET","GetTestcase.php?prod="+str,true);
        xmlhttp.send();
    }
    </script>
  </head>
  <body>
    <div id="page-wrapper">
        <div class="row">
            <div class="col-lg-12">
                <h1 class="page-header">Add Circle to Product</h1>
            </div>
            <!-- /.col-lg-12 -->
        </div>
        <div class="row">
            <div class="col-lg-12">
                <div class="panel panel-default">
                    <div class="panel-heading">
                        Select the circles where you want your product to test:
                    </div>
                    <div class="panel-body">
                        <div class="dataTable_wrapper">
                            <form action="circlrproduct.php" method="GET">
                                <div class="col-lg-6">
                                    <div class="form-group">
                                    <label>Project Name</label>
                                    <?php
                                        $sqll="SELECT Project_id,Project_title FROM project_master";
                                        echo "<select class='form-control' name='project' onchange='showCircle(this.value)'>";     
                                        echo "<option value=''>Select One</option>";  
                                        foreach ($conn->query($sqll) as $row){
                                        echo "<option value=$row[Project_id]>$row[Project_title]</option>";  
                                        }
                                        echo "</select>";                                                                                                                                                
                                    ?>
                                    </div>
                                    <div class="form-group">
                                        <div id="txtHint"></div>
                                        <!--<div id="txtHint1"></div>-->
                                    </div>                                                                                                                                                                               
                                    <div class="form-group">
                                        <button type="submit" class="btn btn-primary" name="submit" value="submit">Submit</button>
                                        <button type="reset" class="btn btn-primary">Reset</button>
                                    </div>
                                </div>                                                                                  
                            </form>
                        </div>
                    </div>                                
                </div>
            </div>                        
        </div>   
    </div> 
</body>
</html>

First i select the project from the drop down list and at at the time of selection a ajax call will be made to get the product. 首先，我从下拉列表中选择项目，在选择时，将进行ajax调用以获取产品。

GetProduct.php GetProduct.php

 if(isset($_GET['q'])){ $q = (string)filter_input(INPUT_GET,'q'); }else{ // header('Location:AddCircleToProduct.php'); } $sql2="SELECT Product_id,Product_title FROM product_master where Project_id = '$q'"; $support = $q; $result = $conn -> query($sql2); echo "<label>Product Name</label>"; echo "<select class='form-control' name='product' onchange='product1(this.value)'>"; echo "<option value=''>Select One</option>"; while ($row = mysqli_fetch_array($result)){ echo "<option value={$row['Product_id']}>{$row['Product_title']}</option>"; } echo "</select>";

At the time of product selection one another call is made through Ajax to get the test case. 在产品选择时，通过Ajax再次调用以获取测试用例。

3.GET Test case 3.GET测试用例

    if(isset($_GET['prod'])){
        $product = filter_input(INPUT_GET, 'prod');
        $query = "select * from testcase_master where Product_id = '$product'";
        //echo $query;
       echo '<div class="panel-body"><div class="dataTable_wrapper"><table class="table table-striped table-bordered table-hover" id="dataTables-example"">';
       echo '<thead><tr><th>SELECT</th><th>Testcase Id</th><th>Testcase Title</th><th>Description</th><th>Created by</th><th>Subscriber Type</th><th>Priority</th></tr></thead><tbody>'; 
        if($results = $conn -> query($query) or die(mysqli_errno($conn))){
            while ($row =  mysqli_fetch_array($results)) {
                echo '<tr class="odd gradeX">';
                echo "<td><input type=\"checkbox\" name=\"checkbox[]\" id=\"checkbox[]\"  value=\"".$row['Testcase_id']."\" class='form-control'/></td>";
                echo "<td>{$row['Testcase_id']}</td><td>{$row['Testcase_title']}</td><td>{$row['Testcase_desc']}</td><td>{$row['Created_by']}</td><td>{$row['Subscriber_type']}</td><td>{$row['Priority']}</td>";
                echo '</tr>';
            }
           echo '</tbody></table></div></div>';
        } else {
             echo 'ss' . mysqli_error($conn);
        }
     }
      else{
         echo'Sorry You cant access this page Directly.';
    }
    ?>

After getting the list of test case selection of checkbox is made. 获取复选框的测试用例选择列表后。

4.circledata.php 4.circledata.php

    <?php
    include './config.php';
    if(isset($_GET['submit'])){
    if(isset($_GET['checkbox'])){
      if (is_array($_GET['checkbox'])) {
       foreach($_GET['checkbox'] as $value){
        $q1 = "INSERT INTO tmtool.assigned_circle_product (`Circle_id`, `Product_id`) VALUES ('.$value.','$_SESSION[prod]')";
        if($conn -> query($q1) == TRUE){
              echo "Data Entered successfully\n";
        } else {
            echo 'Error' . mysqli_error($conn);
        }
      }
     } else {
      $value = $_GET['checkbox'];
      echo $value;
     }
    }        

   }

   ?>

And in the last file all the data are commited into the database. 在最后一个文件中，所有数据都被提交到数据库中。

Please tell me if anything i can do to improve my code. 请告诉我是否可以采取任何措施来改进我的代码。 Thank you. 谢谢。

创建一个下拉列表链，该列表基于前一个输入，形成php中的数据库

问题描述

1 个解决方案

解决方案1
0 已采纳 2015-03-04 12:42:51

创建一个下拉列表链，该列表基于前一个输入，形成php中的数据库

问题描述

1 个解决方案

解决方案1 0 已采纳 2015-03-04 12:42:51

解决方案1
0 已采纳 2015-03-04 12:42:51