Question

我想从下拉列表中获取数据。就像我从下拉列表中选择员工ID 40一样，它将从该员工的数据库中获取数据并显示在文本框中。

这是我的下拉代码。请帮助我如何获得所选值。

<?php
    $con=mysqli_connect("localhost","root","","hct_db");
    // Check connection
    if (mysqli_connect_errno()) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
?>

 <label>Select Employee ID</label>
     <select class="form-control" name="employee_id">
         <?php 
         $result = mysqli_query($con,"SELECT employee_id FROM employee order by employee_id");

         while($row = mysqli_fetch_array($result)) 
             echo "<option value='" . $row['employee_id'] . "'>" . $row['employee_id'] . "</option>";
         ?>
     </select>

Answer 1

<?php
if(isset($_REQUEST['submit']))
{
  $value=$_POST['employee_id'];
  $query = mysql_query($con,"SELECT employee_name FROM employee where employee_id=$value");
  $result=mysql_fetch_array($query);
  $emp_name=$result['employee_name'];
}
?>

<form action="" method="post" name="form">
 <label>Select Employee ID</label>
   <select class="form-control" name="employee_id">
    <?php $result = mysqli_query($con,"SELECT employee_id FROM employee order by employee_id");
 while($row = mysqli_fetch_array($result)) 
   echo "<option value='" . $row['employee_id'] . "'>" .$row['employee_id'] . "</option>";
 ?>
  </select>
<input type="submit" name="submit" value="submit">
</form>
 <input type="text" value="<?=$emp_name?>" name="emp_name"/>

check this code as your need

Answer 2

首先给你一些id选择这样的选项：

<select class="form-control" name="employee_id" id='employee'>

添加如下文本框：

<input type='text' name='emp_name' id='emp_name' />

比使用jquery和ajax这样：

$('#employee').change(function(){
    var selected_id = $(this).val();
    var data = {id:selected_id};
    $.post('getemp_name.php',data,function(data){
         $('#emp_name').val(data);
    });
});

<强> getemp_name.php

if(isset($_POST['id'])){
    //fire query using this id and get the name of employee and echo it
    echo $emp_name;
}

Answer 3

使用此示例，您可以通过单击Here轻松地学习如何从下拉列表中获取数据。该存储库包含所有代码

Answer 4

要获取所选值，您必须在用户提交表单后到达$_GET或$_POST超级用户。

所以，在提交之后，如果你发帖，请把它作为：

<?php 
$employee_id = $_POST['employee_id']; ?>

如果你获得：

<?php 
$employee_id = $_GET['employee_id']; ?>

Answer 5

在onChange

上应用以下功能

function myFunction(mySelect) {
        var x = document.getElementById("mySelect").value;
        document.getElementById("demo").innerHTML = "You selected: " + x;
    }

example

Answer 6

这取决于你想要什么。如果要在查询后使用查询中的所选ID，可以使用以下查询：

    "SELECT   *
     FROM     `employee`
     WHERE    `employee`.`employee_id` = " . (int) $_POST['employee_id'] . "
     LIMIT    1"

假设您在表单中使用post方法。

如何从下拉列表中选择后获取数据

6 个答案: