html向下钻取下拉选定值未插入MYSQL

Posted 2023-03-05

【中文标题】html向下钻取下拉选定值未插入MYSQL

【英文标题】：html drill down drop down selected value does not inserted in MYSQL

【发布时间】：2018-07-07 18:33:38

【问题描述】：

我有两个下拉菜单。首先从数据库下拉填充。根据第一个下拉列表的选定值从数据库填充第二个下拉列表。

$(document).ready(function() 
  $("#c").change(function() 
    var c1 = $('#c :selected').text();
    if(c1 != "") 
      $.ajax(
        url:'getstatw.php',
        data:c:c1,
        type:'POST',
        success:function(response) 
          var resp = $.trim(response);
          $("#c").html(resp);
        
      );
     else 
      $("#c").html("<option value=''>Select state</option>");
    
  );
);

 <form id = "world" method="post" action="insert.php">
<select name="country" id="c" style = "width:200px" class="btn btn-primary dropdown-toggle" ;>
      <option>country</option>
	 
      <?php
       	  
      $sql = "select DISTINCT country from table1";
      $res = mysqli_query($con, $sql);
      if(mysqli_num_rows($res) > 0) 
        while($row = mysqli_fetch_object($res)) 
          echo "<option value='".$row->id."'>".$row->c."</option>";
		  
        
      
      ?>
    </select>
<br><br>
  <label for="s" >State</label>

<select name="State" id="s" style = "width:200px  " ; class="btn btn-primary dropdown-toggle";><option>Select state</option></select><br><br>
<button id = "sub" type="submit" class="btn btn-primary" disabled>Submit</button>
</form>
  

插入.php

$con=mysqli_connect("localhost","root","","world");

// Check connection
if (mysqli_connect_errno()) 
  echo "Failed to connect to MySQL: " . mysqli_connect_error();


// escape variables for security
//home tab
$c = mysqli_real_escape_string($con, $_POST['country']);
$s = mysqli_real_escape_string($con, $_POST['state']);


//query for table_mainast
$sql1="INSERT INTO table1 (Country, State)
VALUES ('$c', '$s',)";
//query for table_dataast


if (!mysqli_query($con,$sql1)) 
  die('Error: ' . mysqli_error($con));


echo "1 record added";

mysqli_close($con); 

获取状态.php

<?php
$con=mysqli_connect("localhost","root","","test");

// Check connection
if (mysqli_connect_errno()) 
  echo "Failed to connect to MySQL: " . mysqli_connect_error();


if(isset($_POST['c'])) 
  $sql = "select DISTINCT `State` from `table2` where `Country`='".mysqli_real_escape_string($con, $_POST['c'])."'";
  $res = mysqli_query($con, $sql);
  if(mysqli_num_rows($res) > 0) 
     echo "<option value=''>------- Select --------</option>";
    while($row = mysqli_fetch_object($res)) 
      echo "<option value='".$row->id."'>".$row->c."</option>";
    
  
 else 
  header('location: ./');



?>

我已经尝试了网上给出的几乎所有解决方案。但不明白我的数据是不是插入mysql数据库。 How to insert HTML select value as text in MySQL via PHP PHP Drop down list selected value not inserted in the database

【参考方案1】：

您需要正确连接它们，而且您不要将, 放在查询的最后一列之后

跟随变化

$sql1="INSERT INTO table1 (Country, State)
VALUES ('$c', '$s',)";

到

$sql1="INSERT INTO table1 (country, state)
VALUES ('".$c."', '".$s."')";

【讨论】：

我也试过了，但是这个数据没有插入

@Sarah_Salar 如果出现任何错误，请告诉我？

不要给出错误。只是不要将值插入表中。提交表格后为空白。没有插入记录

为什么没有人帮助这个论坛

@Sarah_Salar 已经回答了，请您检查一下表格列

【参考方案2】：

如果想在第二个下拉列表中添加选项，您需要使用append 而不是html 并且您使用相同的ID 即#c 以将响应写入ajax 成功，将其更改为第二个下拉ID 即@ 987654325@

你可以试试这个：

$(document).ready(function() 
  $("#c").change(function() 
    var c1 = $('#c :selected').text();
    if(c1 != "") 
      $.ajax(
        url:'getstatw.php',
        data:c:c1,
        type:'POST',
        success:function(response) 
          var resp = $.trim(response);
          $("#s").append(resp);
        
      );
     else 
      $("#c").append("<option selected value=''>Select state</option>");
    
  );
);

然后在插入查询中删除,。

$sql1="INSERT INTO table1 (Country, State)
VALUES ('$c', '$s')";

【讨论】：

请看这个问题***.com/questions/49043561/…