向下钻取下拉选择的值不会插入到MYSQL中 [英] html drill down drop down selected value does not inserted in MYSQL

问题描述

我有两个下降。首先从数据库填充下拉菜单。第二个下拉列表从数据库中填入第一个下拉列表的选定值。

$（document）.ready（function（）{$（＃code>）

<如果（c1！=）{$ .ajax（{url：'getstatw.php'，data（'c'）。 ：{c：c1}，type：'POST'，success：function（response）{var resp = $ .trim（response）; $（＃c）。html（resp）;}}）;} else { $（＃c）。html（< option value ='> Select state< / option>）;}}）;}）;

< form id =worldmethod =postaction =insert.php>< select name =countryid =cstyle =width：200pxclass =btn btn-primary dropdown-toggle;> <选项>国家< /选项> <？php $ sql =从table1中选择DISTINCT国家/地区; $ res = mysqli_query（$ con，$ sql）; if（mysqli_num_rows（$ res）> 0）{while（$ row = mysqli_fetch_object（$ res））{echo< option value ='。$ row-> id。'>。$ row- &℃。 < /选项> 中; }}？> < /选择><峰; br><峰; br> < label for =s>状态< / label>< select name =Stateid =sstyle =width：200px; < / option>< / select>< br>< br>< button id =subtype =submit class =btn btn-primarydisabled>提交< / button>< / form>

insert.php

$ con = mysqli_connect（localhost，root，，world）; // Check connectionif （mysqli_connect_errno（））{echo无法连接到MySQL：。 mysqli_connect_error（）;} //安全转义变量// 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'，）; //查询table_dataastif（！mysqli_query（$ con，$ sql1））{die '错误：'。mysqli_error（$ con））;} echo1 record added; mysqli_close（$ con）;

 <？php $ con = mysqli_connect（localhost，root，，test）; //检查connectionif（mysqli_connect_errno（））{echo 无法连接到MySQL：。 mysqli_connect_error（）;} if（isset（$ _ POST ['c']））{$ sql =从`table2`中选择DISTINCT`State`，其中`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：./');}?>  

b $ b

我已经尝试了几乎所有在网上给出的解决方案。但不明白我的数据没有插入到mysql数据库中。
如何插入HTML通过PHP在MySQL中选择值作为文本

解决方案

如果想要添加第二个下拉选项需要使用 append 而不是 html ，并使用相同的ID，例如 #c 将响应写入ajax成功，将其更改为第二个下拉ID，即 #s

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}，
类型：'POST'，
成功：有趣ction（response）{
 var resp = $ .trim（response）; 
 $（＃s）。append（resp）; 
} 
}）; 
} else {
 $（＃c）。append（< option selected value =''>选择状态< / option>）; 
} 
}）; 
}）; 
  

然后在插入查询中删除，。

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

i have two drop downs. first drop down populate from database. second drop down populated from database based on selected value of first drop down.

$(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>
  

insert.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); 

getstate.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: ./');
}


?>

i have tried almost all solution given on net. but do not understand my data is not inserted into mysql database. How to insert HTML select value as text in MySQL via PHP PHP Drop down list selected value not inserted in the database

解决方案

if want to add options to the second drop down you need to use append not html and your using same ID i.e. #c to write the response in ajax success, change it to second drop down ID i.e. #s

you can try this:

$(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>");
    }
  });
});

then remove , in insert query.

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

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