html下拉下拉选择值未在MYSQL中插入

Question

我有两个下拉菜单。 首先从数据库中下拉列表。 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> 

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: ./'); } ?> 

我已经尝试了网上给出的几乎所有解决方案。 但是不明白我的数据没有插入到mysql数据库中。 如何通过PHP在MySQL中将HTML选择值作为文本插入PHP PHP下拉列表中未插入数据库的选择值

Answer 1

您需要正确concantenate他们，你也不要你放,最后一栏后，在查询

更改以下

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

至

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

Answer 2

如果要在第二个下拉列表中添加选项，则需要使用append not html和您使用相同的ID（即#c以成功在Ajax中写入响应，将其更改为第二个下拉ID，即#s

您可以尝试以下方法：

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

然后取出,在插入查询。

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