Question

我正在尝试在Jquery中创建一个country -state选择器，它应该使用PHP在db中提交值。这是脚本：

<script type="text/javascript">
var state_arr=new Array("Jammu and Kashmir","Delhi","Uttar Pradesh","Tamil nadu","Maharashtra","Karnataka","West Bengal","Punjab","Haryana");
var s_a = new Array();
s_a[0]="";
s_a[1]="Srinagar|Jammu";
s_a[2]="New Delhi";
s_a[3]="Lucknow|Kanpur|Ghaziabad|Noida|Varanasi";
s_a[4]="Chennai";
s_a[5]="Mumbai|Pune|Aurangabad|Thane";
s_a[6]="Bangalore|Mysore";
s_a[7]="Kolkata";
s_a[8]="Chandigarh|Mohali";
s_a[9]="Gurgaon|Chandigarh";
function print_state(state){
    //given the id of the <select> tag as function argument, it inserts <option> tags
    var option_str = document.getElementById(state);
    option_str.length=0;
    option_str.options[0] = new Option('Select State','');
    option_str.selectedIndex = 0;
    for (var i=0; i<state_arr.length; i++) {
    option_str.options[option_str.length] = new Option(state_arr[i],state_arr[i]);
    }
}
function print_city(city, selectedIndex){
    var option_str = document.getElementById(city);
    option_str.length=0;    
    option_str.options[0] = new Option('Select City','');
    option_str.selectedIndex = 0;
    var city_arr = s_a[selectedIndex].split("|");
    for (var i=0; i<city_arr.length; i++) {
    option_str.options[option_str.length] = new Option(city_arr[i],city_arr[i]);
    }
}

以下是显示选择器的PHP代码：

<?php if (!isset($_POST['submit_val3'])) { ?>
Select the State:
<select onchange="print_city('city',this.selectedIndex);" id="state" name ="state"></select>
<br />
City:
<select name ="city" id ="city"></select>
<script language="javascript">print_state("state");</script>
<input type="submit" name="submit_val3" value="Submit" />
<?php } ?>

以下是提交的PHP代码：

if(isset($_POST['submit_val3']))
{
  $city=$_POST['city'];
  $state=$_POST['state'];
 $dbc = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD) or die(mysql_error());
 mysql_select_db(DB_NAME);
  $query="UPDATE register_emp SET state='$state',city='$city' WHERE ID = '" . $_SESSION['ID'] . "'";
  mysql_query($query,$dbc) or die('query failed');
  mysql_close();
}

我的问题是，点击“提交”按钮时，没有任何值提交给数据库。我在哪里错了？

Answer 1

请添加表单标签和操作，这就是为什么它没有提交到php页面。


    <?php if (!isset($_POST['submit_val3'])) { ?>
    <form method="post" action="yourphppage.php">  <!-- form tag and action page -->
    Select the State:
    <select onchange="print_city('city',this.selectedIndex);" id="state" name ="state">
    

    City:
    <select name ="city" id ="city">
    <script language="javascript">print_state("state");
    <input type="submit" name="submit_val3" value="Submit" />
    </form>
    <?php } ?>

如何使用PHP，MySqL和JQuery将动态字段的值插入到数据库中

1 个答案: