如何使用ajax将值传递到另一个页面php

I am trying to pass a value of a select option input using Ajax into another page PHP and get the database and fetch data into select option. However, my problem is there is nothing to show in second select option.

I have the select option called Country which contains Country 1, Country 2, Country 3, Country 4 with cid 1,2,3,4.

what I wanted is past Country ID (cid) into getdata.php and query the database of City which contains City a, city b, city c, etc. and fetch city's data into select option using Ajax.

I need to make city select option dynamically change when I select the country.

here is my database structure :

Country.db

enter image description here

City.db

enter image description here

Here is my script :

dropdown.php

<?php

    require_once "connection.php";
?>

<html>
<head>
    <title>Dropdown ajax</title>
</head>
<body>


<div class="country" >
    <label>Country</label>
    <select name="country" onchange="getId(this.value);">
    <option value=""> -- Select Country -- </option>

    <!-- populate value using php -->

    <?php
        $query = "SELECT * FROM country";
        $result = mysqli_query($con,$query);

        //loop

        foreach ($result as $country) {
        ?>
            <option value="<?php echo $country["cid"]; ?>"> <?php echo $country["country"]; ?> </option>

        <?php 
            } 
        ?>

</select>
</div>
<div class="city">
    <label>City</label>
    <select name="city" id="cityList">

        <option value=""></option>

    </select>
</div>
<script src="jquery-3.2.1.min"></script>
<script type="text/javascript">
    function getId(val){
        // ajax function

        $.ajax({
            type:"POST",
            url:"getdata.php",
            data:"cid="+val,
            success:function(data){
                $(#cityList).html(data);
            }
        });
    }
</script>

</body>
</html>

from the dropdown, I am trying to past cid into getdata.php using ajax and fetch the database into select option inside drop-down.php

getdata.php

<?php

require_once "connection.php";

if(!empty($_POST["cid"])){
    $cid = $_POST["cid"];
    $query = "SELECT * FROM city WHERE cid = $cid";
    $result = mysqli_query($con,$query);

    foreach ($result as $city) {

    ?>
    <option value="<?php echo $city["cityId"];?>"><?php echo $city["city"];?></option>

    <?php       
    }
}

?>

here is my connection.php

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

    //check connection
    if(mysqli_connect_errno()){
        echo "Failed to connect :".mysqli_connect_errno();
    }
?>

the result is like this :

enter image description here

so how to fix this problem?

查看全部
dqh19413
dqh19413
2017/09/08 23:52
  • php
  • html
  • mysql
  • ajax
  • javascript
  • 点赞
  • 收藏
  • 回答
    私信
满意答案
查看全部

1个回复