Question

我似乎有一个问题。我有一个输入字段和一个<select>字段。我需要在输入字段中键入一个位置，如果该单词与数据库中的记录匹配，则应该将<select>下拉列表中的人员姓名包括在内。这是我的index.php文件：

<html>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<body>
<input type="text" name="location_input" id="location_input">
Tutor:<select name="locations" id="locations">
    <script> 
        $("#location_input").keyup(function(){
            const location = $("#location_input").val();
            $("#locations").html(''); //reset dropdown
            // do ajax call to get locations
            $.ajax({
                url: 'search.php',  //replace this with your route of the search function
                data: {location}, //pass location as body data
                dataType: 'json', //expect a json response back
                success: function(data) {
                    data.forEach(function(el) { //loop over the json response
                        let option = `<option id=${el.id} value=${el.name}>${el.name}</option>`
                        $("#locations").append(option); //append locations to select dropdown
                    });
                },
                error: function(err) {  //error functions
                    console.log(err);
                    alert("Error")
                }
            });
        });
    </script>

</select>
</body>
</html>

这是我的search.php文件：

   <?php
function SearchLocations() {
    $conn = new mysqli('localhost', 'root', '', 'tutors') or die ('Cannot connect to db');
    $result = $conn->query("select * from tutor_location where Location_tags LIKE ='%". $_GET['location']."%'");

    $locations = [];

    while ($row = $result->fetch_assoc()) {
        $locations[] = $row;    

    }

    return json_encode($locations);

}

?>

这是我的数据库的屏幕截图：我面临的问题是它给了我localhost says error，而控制台没有显示错误。

Answer 1

您可以按照以下要求尝试使用ajax，

$.ajax({
      type:"POST",
      dataType:"json",
      url: 'search.php',
      data: {order_numbers: values, order_state: status},
      success: function(response){ },
      error: function(response){}
    });

下拉列表未从数据库填充

1 个答案: