Question

我想在codeigniter框架中使用jQuery自动完成多个字段我提到了this tutorial.But我的搜索文件没有工作。文本字段没有显示自动完成列表这是我的代码，请帮我解决这个问题。

视图 .....................

<div class="row">
    <form action="" name="students" method="post" id="students">
        <input type="text" name="patientId" id="patientId_1" class="ui-autocomplete-input">
        <input type="text" name="firstname" id="firstname_1" class="ui-autocomplete-input">
        <input type="text" name="nic" id="nic_1" class="ui-autocomplete-input">
        <input type="text" name="telephone" id="telephone_1" class="ui-autocomplete-input">
    </form>
</div>

的jQuery .......

$('#patientId_1').autocomplete({
        source: function( request, response ) {
            $.ajax({
                url : 'http://localhost/cafdc/BillingController/test',
                dataType: "json",
                data: {
                    name_startsWith: request.term,
                    type: 'patient_table',
                    row_num : 1
                },
                success: function( data ) {
                    response( $.map( data, function( item ) {
                        var code = item.split("|");
                        return {
                            label: code[0],
                            value: code[0],
                            data : item
                        }
                    }));
                }
            });
        },
        autoFocus: true,
        minLength: 0,
        select: function( event, ui ) {
            var names = ui.item.data.split("|");
            $('#firstname_1').val(names[1]);
            $('#nic_1').val(names[2]);
            $('#telephone_1').val(names[3]);
        }
    });

控制器 .......................

public function test()
        {
            $data=$this->Billing_Model->get_data();
            echo json_encode($data);
        }

模型 ....................

 public function get_data()
    {
        if($_POST['type'] == 'patient_table'){
            $row_num = $_POST['row_num'];
            $result =$this->db->query( "SELECT  patientId, fname, nic, tpnumber FROM tblpatient where name LIKE '".strtoupper($_POST['name_startsWith'])."%'");
            $data = array();
            while ($row = mysqli_fetch_assoc($result)) {
                $name = $row['patientId'].'|'.$row['fname'].'|'.$row['nic'].'|'.$row['tpnumber'].'|'.$row_num;
                array_push($data, $name);
            }

        }
    }

Answer 1

您的代码无法正常工作因为您不是return来自型号的任何数据

public function get_data()
    {
        if($_POST['type'] == 'patient_table'){
            $row_num = $_POST['row_num'];
            $result =$this->db->query( "SELECT  patientId, fname, nic, tpnumber FROM tblpatient where name LIKE '".strtoupper($_POST['name_startsWith'])."%'");
            $data = array();
            while ($row = mysqli_fetch_assoc($result)) {
                $name = $row['patientId'].'|'.$row['fname'].'|'.$row['nic'].'|'.$row['tpnumber'].'|'.$row_num;
                array_push($data, $name);
            }
            return $data;// return your data

        }
    }

Answer 2

您必须在ajax调用中设置$('#topSearchButton').on("click", function () { $('li:not([type="Documents"],:contains(Documents))',$(this).next()).toggle("fast"); });，否则会将其设为type:"POST"，因此请在GET调用中添加type，例如，

$.ajax

Codeigniter自动完成多个字段无法正常工作

2 个答案: