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Javascript 如何提取/分割数据的一部分_Javascript_Php_Jquery_Ajax - Fatal编程技术网

Javascript 如何提取/分割数据的一部分

Javascript 如何提取/分割数据的一部分,javascript,php,jquery,ajax,Javascript,Php,Jquery,Ajax,我使用以下PHP代码从数据库中选择数据: <?php include './database_connect.php'; $ppid=$_POST['selectPatientID']; $query="SELECT * FROM patient WHERE p_Id='$ppid'"; $result= mysqli_query($conn, $query); while($row= mysqli_fetch_array($result)){ echo ($row['p

我使用以下PHP代码从数据库中选择数据:

<?php
include './database_connect.php';

$ppid=$_POST['selectPatientID'];

$query="SELECT * FROM patient WHERE p_Id='$ppid'";

$result=  mysqli_query($conn, $query);
while($row=  mysqli_fetch_array($result)){
    echo ($row['p_fname']); 
    echo ' ';
    echo ($row['p_lname']);
}    
?>  

您必须
json\u编码

$row =  mysqli_fetch_array($result)){

header('Content-Type: application/json');// header for json encode response
echo json_encode($row);
exit;
您的jQuery ajax代码将是:

$.ajax({
   url: "UDPatient.php",
   type: 'POST',
   data: {selectPatientID: $('#selectPatientID').val()},
   dataType: 'json',// to receive data into json format
   success: function (result) {
       alert(result.p_fname);
       alert(result. p_lname);
   }
}); 

你应该试着做一些类似的事情

<?php
include './database_connect.php';

$ppid=$_POST['selectPatientID'];

$query="SELECT * FROM patient WHERE p_Id='$ppid'";

$result=  mysqli_query($conn, $query);


$row=  mysqli_fetch_array($result);
echo json_encode($row);
?>  

可以解析json结果(或者将数据类型设置为“json”)

json\u encode()
php中的结果数组,以及
json.parse()
php中的结果javascript@chris85:是的,在我的
php
文件中,有两个值是从数据库中获取的。我怎样才能把它们分开?@Manav:你能帮我吗,先生???从文件中提供XML或JSON。或者,如果您对SQL注入持开放态度,则应该参数化该查询。
<?php
include './database_connect.php';

$ppid=$_POST['selectPatientID'];

$query="SELECT * FROM patient WHERE p_Id='$ppid'";

$result=  mysqli_query($conn, $query);


$row=  mysqli_fetch_array($result);
echo json_encode($row);
?>  
<script>
    $(document).ready(function () {
        $('#selectPatientID').on('change', function (event) {
            event.preventDefault();
            $.ajax({
                url: "UDPatient.php",
                type: 'POST',
                data: {selectPatientID: $('#selectPatientID').val()},
                dataType: 'html',
                success: function (result) {
                    var jsonResult = JSON.parse(result);
                    var p_fname = jsonResult.p_fname;
                    var p_lname = jsonResult.p_lname;
                    // whatever else
                }
            });
        });
    });
</script>