javascript - Ajax提交返回错误但更新数据库正常

Question

<script>
        function addprescription() {
            var Case_Histroy=$('#Case_Histroy').val();
            var Medication=$('#Medication').val();
            var Note=$('#Note').val();
            var pname="<?php echo($patient->getUsername()); ?>";
            var dname="<?php echo($doctor->getUsername()); ?>";
            var id="<?php echo($id); ?>";
            frmData={Case_Histroy:Case_Histroy,Medication:Medication,Note:Note,pname:pname,dname:dname,id:id}
            console.log( frmData);
            $.ajax({
                    type: "POST",
                    url: "loadfiles/AddAppointmentSubmit.php",
                    data: frmData,
                    success: function (msg) {
                        alert(msg);
                        $("#alert").html(msg)
                    }
                    ,
                error : function () {
                alert("failure");
            }
        });
        }
</script>

我有一个提交表单的功能。但ajax功能警告其失败。但数据库似乎已更新。当我点击按钮时。我在控制台中找不到原因。

这是php文件

<?php
echo "I'm in";
include "../../Adaptor/mysql_crud.php";
include ("Prescription.php");
$prescription=new Prescription();
if(isset($_POST)){
    $Note=htmlspecialchars($_POST['Note']);
    $Case_Histroy=htmlspecialchars($_POST['Case_Histroy']);
    $medication = htmlspecialchars($_POST['Medication']);
    $pname=$_POST['pname'];
    $danme=$_POST['dname'];
    $id=$_POST['id'];
    $prescription->insert($pname,$danme,$Case_Histroy,$medication,$Note,$id);
    ?>
    <div class="alert alert-success" id="alert"><strong><?php echo "Submitted succesfully"; ?></strong></div>
<?php
}
?>

Answer 1

尝试添加 else给您的声明if :

插入($pname,$danme,$Case_Histroy,$medication,$Note,$id); ?> <强> <强> }?>

此外，没有必要将 php 粘贴在 <div> 的中间。你可以只使用 echo在开始时，因为您没有向其中引入任何变量:

echo '<div class="alert alert-success" id="alert"><strong>Submitted successfully</strong></div>';