如何在php

Question

这是我要通过ID在table中显示数据的td。

<tbody>
<td id="show_file"></td>
</tbody>

这是向EditDeleteLecture.php发送请求以从数据库检索数据的功能。

function readRecords(){
    var readrecords = "readrecords";
    $.ajax({
        url:"ajax/EditDeleteLecture.php",
        type:"POST",
        data:{readrecords:readrecords},
        success:function(data,status){
            var user1 = JSON.parse(data);           
            $("#show_file").val(user1.file_name);
        }
    });
}

此代码写在EditDeleteLecture.php中。

if(isset($_POST['readrecords'])){
    $displayquery = "select content.file_name,content.cont_id AS ID,
                            content.description,content.course_code,
                            content.upload_date,course.course_name 
                    from content 
                        join class on class.course_code=content.course_code 
                        join course on class.course_code=course.course_code 
                        join faculty on faculty.faculty_name=class.faculty_name 
                    where faculty.faculty_phone='$single_user' 
                    order by ID desc"; 

    $result = sqlsrv_query($conn,$displayquery);
    while ($row = sqlsrv_fetch_array($result)) {            
        $response1 = $row;
    } 
    echo json_encode($response1);
}

console.log（数据）显示

this.{"0":"waqsa.pdf",
    "file_name":"waqsa.pdf",
    "1":1193,
    "ID":1193,
    "2":"aaaa",
    "description":"aaaa",
    "3":"CS-311",
    "course_code":"CS-311",
    "4":"2018-10-22",
    "upload_date":"2018-10-22",
    "5":"Programming Fundamental",
    "course_name":"Programming Fundamental"
} 

Answer 1

谢谢大家。问题已经解决了。我刚刚在找$("#show_file").text(user1.file_name); 代替

$("#show_file").text(user1.file_name);