Question

Am使用ajax和json从MySQL中获取一些表列数据，需要将这些值显示给相应的输入id标记。但是它没有显示值和数组长度显示超出预期的值。预期长度值为8但显示为101在控制台中。并在控制台中排列数据

    [{"name":"adasdas",
      "gen":"F",
      "add":"asdasd",
      "dob":"2019-09-25",
      "ag":"25",
      "mob":"65165165",
      "tel":"1"}]

我的 ajax请求是

$('#add').click(function(event){
            event.preventDefault();

                    var patient_no  = $('.patient_no').val();

                    $.ajax({
                        url : 'get_visit_details.php',
                        type : 'post',
                          data :{patient_no:patient_no},
                          datatype : 'json',
                        success:function(data){

                            var len = data.length;

                            console.log(len);
              console.log(data);

                  for(var i = 0;i < len; i++){
                                var pat_name    = data[i]['name'];
                                var gender      = data[i]['gen'];
                                var address     = data[i]['add'];
                                var d_birth     = data[i]['dob'];
                                var age           = data[i]['ag'];
                                var mobile      = data[i]['mob'];
                                var tele          = data[i]['tel'];

                                $('.pat_name').val(pat_name);
                                $('.gender').val(gender);
                                $('.address').val(address);
                                $('.d_birth').val(d_birth);
                                $('.age').val(age);
                                $('.tele').val(tele);
                                $('.mobile').val(mobile);

                            }
                        }
                });
          });

和 php代码是

$patient_no =   $_POST['patient_no'];

$sql = "SELECT pat_name,gender,address,d_birth,age,mobile,tele,reg_no 
        FROM patient 
        WHERE reg_no = '$patient_no' ";
$result = query($sql);
confirm($result);

if($row = fetch_array($result)) {
        $pat_name   =   $row['pat_name'];
        $gender     =   $row['gender'];
        $address    =   $row['address'];
        $d_birth    =   $row['d_birth'];
        $age        =   $row['age'];
        $mobile     =   $row['mobile'];
        $tele       =   $row['tele'];
    }

        $patient_data = array();
        $patient_data[] = array("name" => $pat_name,"gen" => $gender, "add" => $address ,"dob" => $d_birth ,
                                "ag" => $age , "mob" => $mobile,"tel" => $tele);

    echo json_encode($patient_data);

Answer 1

101是字符串的长度

[{"name":"adasdas","gen":"F","add":"asdasd","dob":"2019-09-25","ag":"25","mob":"65165165","tel":"1"}]

这的预期结果是“ 1”，因为您在json数组中有7个json objtect

JavaScrip无法解码de Json吗？尝试这样做

var obj = jQuery.parseJSON( '[{"name":"adasdas","gen":"F","add":"asdasd","dob":"2019-09-25","ag":"25","mob":"65165165","tel":"1"}]' );
console.log(obj.length);

这必须返回1

console.log(obj[0].name)

这必须返回“ adasdas”

我测试在开发人员控制台中，如果不起作用，则可能的php输出包含回教字符，请在“原始”中显示php输出（在webbrowser中，单击“查看源代码”）

Answer 2

我可以看到数据是数组对象。因此获得如下长度

var data_count = Object.keys(data).length;
console.log(data_count);

如果要获取对象的长度，请使用

Object.getOwnPropertyNames(data).length

数组长度超出预期

2 个答案: