douyakan8924 2015-10-18 17:17
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AJAX SQL JSON不起作用

My AJAX seems like not working. It should load patient details into a dropwdown.

this is my ajax code:

function PopulatePatient()
{
        $("#PatientDropDown").empty();
        $("#PatientDropDown").append("<option>Loading.....</option>");

        $.ajax({
                type:"POST",
                url:"../listener/populatePatientName.php",
                contentType:"application/json; charset=utf-8",
                dataType:"json",
                success: function(data){
                        $("#PatientDropDown").empty();
                        $("#PatientDropDown").append("<option value=''>-Select Patient-</option>");

                $.each(data,function(i,item)
                {
                    $("#PatientDropDown").append('<option value="'+ data[i].patientUserId +'">'+ data[i].patientFirstName +'</option>');

                });

        },

        complete: function()
        {

        }

        });
}

This is my PHP Code (populatePatientName.php):

<?php

include '../core/init.php';

$sql = mysql_query("SELECT patientUserId,patientFirstName FROM patientdetails");

if(mysql_num_rows($sql))
{
  $data=array();
  while($row=mysql_fetch_array($sql))
  {
$data[]=array
(
  'patientUserId' => $row['patientUserId']
  'patientFirstName' => $row['patientFirstName']
);
}
  header('Content-type: application/json');
  echo json_encode($data);
}

 ?>

This is my HTML Code:

<select class="form-control" id="PatientDropDown">
</select>

This code works in my friend PC. My database and query are running well. Table contained data.

My code stop at this point : $("#PatientDropDown").append("Loading.....");

It does not enter ajax. Any solution?

  • 写回答

6条回答 默认 最新

  • duaabhuv188411 2015-10-18 17:59
    关注

    I found the solution, I forgot the comma,

    <?php
    
    include '../core/init.php';
    
    $sql = mysql_query("SELECT patientUserId,patientFirstName FROM patientdetails");
    
    if(mysql_num_rows($sql))
    {
      $data=array();
      while($row=mysql_fetch_array($sql))
      {
    $data[]=array
    (
      'patientUserId' => $row['patientUserId'],
      'patientFirstName' => $row['patientFirstName']
    );
    }
      header('Content-type: application/json');
      echo json_encode($data);
    }
    

    ?>

    $data[]=array
        (
          'patientUserId' => $row['patientUserId'],
          'patientFirstName' => $row['patientFirstName']
        );
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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