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jQuery array issue

开发者 https://www.devze.com 2022-12-21 03:49 出处:网络
<?php include_once(\'db.php\'); $location = $_POST[\'location\']; $doctor = $_POST[\'doctor\']; $patient_id = $_POST[\'patient_id\'];
<?php
include_once('db.php');
$location = $_POST['location'];
$doctor = $_POST['doctor'];
$patient_id = $_POST['patient_id'];

if(($location != "") && ($doctor != "")) {
  $sql = "select Name,Age,Gest_age,Weight from rop_form where Location = '".$location."' and Doctor = '".$doctor."' and Patient_id = '".$patient_id."'";
  $result = mysql_query($sql);
  $myresult  = "";
  while($row = mysql_fetch_array($result)) {
  开发者_运维百科  $myresult1['Patient_id'] = 'R'.$patient_id; 
    $myresult1['Name'] = $row['Name']; 
    $myresult1['Age'] = $row['Age']; 
    $myresult1['Weight'] = $row['Weight']; 
    $myresult1['Gest_age'] = $row['Gest_age']; 
  }
  $myresult = json_encode($myresult1); 
}
else {
  $myresult .= "";
}
echo $myresult;
?>

This is my PHP code.

This is the jQuery code.

$("#patient_id").change(function() {
  $.post("/diabetes/patient_detail_change.php",{ location:$("#location").val(),doctor:$("#doctor").val(),patient_id:$("#patient_id").val()} ,function(json_data) {
    alert(json_data);
    //var my_json = //{"Patient_id":"R00020","Name":"admin","Age":"12","Weight":"67","Gest_age":"2"//};
    $.each(json_data, function(key,value) {
      alert(key + ': ' + value);
      if(key == 'Name'){ $("#name").val(value); }
      if(key == 'Age'){ $("#age").val(value); }
      if(key == 'Weight'){ $("#ropweight").val(value); }
      if(key == 'Gest_age'){ $("#gest_age").val(value); }
    });
  });
});

alert(json_data); this line prints properly like

{"Patient_id":"R00006","Name":"admin","Age":"12","Weight":"67","Gest_age":"2"} which is the fomat required for jquery

But the .each loop statement which is present like: alert(key + ': ' + value); does not print like Patient_id : R00006 and all .but it prints like 0:{ 1:P 2:a 3:t 4:i ..what might be the problem?


In addition to Matt Ellen's answer, the $.each() method is for looping over JavaScript arrays and array-like objects (that have a length property). PHP's associative arrays (keyword->value) are converted into a native JavaScript object instead. You could use a for...in loop instead:

for (var key in json_data) { 
   alert(key + ': ' + json_data[key]); 
   if(key == 'Name'){ $("#name").val(json_data[key]);} 
   if(key == 'Age'){ $("#age").val(json_data[key]);} 
   if(key == 'Weight'){ $("#ropweight").val(json_data[key]);} 
   if(key == 'Gest_age'){ $("#gest_age").val(json_data[key]);} 
}

But you probably don't need the loop. You can just use:

$.post (
  "/diabetes/patient_detail_change.php",
  { 
    location:$("#location").val(),
    doctor:$("#doctor").val(),
    patient_id:$("#patient_id").val()
  }, 
  function (json_data){
    if ("Name" in json_data) { $("#name").val(json_data.Name);}
    if ("Age" in json_data) { $("#age").val(json_data.Age);}
    if ("Weight" in json_data) { $("#ropweight").val(json_data.Weight);}
    if ("Gest_age" in json_data) { $("#gest_age").val(json_data.Gest_age);}
  }, 
  "json"
);


In your post statment you need to specify that you are returning JSON.

$.post("/diabetes/patient_detail_change.php",{location:$("#location").val(),doctor:$("#doctor").val(),patient_id:$("#patient_id").val()} ,function(json_data){

alert(json_data);

$.each(json_data, function(key,value){
alert(key + ': ' + value);
if(key == 'Name'){ $("#name").val(value);}
if(key == 'Age'){ $("#age").val(value);}
if(key == 'Weight'){ $("#ropweight").val(value);}
if(key == 'Gest_age'){ $("#gest_age").val(value);}

});
}, "json");

Like so.

At the moment your returned data is being treated as a string, so the each statement is outputting each character.

See the definition here jQuery post definition


You should use $.post ("/diabetes/patient_detail_change.php",{ location:$("#location").val(),doctor:$("#doctor").val(),patient_id:$("#patient_id").val()}, function (json_data){ //blah blah },"Json" );

please check the last argument "Json"


Use $.getJSON instead of $.post.
This will return an object (parsed JSON), rather than a string

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