jQuery array issue

Question

<?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)) {
    $myresult1['Patient_id'] = 'R'.$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?

Andy E · Accepted Answer · 2010-02-25 09:05:13Z

2

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"
);

edited Feb 25, 2010 at 9:05

answered Feb 23, 2010 at 12:05

Andy E

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3 Comments

Hacker Over a year ago

$("#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); alert(json_data.Name); if (json_data.Name) { $("#name").val(json_data.Name);} if (json_data.Age) { $("#age").val(json_data.Age);} if (json_data.Weight) { $("#ropweight").val(json_data.Weight);} if (json_data.Gest_age) { $("#gest_age").val(json_data.Gest_age);} }, "Json" ); });

Hacker Over a year ago

I did this its still prints the alert(json_data); properly like {"Patient_id":"R00006","Name":"admin","Age":"12","Weight":"67","Gest_age":"2"} but alert(json_data.Name); prints as undefined :(

Matt Ellen Over a year ago

Hi pradeep. There is a very minor typo in Andy's code you need to change "Json" to "json".

Matt Ellen · Accepted Answer · 2010-02-23 11:58:07Z

2

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

answered Feb 23, 2010 at 11:58

Matt Ellen

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Comments

Tinku · Accepted Answer · 2010-02-23 12:00:12Z

1

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"

answered Feb 23, 2010 at 12:00

Tinku

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2 Comments

Hacker Over a year ago

I did that and added json at the end .but then also its printing in same way as earlier :(

Tinku Over a year ago

Open firebug goto Net -> xhr tab and check that json data is coming in proper format

meouw · Accepted Answer · 2010-02-23 12:00:54Z

1

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

answered Feb 23, 2010 at 12:00

meouw

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1 Comment

Matt Ellen Over a year ago

That uses the GET method, but pradeep is POSTing the data.

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