nope man it did not work :(
see the code
jquery code:
the alert(json_data ); line prints this
{"Patient_id":" R00008","Name": "","Age":"12"," Weight":"67","G est_age":"2"}
but the alert(key + ': ' + value); is printing like
0:{ 1:P 2:a and so on .
the php code is like this and as u pointed out i have corrected the variable and array issue
see the code
jquery code:
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);
$.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);}
});
});
});
{"Patient_id":" R00008","Name": "","Age":"12"," Weight":"67","G est_age":"2"}
but the alert(key + ': ' + value); is printing like
0:{ 1:P 2:a and so on .
the php code is like this and as u pointed out i have corrected the variable and array issue
Code:
<?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;
?>
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