error in json function saving data into database

Question

I have tried a lot but I have not been able to find out what is wrong with this function to save two values into database. It has been working fine for another function to save one value. It behaves very strange here. Sometimes send 'parent' value & sometimes stop sending it but never send msg value. Here is function. It works fine for one input i.e. parent but problems start with the addition of 2nd input.

<script> 
function ADDLISITEM(form)
{ 
var parent = form.txtInput.value;
    var msg    = form.msgInput.value;
    form.txtInput.value = "";
    form.msgInput.value = "";
    var url = "send_mysql.php"
var request = null;
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
request=new XMLHttpRequest();
 }
else
{// code for IE6, IE5
request=new ActiveXObject("Microsoft.XMLHTTP");
}
request.open("POST", url, true);
request.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
request.setRequestHeader("Connection", "close");
request.onreadystatechange = function(){
    if (request.readyState == 4) {
        if (request.status == 200) {
            //alert('POST');
    } else {
        alert(request.status); // fails here
    }
    }
}
request.send("parent=" + encodeURIComponent(parent).replace(/%20/g, '+')+"&msg=" + 
    encodeURIComponent(msg).replace(/%20/g, '+'));
}
</script>

This is send.php

$username = "babar";
$password = "k4541616";
$hostname = "localhost";
$dbh = mysql_connect($hostname, $username, $password) or die("Unable to connect
to MySQL");
$selected = mysql_select_db("spec",$dbh) or die("Could not select first_test");

  //die(var_export($_POST,TRUE));

  $parent = $_POST['parent'];
  $msg = $_POST['msg'];
  $name   = 'Akhtar Nutt';
  //$parent2 = json_decode($parent);
  $msg_ID = '2q7b2sfwwe';
  //$msg2    = json_decode($msg);
  $query  =  "INSERT INTO msg2_Qualities(id,name,msg,msg_id,parent) VALUES  
  ('','$name','$msg','$msg_ID','$parent')";
    if(!mysql_query($query, $dbh))
        {die('error:' .mysql_error())
      ;}

?>

Linus Kleen · Accepted Answer · 2010-11-30 16:16:23Z

2

Alter

request.send("parent=" + encodeURIComponent(parent).replace(/%20/g, '+')+"msg=" + encodeURIComponent(msg).replace(/%20/g, '+'));

to:

request.send("parent=" + encodeURIComponent(parent).replace(/%20/g, '+')+"&msg=" + encodeURIComponent(msg).replace(/%20/g, '+'));

You're missing the argument separator & in your query string...

You also might want to refrain from using values in $_REQUEST as they aren't reliable. If your script expects data from a POST then retrieve these values from $_POST.

answered Nov 30, 2010 at 16:16

Linus Kleen

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5 Comments

XCeptable Over a year ago

I tried both but it make no difference, Yes, that was missing I add it but it also does not make an difference.

Linus Kleen Over a year ago

OK. Then let send.php do a "die(var_export($_POST,TRUE))" right at the start. What's the result when sending data via XHR then?

XCeptable Over a year ago

@ goreSplatter, can you explain a bit. I put die(var_export($_POST,TRUE)); right in the start of send_mysql.php i.e after database connection. Whats do you mean by 'sending data via XHR '?

XCeptable Over a year ago

@ goreSplatter, OK, I dont't know what was wrong but it started working eventually though I tried this way before, I removed json encode/decode & send it as a string. thanX for your '&' correction as that was also an issue.

Linus Kleen Over a year ago

XHR is the XMLHttpRequest object. In onreadystatechange you can check the response of your script. The output of var_export which would have been captured could have given you more clues. Anyways... Thanks for accepting and happy coding.

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