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Pardon me if I am making a silly mistake here... But I am little new to PHP... I have been stuck at this for long time. I have tried several methods but I am only able to pass ONe table row , not multiple rows

I've even tried with the following approach, but I landed with the same error. http://www.electrictoolbox.com/json-data-jquery-php-mysql/

Here's what I am trying to do :

1) Make Ajax call using JQuery, to fetch data from server.

  $.ajax({                                      
  url: 'test.php',                  //the script to call to get data          
  data: "",                        
  dataType: 'json',                //data format      
  success: function(data)          //on recieve of reply
  {
    //use "data"
  }

2) fetch table data from MySql database, using PHP.

$result = mysql_query("SELECT username,characterType FROM usertable");

3) pass ALL of the rows as JSON data, back to the calling function.

//This works , Returns one row and I am able to get the result back at AJAX end
$array = mysql_fetch_row($result);                          
echo json_encode($array);

mysql_fetch_assoc($result) FAILS with the following error : 

XML Parsing Error: no element found Location: moz-nullprincipal:{6b1***-****somecrap numbers****-***86} Line Number 17, Column 3:

$array = mysql_fetch_assoc($result);                          
echo json_encode($array);

I have even tried using mysql_fetch_array($result, MYSQL_NUM) as suggested in the other relevant questions, but I am not able to workaround the problem.

Any Help is greatly appreciated.

============

UPDATE:

I may be wrong but just wanted to know, if this could be the possibility or not... Can this be a local setup related issue ? A Configuration mistake .. ?

I have done my localhost setup using "XAMPP installation" [ Apache / MySQL / Tomcat ... bundled in one package ] ... When I run the file as "PHP application" it runs just fine ... I am able get "all rows" But when I deploy the code on my apache servers it doesn't run ... The whole php file comes as a response , with "XML parsing Error" [ I am using firebug to track the response ]

Thanks Pranav

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Have you looked at the JSON output from your PHP? What does it look like? Specifically, what does line 17 character 3 look like? In any case it doesn't look like _assoc returns multiple rows - it returns one row in a different form. You may have to loop through the results fetching a row at a time to build an array in PHP then serialize that. –  Rup Aug 2 '12 at 23:17
 
Can you post more of the script that's giving the error? –  liz Aug 2 '12 at 23:19
 
I have posted all of the scripts that's causing the issue .. $con = mysql_connect($host,$user,$pass); $dbs = mysql_select_db($databaseName, $con); $result = mysql_query("SELECT username,characterType,totalPayoff FROM $tableName"); $array = mysql_fetch_assoc($result); echo json_encode($rows); –  Pranav Aug 2 '12 at 23:22
1  
If you've got more code you should edit it into your question rather than post as a comment - thanks! –  Rup Aug 2 '12 at 23:24
 
It sounds like the server you're deploying to isn't setup to process PHP files correctly, and is sending the source code, which the browser is trying to interpret as XML. Do you have any PHP successfully running on the server you're trying to deploy to? Can you elaborate on what you mean by 'When I run the file as "PHP application"' -- are you just talking about running it locally in XAMPP vs. running it on your deployment server? –  JMM Aug 3 '12 at 22:13
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3 Answers

up vote 0 down vote

A few things to note here:

As Rup mentioned, $array = mysql_fetch_row($result); only returns a single row.

The other thing is that test.php does not currently return a JSON object; to see whats going on, it helps to comment out dataType: 'json', and use console.log(data); to see what is being returned. Your response probably looks something like this:

["someUsername","someChartype"]

The format of the object you are expecting is probably something more like:

[{"username":"foo","chartype":"admin"},
 {"username":"bar","chartype":"user"},
 {"username":"fum","chartype":"guest"}]

To get the format of the associative array, use mysql_fetch_assoc($result) instead; that will give

{"username":"someUsername","chracterType":"someChartype"}

To summarize, update your php code to:

  • Loop through your results, pushing them onto an array.

  • Encode the entire array at the end

    $outputArray = array();
while( $row = mysql_fetch_assoc( $result ) ){
    array_push($outputArray, $row);
}
echo json_encode($outputArray);

Take a look at How to build a JSON array from mysql database for more info. Also, as one of the answers mention, its worth using PDO rather than mysql_*

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up vote 0 down vote

Don't start manually composing a JSON string as @Tuanderful indicated. Let json_encode() handle that -- just give it the right data. Try something like this:

$records = array();

while ( $record = mysql_fetch_assoc( $result ) ) {

  $records[] = $record;

}

echo json_encode( $records );
share|improve this answer
 
I have tried this as well.. Please see the updated info about the problem.. I think I am missing out on something else –  Pranav Aug 3 '12 at 17:55
 
Do you have an XML declaration -- <?xml version="1.0"?> -- or something like it (e.g. <?) in your PHP script, and the short_open_tag configuration option enabled? (Run phpinfo() or ini_get( 'short_open_tag' ) to find out.) –  JMM Aug 3 '12 at 22:07
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up vote 0 down vote

Well I would to like this:

$.ajax({                                      
  url: 'test.php',                  //the script to call to get data          
  //data: "",                        
  dataType: 'json',                //data format      
  success: somevar
})



var somevar = function(data)          //on recieve of reply
  {
    //use "data"
    somevar2 = jQuery.parseJSON(data)
    //now You can enter each row/cell from db(/php) like this
    // somevar2[0] would be the first row from php
    //if You want to do something with every line You can use:
    //$.each(somevar2, function(key,value){
    //$('#divname').html(value.cellnameoftable)
  }

I hope it will help ;)

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