json_encode(); HALP!

[RP]

So i was working on this script last night and had it working fairly well, i came back in the morning and made some alts, only to have my computer crash and as per usual i had saved a unworking version and couldnt revert. My last version of the page was at 4pm, however its not working. It is constantly giving me an "Fatal error: Call to undefined function:  json_encode() on line 64". Last night it was delivering me a Json array like it should be.

 

<?php 
$code = "user 2";

if (!$db = mysql_connect($host, $user, $pass)) {
echo 'Could not connect to mysql';
exit;
}
if (!mysql_select_db($dbname, $db)) {
echo 'Could not select database';
exit;
}



$query = "SELECT * FROM graph WHERE username = '$code'";
$result = mysql_query($query);
$num = mysql_numrows($result);

mysql_close();

$i=0;
while ($i < $num) {
$value0=mysql_result($result,$i,"value0");
$value1=mysql_result($result,$i,"value1");
$value2=mysql_result($result,$i,"value2");
$value3=mysql_result($result,$i,"value3");
$value4=mysql_result($result,$i,"value4");
$i++;
}

$dataset = array(
        'ds1'=>array(
		   array(0,$value0),
               array(1,$value1),
               array(2,$value1),
               array(3,$value3),
               array(4,$value4)
        )
);

$options = array(
        'xTicks'=>array(
                array('v'=>0,'label'=>'Mon'),
                array('v'=>1,'label'=>'Tue'),
                array('v'=>2,'label'=>'Wed'),
                array('v'=>3,'label'=>'Thur'),
			array('v'=>4,'label'=>'Fri')
        )
);

$return = array(
        'dataset'=>$dataset,
        'options'=>$options
);


header("Content-type: text/json");
echo json_encode($return);
?>

 

Doing some reaserach i tried using the default php example for json_encode();

<?php
$arr = array ('a'=>1,'b'=>2,'c'=>3,'d'=>4,'e'=>5);

echo json_encode($arr);
?>

 

this gave me a fatal error as well. Nothing has changed on my server at all.

 

So i ask you PhpFreaks.com, WTF is going on? and furthermore, how do i fix it?

[StormTheGates]

Maybe try something like this?

 

$return = json_encode($return);
echo "$return";

 

 

[RP]

no dice : (

 

still getting a fatal error on line 64 - $return = json_encode($return);

[chronister]

If it is telling you Call To Undefined Function, then that means that the function is not defined. Where is this json_encode()  function actually set at?

 

You need to ensure that if it is in an external file like an include, that the include is being well ... included.

 

anytime you get a Call To Undefined Function, that means that the function is not available to that part of the script.

 

If xyz(); is defined in included_stuff.php this would return an error

 

xyz('pass_var');

include('included_stuff.php');

 

Here I am calling the function before it is available in the script as it is defined in included_stuff.php

 

Find the function and make sure it's defined before calling it

 

nate

[RP]

The Dataset links into this piece of code:

<script>

 

 

Object.extend(Plotr.LineChart.prototype,{

        getRemoteDatasetAndRender: function(url) {

                var xhr_options = {

                        method: 'get'                               

                }

                xhr_options.onComplete = function(xhr) {

                        var json = eval('('+xhr.responseText+')');

                        this.addDataset(json.dataset);

                        this.setOptions(Object.extend(this.options,json.options));

                        this.render();

                }.bind(this);

                new Ajax.Request(url,xhr_options);               

        }       

});

//we won't include the axis details in the options,

// as the remote call will provide them

var options = {

        padding: {left: 30, right: 0, top: 10, bottom: 30},

        backgroundColor: '#dbdbdb',

        colorScheme: 'red'

};

var line = new Plotr.LineChart('plotr',options);

//then just pass the URL to the function

line.getRemoteDatasetAndRender('dataset4pm.php');

 

 

</script>

 

Would this undefiend function apply even for the default example i listed in my first post?

[RP]

bump becasue i still cant make this work and i dont have much more hair to pull out.

[chronister]

I did not realize that json_encode() is a built in php function.

 

Looks like the PECL extension has to be installed.

 

http://www.php.net/manual/en/ref.json.php

 

If your getting call to undefined, then that means that PECL extensions are not installed.

 

Gonna have to install that before it will work.

 

 

[RP]

if this was the problem how was it generating strings before i lost my working version?

[chronister]

did you move servers or make changes to the configuration?

 

When I tried to use it I get call to undefined function. I know that I don't have the PECL extension installed.

 

Do a phpinfo() and see if they are installed on your system.

[RP]

I havent switched server or touched anything, i almost wish i had so i could find the problem.

 

Running phpinfo() and doing a find for json came up with only one result under Configure Command;

 

'--with-zip=/usr' '--with-json' '--with-pear' '--enable-utf8' '--enable-track-vars' '--enable-trans-sid'

 

Im not sure exactly what this means but i do know that i was perviously able to watch firebug display my Json string and in turn watch it build my graph.

[chronister]

hmmmmm sorry man, I am out of Ideas.

 

Does PHPINFO show anything about the PECL extensions?

[RP]

not a thing :\

[chronister]

Then I would start there. From what I saw in the manual, the PECL extension needs to be installed. But I am just grasping at straws here.