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Sep 2nd, 2009

How do I use PHP to create a JSON object?

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I am trying to use a Dojo (ajax) table on my page that has detail rows in which I want to draw detail tables. The tables have a format event which calls formatDetail. In this event I would like to build a simple table using JSON data.

My problem is how do I use PHP to create the JSON data object so that my formatDetail javascript can use it?

My test code so far is, in the header:

javascript Syntax (Toggle Plain Text)
function formatDetail(inWatchId, inRowIndex) {
	alert("Length is: " + myJSON.length);
}
function formatDetail(inWatchId, inRowIndex) {
	alert("Length is: " + myJSON.length);
}

And in the body of the page:

php Syntax (Toggle Plain Text)
$stmt = sqlsrv_query($connection,$query);
while ($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC))
{
	$arr[] = $row;
}
$objJSON = '{"sample":'.json_encode($arr).'}';
echo "<script>var myJSON = $objJSON;</script>";
$stmt = sqlsrv_query($connection,$query);
while ($row = sqlsrv_fetch_array($stmt, SQLSRV_FETCH_ASSOC))
{
	$arr[] = $row;
}
$objJSON = '{"sample":'.json_encode($arr).'}';
echo "<script>var myJSON = $objJSON;</script>";

When I try this, the js alert displays "Length is undefined"

What am I doing wrong?

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agrarian

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Sep 2nd, 2009

Re: How do I use PHP to create a JSON object?

Don't try to mix creating your own json strings

php Syntax (Toggle Plain Text)
$objJSON = array('sample' => null);
// whatever
$objJSON['sample'] = $arr;
$objJSON = json_encode($objJSON);
echo '<script type="text/javascript">var myJSON = ' . $objJSON . ';</script>
 
$objJSON = array('sample' => null);
// whatever
$objJSON['sample'] = $arr;
$objJSON = json_encode($objJSON);
echo '<script type="text/javascript">var myJSON = ' . $objJSON . ';</script>

Last edited by ShawnCplus; Sep 2nd, 2009 at 4:13 pm.

ShawnCplus

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Code Monkey

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Sep 2nd, 2009

Re: How do I use PHP to create a JSON object?

Thanks! That enabled me to get it working!

agrarian

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since Mar 2008

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