writing JavaScript with PHP

Question

I am trying to right some some JavaScript to define a variable using PHP. The variable has a little jQuery in it which should get rendered by the client. I say "should" but really mean I want it to. What should I do to make o2 the same as o1 (i.e. covert the jQuery)?

Thank you!

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 
<html xmlns="http://www.w3.org/1999/xhtml"> 
<head> 
<meta http-equiv="content-type" content="text/html; charset=ISO-8859-1" /> 
<title></title> 
<script src="http://code.jquery.com/jquery-latest.js" type="text/javascript"></script> 
<script>
$(function() {
    var o1={"foo":{"bar": "abc/index.php?="+$("#id").val()}};
    console.log(o1);
<?php
$d='abc/index.php';
$json='{"foo":{"bar": "'.$d.'?id=\"+$(\"#id\").val()"}}';
//echo($json);
$json=json_decode($json);
//echo(print_r($json,1));
$json=json_encode($json);
//echo($json);

echo("var o2=".$json.";");
?>

console.log(o2);
});

</script>

</head>

<body>
<input type="hidden" id="id" value="123" />
</body> 
</html>

PHP is rendered server side while javascript is rendered by the client (i.e. browser). You'll need to send the json to a php script using ajax. — Ryan
– Ryan, Commented May 20, 2012 at 23:09
Are you getting this "123" value in your hidden input from PHP side? How are you printing this id? from something like $row['id']? — Oscar Jara
– Oscar Jara, Commented May 20, 2012 at 23:25
@Ryan. Understand the ajax stuff. Just want to use PHP to pre-write a variable to be available to the client. — user1032531
– user1032531, Commented May 20, 2012 at 23:41
@Oscar. Yes, from PHP side. I guess I can just write this variable directly into the json, but it is bugging me why this doesn't work. — user1032531
– user1032531, Commented May 20, 2012 at 23:44
@user1032531 Ok, then I think it will be easier, take a look at my answer maybe helps. — Oscar Jara
– Oscar Jara, Commented May 20, 2012 at 23:54

Oscar Jara · Accepted Answer · 2012-05-20 23:53:59Z

0

You have this:

<input type="hidden" id="id" value="123" />

So I can assume that you are getting that id from a db right? Maybe like this:

<input type="hidden" id="id" value="<?php echo $row[id];?>" />

If this is the case do the following (I tested and works):

You don't need hidden inputs to do it.

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 
<html xmlns="http://www.w3.org/1999/xhtml"> 
<head> 
<meta http-equiv="content-type" content="text/html; charset=ISO-8859-1" /> 
<title></title> 
<script src="http://code.jquery.com/jquery-latest.js" type="text/javascript"></script> 
<script>
$(function() {

    <?php
        //From your SQL query you obtained
        //$row['id'] = 123;
        $id = $row['id'];
    ?>

    var o1={"foo":{"bar": "abc/index.php?=" + <?php echo $id ?>}};
    console.log(o1);

    <?php
    $d='abc/index.php';
    $json='{"foo":{"bar": "'.$d.'?='.$id.'"}}';
    //echo($json);
    $json=json_decode($json);
    //echo(print_r($json,1));
    $json=json_encode($json);
    //echo($json);
    echo("var o2=".$json.';');
    ?>

    console.log(o2);
});
</script>
</head>
<body>
</body> 
</html>

edited May 20, 2012 at 23:53

answered May 20, 2012 at 23:48

Oscar Jara

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4 Comments

user1032531 Over a year ago

Thanks Oscar, I'm with you. Just still curious why my original implementation didn't work. I spent hours trying to get it to do so. Thinking it might have been a striptag type of thing.

Oscar Jara Over a year ago

@user1032531 No problem :-) sometimes happens.

user1032531 Over a year ago

Thanks again Oscar. Your solution works but still curious why my earlier attempt didn't. Any ideas why not?

Oscar Jara Over a year ago

@user1032531 Well, at first sight, jQuery can't be rendered in a simple PHP string as you expect when doing this: "$json='{"foo":{"bar": "'.$d.'?id=\"+$(\"#id\").val()"}}';" I think that was the error :-) but maybe someone around here will argue lol

ilanco · Accepted Answer · 2012-05-20 23:14:02Z

0

Untested, may contain errors:

<script>
$(function() {
    var o1={"foo":{"bar": "abc/index.php?="+$("#id").val()}};
    console.log(o1);

    <?php
    $d='abc/index.php';
    $json='{"foo":{"bar": "' . $d . '?="+$("#id").val()}}';

    echo("var o2=".$json.';');
    ?>

    console.log(o2);
});
</script>

answered May 20, 2012 at 23:14

ilanco

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1 Comment

user1032531 Over a year ago

While this works, it doesn't if $json is first json_decode() and then json_encode().

Roger Russel · Accepted Answer · 2012-05-20 23:47:09Z

0

I prefer make json with the json_encode, to avoid unexpected problems with parse and validation of json.

the json_encode and json_decode only work with UTF-8, another charset will result in a blank string, if it has special chars.

<script>
<?php

$d = "abc/index.php";
$json['foo']['bar'] = $d . '?id=$("#id").val()';

echo "var o2 = ".json_encode($json);

?>
console.log(o2);
</script>

answered May 20, 2012 at 23:47

Roger Russel

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1 Comment

user1032531 Over a year ago

Thanks Roger, My definition of $json was very small, but the real one had many more variables and sub-variables. Also, the original definition is based on a client only side application, and I wanted to just cut-and-past it with as little modification as possible. Maybe a limitation of json_decode?

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