在PHP的json_encode（）中包含JavaScript

Question

I need to generate the following script using PHP and json_encode().

var obj= {
    o1:123,
    o2: {
        o3: 456,
        o4: function() {return $( "#myID" ).val();}
    }
}

My attempt to do so is as follows.

<?php
$a=array(
    'o1'=>123,
    'o2'=>array(
        'o3'=>456,
        'o4'=>'function() {return $( "#myID" ).val();}'
    )
);
$json=json_encode($a);
?>
<script type="text/javascript"> 
    <?php echo("var obj={$json};");?>
    console.log(obj);
</script>

The resultant output is as follows. The quotes around the properties poses no issues, however, the quotes around the JavaScript renders it as a string and not JavaScript. I obviously can't not quote the JavaScript in the array as it will result in malformed JSON.

How can I include JavaScript in PHP's json_encode()?

var obj={
    "o1":123,
    "o2":{
        "o3":456,
        "o4":"function() {return $( \"#username\" ).val();}"
    }
};

Answer 1

JSON does not support functions. Its a data interchange format. Perhaps you should send some configuration data down and have a function change its behavior on that. Or since you aren't doing it as an AJAX call at this point, just echo it out:

<script type="text/javascript"> 
<?php echo("var obj={$json};");?>
<?php echo("obj.myFunc  = function..."); ?>
console.log(obj);
</script>

Answer 2

How about removing the quotations that surround the function?

<?php

$obj = array(
    'o1' => 123,
    'o2' => array(
        'o3' => 456,
        'o4' => 'function() {return $( "#myID" ).val();}',
        'o5' => 'function(param) {return $( "#myID" ).val();}'
    )
);
$json = json_encode($obj);

while ($func = strpos($json, '":"function(')) {
    $json = substr($json, 0, $func + 2) . substr($json, $func + 3);
    while ($quote = strpos($json, '"', $func + 2)) {
        $func = $quote + 1;
        if (substr($json, $quote - 1, 1) == "\\") {
            $json = substr($json, 0, $quote - 1) . substr($json, $quote);
            continue;
        }
        $json = substr($json, 0, $quote) . substr($json, $quote + 1);
        break;
    }
}

echo $json;

This checks if the string starts with function( , and if it is, removes the double quotes.

The result is a JSON (but can still be used as a JavaScript object):

{"o1":123,"o2":{"o3":456,"o4":function() {return $( "#myID" ).val();},"o5":function(param) {return $( "#myID" ).val();}}}

Upon setting this object to a variable, you can see that the function registered fine.

With that, you can still use the same technique you used before:

<script type="text/javascript"> 
    <?php echo("var obj={$json};");?>
    console.log(obj);
</script>

Answer 3

This is what I ended up doing. Kind of hackish, and I expect Dave's is a better solution, but haven't yet tested it. I will do so.

<?php

class fubar {
    public function json_encode_JS($array)
    {
        $array=$this->_json_encode_JS($array);
        $json=json_encode($array);
        $json=str_replace(array('"%','%"'), '', $json);
        $json=str_replace('\"', '"', $json);
        return $json;
    }
    private function _json_encode_JS($array)
    {
        foreach($array as $key => $value){
            if(is_array($value)){$array[$key]=$this->_json_encode_JS($value);}
            else {
                if(strpos($value, 'function(')===0){
                    $array[$key]='%'.$value.'%';
                }
            }
        }
        return $array;
    }
}

$array=array(
    'o1'=>123,
    'o2'=>'function() {return $("#myID").val();}',
    'o3'=>array(
        'o4'=>456,
        'o5'=>'function() {alert("Hello why don\'t you answer "+$("#myID").val());}'
    )
);

?>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.js" type="text/javascript"></script>
<script type="text/javascript"> 
    <?php
    $obj=new fubar();
    echo('var obj='.$obj->json_encode_JS($array).';');
    ?>
    console.log(obj);
    obj.o3.o5();
</script>
<input id="myID" name="myID" type="text" value="hello">