jQuery中未捕获的SyntaxError：意外令牌}-为什么？

Question

This is my php code

$p1 = explode(',',$pprod_material->p1);
$p11 = json_encode($p1);

This is my Html code

<input  style="margin-left:-10px;"   type="checkbox" onClick="qst_check_ingri(<?php echo $p11; ?>);"  id="<?php echo $id44;  ?>" value="<?= $id44; ?>" name="extra_check" >

This is my click function in script

function qst_check_ingri (val) 
    {
        alert(JSON.stringify(val));
    }

This is my view-source code output

<input style="margin-left:-10px;" class="badgebox" type="checkbox" onclick="qst_check_ingri('[" 1","4","7","10","13","16","18","21","24","27","30","33","36",""]');"="" id="4-e1" value="4-e1" name="extra_check"> 

I don't why I'm getting this error from my view code is fine still it shows

Uncaught SyntaxError: Unexpected token }

Answer 1

通过单引号传递您的php值

onClick="qst_check_ingri('<?php echo $p11; ?>');"