在脚本中传递php的变量

Question

$row2 = mysql_fetch_array($result2); 
$row3 = mysql_fetch_array($result3); 
$row4 = mysql_fetch_array($result4); 
echo $row4; 
$row5=10; 


?> 
<script> 

var t = new PieChart({ 
    container: document.getElementById("demo"), 
    area: { height: 350 }, 
    data: { 
     preloaded: { 
      subvalues: [ 
       { id: "pun", value:x, style: { expandable: false } }, 
       { id: "tec", value: 20, style: { expandable: false } }, 
       { id: "hel",value:50,style:{expandable:false}}, 
       { id: "int",value:10,style:{expandable:false}}, 
       { id: "stu",value:10,style:{expandable:false}} 
      ] 
     } 
    } 
}); 
</script> 

如何在php.if中传递“x”的值;它不显示饼图。使用json编码也不起作用

Answer 1

你可以在里面注入一个PHP代码。

<script> 
var t = new PieChart({ 
    container: document.getElementById("demo"), 
    area: { height: 350 }, 
    data: { 
     preloaded: { 
      subvalues: [ 
       { id: "pun", value: <?php echo (int)$x; ?>, style: { expandable: false } }, 
       { id: "tec", value: 20, style: { expandable: false } }, 
       { id: "hel",value:50,style:{expandable:false}}, 
       { id: "int",value:10,style:{expandable:false}}, 
       { id: "stu",value:10,style:{expandable:false}} 
      ] 
     } 
    } 
}); 
</script>