php变量插入在json中格式出错

Question

 $pdata='{
"touser":"$openid",
"template_id":"Tcj_l-RgdEQ_ppjE7SE_Uo-G-ME51HIzbZsjLqp5mso",
"url":"https://jc.sdp178.com/web/dgoods.html?oddnumtracking_number=886594027291530556&logistics_company=YTO",
"topcolor":"#FF0000",
"data":{
  "name":{
  "value":"商品名字",
  "color":"#173177"
  },
  "remark":{
  "value":"测试",
  "color":"#173177"
  }
}
}';

不用json_decode()的情况该怎么写，才符合json格式又可以输出php的变量

Answer 1

方法1
把数据放数组里然后json_encode一下

方法2

$pdata='{
"touser":"'.$openid.'",
"template_id":"Tcj_l-RgdEQ_ppjE7SE_Uo-G-ME51HIzbZsjLqp5mso",
"url":"https://jc.sdp178.com/web/dgoods.html?oddnumtracking_number=886594027291530556&logistics_company=YTO",
"topcolor":"#FF0000",
"data":{
  "name":{
  "value":"商品名字",
  "color":"#173177"
  },
  "remark":{
  "value":"测试",
  "color":"#173177"
  }
}
}';

方法3

$pdata=
<<<eof
{
"touser":"$openid",
"template_id":"Tcj_l-RgdEQ_ppjE7SE_Uo-G-ME51HIzbZsjLqp5mso",
"url":"https://jc.sdp178.com/web/dgoods.html?oddnumtracking_number=886594027291530556&logistics_company=YTO",
"topcolor":"#FF0000",
"data":{
  "name":{
  "value":"商品名字",
  "color":"#173177"
  },
  "remark":{
  "value":"测试",
  "color":"#173177"
  }
}
}
eof;