Question

我看过一些与此有关的帖子，但在那里得不到多少帮助。

我试图将数组从php文件传递到js文件这是php文件中的代码

＆＃13;

<!DOCTYPE html >
<html >
<head> ------ </head>

<body onload="gotofunction()">

<?php
  session_start();
  $_SESSION['street']= $this->street;
  $_SESSION['city']= $this->city;
  $_SESSION['state']= $this->state;
  $_SESSION['zip']= $this->zip;
?>

<script type="text/javascript">
  function voodoo(){
  	alert('alert in voodoo');  // 1st alert
   document.contents.submit();
}

</script>
</body>
</html>

<?php echo "php text";   //1st echo
  $var= json_encode($this->variable);
  ?>

<script type="text/javascript">
  var data = {
	var1:'<?php echo $this->variable;?>',
	path:'<?php echo $this->path;?>'
	}

</script>

＆＃13;

和js文件

＆＃13;

function gotofunction(){

    alert('alert number1');  	//1st alert
    alert(' data '+data.var1 );	//2nd alert
    $.post(data.path+'res/data_controller/tablog1', {'var':data.var1},function(data){
     $('#pos1').html(data);
    });
	}

＆＃13;

我面临的问题是，在js文件中，错误来自＆＃34;数据未被声明＆＃34;并且js文件中的第二个警报也不起作用。在php文件中可以看到php echos但不是警报。

提前致谢

Answer 1

在身体之前添加JS就像这样。

<?php
  session_start();
  $_SESSION['street']= $this->street;
  $_SESSION['city']= $this->city;
  $_SESSION['state']= $this->state;
  $_SESSION['zip']= $this->zip;
?>
<!DOCTYPE html >
<html >
<head> ------ </head>
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>

<script type="text/javascript">
      var data = {
    var1:'<?php echo $this->variable;?>',
    path:'<?php echo $this->path;?>'
    }
function gotofunction(){

    alert('alert number1');     //1st alert
    alert(' data '+data.var1 ); //2nd alert
    $.post(data.path+'res/data_controller/tablog1', {'var':data.var1},function(data){
     $('#pos1').html(data);
    });
    }

</script>

<body onload="gotofunction()">



<script type="text/javascript">
  function voodoo(){
    alert('alert in voodoo');  // 1st alert
   document.contents.submit();
}

</script>
</body>
</html>

<?php echo "php text";   //1st echo
  $var= json_encode($this->variable);
  ?>

脚本标记在php文件中不起作用

1 个答案: