Question

我在PHP中有一个图像链接，它将一个变量传递给另一个文件（functions.js）中的脚本，即：

"img src=\"images/del.jpg\" onclick='delete_user_program(".$row1['program_name'].")' onmouseover=\"this.style.cursor='pointer'\" /"

脚本是：

   function delete_user_program(program_name){

    var confirmed = confirm("Are you sure;");
    if (confirmed == true){
        var str="./delete_user_program.php?p1="+program_name;
        window.location=str; 
    }
}

我尝试传递“p1”。

然后delete_user_program.php是：

<?php
    session_start();
    include("connect_db.php"); 
    $con = $_SESSION['connection'];


    $select_query ="SELECT * FROM user_program WHERE program_name='".$_GET['p1']."'";
    $result=@mysqli_query($con,$select_query) or die('Error, query failed');
    $num_result=mysqli_num_rows($result);


    if($num_result>0) {

        //some code
    }


 else {
        echo '<html><script language="javascript">alert("Program not exist.");</script></html>';

调用delete_user_program.php时，我收到p1未定义的错误和“Program not exists。”消息。有小费吗？提前谢谢。

Answer 1

你应该使用ajax。

$(document).ready(function(){
    function delete_user_program(program_name){

        $.get("./delete_user_program.php",
        {
          p1: program_name
        },
        function(data,status){
            alert("Data: " + data + "\nStatus: " + status);
        });
    }
});

Answer 2

<?php
  // file A.php
  echo 'A';
?>


<?php
  // file B.php;
  $a = file_get_contents('A.php');
  // thats synchron 
  echo $a;
?>

php变量到另一个php通过javascript

2 个答案: