Question

我使用$.post将值从当前页面传递到另一个页面以删除一些数据。但价值没有发布。我也试过了.get，但仍然无法正常工作。请提出解决问题的方法。

<i title="Delete" class="icon-remove-sign icon-2x" onClick=" myFunction('<?php echo $val['id']?>','delete_pahse')"></i>
<script>
    function myFunction(rowId,delet) {
        if (confirm("Are You Sure to Delete ") == true) 
        {
            $.get("?action=studentDetail&student_id=<?php echo $_GET['student_id'];?>&rowId="+rowId+"delet="+delet),(function(data){
                document.getElementById('progress').style.display = "none";
                    if(data){
                            alert('The Selected Record Is Successfully Deleted');
                    }else{
                            alert('The Selected Record Is Not Successfully Deleted');
                    }   ;
            });

        }

        document.getElementById("demo").innerHTML = x;
    }
</script>

Answer 1

在send.php中

  <?php
  $mydata="This is my string being passed to another php";
  header("Location: http://xyzphp.in/Receive.php?msg=$mydata");
  ?>

在Receive.php中

<?php
$receive = $_GET['msg'];
echo "Data Received";    
echo $receive;
?>

Answer 2

看起来你错过了＆＃39;＆amp;＆＃39;之前＆＃39; delet =＆＃39;在您的$ .get网址中。这会导致它弄乱rowId - 而不是rowId = 5，它会看到rowId = 5delet =。

Answer 3

@ user3300358，你可以试试下面的JS代码：

$.ajax({
    type:"POST",
    url:"YOUR URL",
    data: { student_id: student_id, rowId: rowId, delet: delet},
    success : function(data) {
        if(data){
            alert('The Selected Record Is Successfully Deleted');
        } else {
            alert('The Selected Record Is Not Successfully Deleted');
        }
    },
    error : function() {
        alert('Something went wrong. Please try again');
    }
 });

如果您还有任何问题，请告诉我。

$ .post未将数据发布到指定页面

3 个答案: