[从JS发送到PHP的函数

Question

我已尝试传递“隐藏”值以删除记录。但是JS函数发送数据，但是mysql代码不起作用。使用“插入到”，它可以工作，为此，奇怪的是，相同的代码不起作用。

This is my code.

    <script type="text/javascript">
        function invia()
     {
        var hides = document.getElementById('hide').value;

          $.ajax({
              type: 'POST',
              url: "note_db_php/delete_db_note.php",
              data: {"hide": hides},
              success: function(data){
                console.log("Dati inviati");
              },

              error: function(data) {
                console.log("Dati non inviati");
              }
          });
      };

    </script>

这是删除页面；

    <?php
    $servername = "";
    $username = "";
    $password = "";
    $dbname = "";

    $hide = $_POST["hide"];



// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

// sql to delete a record
$sql = "DELETE FROM note WHERE id='$hide'";

if ($conn->query($sql) === TRUE) {
echo "Record deleted successfully";
} else {
echo "Error deleting record: " . $conn->error;
}

$conn->close();
?>

Answer 1

缺少属性dataType，type应该为method。

$.ajax({
  url:"note_db_php/delete_db_note.php",
  method:'POST',
  dataType:'json',
  data:{
    "hide": hides
  },
  success:function(data) {
    console.log("Dati inviati"); 
  },
  error: function (request, status, error) {
    console.log("Dati non inviati");
  }
});

Answer 2

替换您的查询-

$sql = "DELETE FROM note WHERE id='$hide'";

有一个以下-

$sql = "DELETE FROM note WHERE id = ".$hide;

并且在ajax端，请控制台data。如果出现错误，请回显您的查询（回显$ sql），然后在PHPMyAdmin中复制并运行查询。

Answer 3

我认为问题出在SQL查询中-$ sql =“从注释中删除ID = $ hide的注释”；

您可以将$ hide放在单引号中并尝试吗？

$ sql =“从注释WHERE id ='$ hide'中删除”; //这会起作用

Answer 4

这里是您在sql语法中的问题，和一行

在此处输入代码否则在您的代码上打开会导致错误的代码

更新的查询在这里，它的工作原理

// sql to delete a record
$sql = "DELETE FROM note WHERE id='$hide'";

if ($conn->query($sql) === TRUE) {
echo "Record deleted successfully";
} else {
echo "Error deleting record: " . $conn->error;
}